Rewrite each of the following numerals with proper commas in the international system of numeration
i) 513625
ii) 4035672
iii) 65954923
iv) 70902005
Solution:
i) 513,625 or Five hundred thirteen thousand six hundred twenty five.
ii) 4,035,672 or Four million thirty five thousand six hundred seventy two.
iii) 65,954,923 or Sixty five million nine hundred fifty four thousand nine hundred twenty three
(iv) 70,902,005 or Seventy million nine hundred two thousand five
Question: 3
Write each of the following numbers in the international system of numeration :
i) Forty three lakh four thousand eighty four.
ii) Six crore thirty four lakh four thousand forty four.
iii) Seven lakh thirty five thousand eight hundred ninety nine only.
Solution:
i) 4,304,084 or Four million three hundred four thousand eighty four.
ii) 63,404,044 or Sixty three million four hundred four thousand forty four.
iii) 735,899 or Seven hundred thirty five thousand eight hundred ninety nine.
Question: 4
Write the following numbers in the Indian system of numeration :
i)Six million five hundred forty three thousand two hundred ten.
ii)Seventy six million eighty five thousand nine hundred eighty seven
iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight.
Solution:
i) 65, 43,210 or Sixty five lakh forty three thousand two hundred ten.
ii) 7, 60,85,987 or Seven crore sixty lakh eighty five thousand nine hundred eighty seven.
iii) 32, 54,79,838 or Thirty two crore fifty four lakh seventy nine thousand eight hundred thirty eight.
Question: 5
A certain nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period. Write this number in words in the Indian system.
Solution:
The number is 333,222,111
In Indian system , the number is written as 33,32,22,111 thirty – three crore thirty – two lakh twenty thousand one hundred and eleven.
Question: 6
How many thousands make a million?
Solution:
1,000 thousands makes a million
Question: 7
How many millions make a billion?
Solution:
1,000 millions make a billion
Question: 8
i) How many lakhs make a million?
ii) How many lakhs make billion?
Solution:
i) Ten lakhs make a million
ii) Ten thousand lakhs make a billion
Question: 9
Write each of the following in numerical form:
i) Nighty-Eight million seven hundred eight thousand four.
ii) Six hundred seven million twelve thousand eighty four.
iii) Four billion twenty five million forty five thousand.
Solution:
i) 98,708,004
ii) 607,012,084
iii) 4,025,045,000
Question: 10
Write the number names of each of the following in international system of numeration :
i) 435,002
ii) 1,047,509
iii) 59,064,523
iv) 25,201,905
Solution:
i) Four hundred thirty-five thousand and two
ii) One million, forty-seven thousand, five hundred and nine
iii) Fifty-nine million, sixty-four thousand, five hundred and twenty-three
iv) Twenty-five million, two hundred one thousand, nine hundred and five
Exercise 1.3
Question: 1
How many four – digit numbers are there in all?
Solution:
There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We cannot use ‘0’ at thousand’s place.
So, we can use only 9 digits at thousand’s place.
Also, we can use 10 digits at hundred’s, 10 digits at ten’s and 10 digits at unit’s place.
So, total numbers of four-digit numbers = 9 × 10 × 10 × 10 = 9000
Question: 2
Write the smallest and the largest six digit numbers. How many numbers are between these two.
Solution:
The smallest digit is 0. But we cannot use 0 at the place having the highest place value in six digit numbers. So, we will use the second smallest digit i.e., 1. All other places are filled by 9.
Hence, the required number = 100000
Smallest six digit number will be 100000.
The largest digit is 9.
We can use 9 at any place. In fact , we can use 9 in all places in six digit numbers.
Hence, the required number = 999999
Largest six digit number will be 999999
Required difference = 999999 – 100000 = 899999
So, the total numbers between 999999 and 100000 will be 899998.
Question: 3
How many 8 – digit numbers are there in all ?
Solution:
There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We cannot use ‘0’ at the place having the highest place value in 8 digit numbers.
So, we can use only 9 digits at the place having the highest place value in 8 digit numbers.
Also, we can use 10 digits at the remaining places in 8 digit numbers So, total numbers of 8-digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000
Question: 4
Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.
Solution:
One crore seventy-five thousand three hundred two.
In order to write the smallest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the smallest digit 1 ( Except 0 ) at the place having the highest place value. The largest digit 7 is put at the rightmost place i.e. at unit’s place, the digit 5 is put at the ten’s place, the digit 3 is put at the hundred’s place and the digit 2 is put at the thousand’s place. All other places are filled by 0. Hence, the required largest number is 10002357.
In order to write the largest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the largest digit 7 at the place having the highest place value. The smallest digit 5 is put at the place after the highest place value. We put the next smallest digit ( i.e., 3 ) after the previous one. After it we place the next smallest digit ( i.e., 2 ) and after that we put the digit 1. All other places are filled by 0. Hence, the required largest number is 75321000.
Question: 5
What is smallest 3-digit number with unique digits?
Solution:
The smallest three-digit number with unique digits is 102.
Question: 6
What is the largest 5- digits number with unique digits?
Solution:
The largest five – digit number with unique digits 98,765.
Question: 7
Write is smallest 3-digit number which does not change if the digits are written in reverse order.
Solution:
The smallest three – digit number that does not change if the digits are written in reverse order is 101.
Question: 8
Find the difference between the number 279 and that obtained on reversing its digits.
Solution:
The number obtained on reversing 279 = 972
Difference = 972 – 279 = 693
Thus, the difference between 279 and that obtained on reversing its digits is 693.
Question: 9
Form the largest and smallest 4- digit numbers using each of digits 7,1,0,5 only once.
Solution:
The largest and smallest four- digit numbers formed using 7,1,0 and and 5are 7,510 and 1,057.
Exercise 1.4
Question: 1
Put the appropriate symbol ( < > ) in each of the following boxes :
i) 102394 ___ 99887
ii) 2507324 ___ 2517324
iii) 3572014 ____ 10253104
iv) 47983505 ____ 47894012
Solution:
i) 102394 > 99887
ii) 2507324 < 2517324
iii) 3572014 < 10253104
iv) 47983505 > 47894012
Question: 2
Arrange the following numbers in ascending order :
Ten lakh or one million (10, 00, 000) milligrams make one kilogram.
Question: 2
A box of medicine tablets contains 2, 00,000 tablets each weighing 20mg. what is the total weight of all the tablets in the box in grams? in kilograms ?
Solution:
Given data: Each tablet weighs = 20 mg
Therefore, The weight of 2, 00,000 tablets = 2, 00,000 × 20 = 40, 00,000 mg
Therefore, The total weight of all the tablets in the box = 40, 00,000 mg
We know 1 g = 1,000 mg
Weight of the box having all tablets = 40,00,000 ÷ 1,000=4000g
And, as 1 kg = 1,000 g
Therefore, Weight of the box having all tablets = 4,000 ÷ 1,000 = 4000g = 4 kg
Question: 3
Population of sundarnagar was 2, 35,471 in the year 1991. In the year 2001 it was found to have increased by 72,958. What was the population of the city in 2001?
Solution:
The population of Sundar Nagar in 2001 = Sum of the population of city in 1991 + Increase in population over the given time period
As given in the question, The population of Sundar Nagar in 1991 = 2, 35,471
As given in the question,
Increase in population over the given time period = 72.958
Therefore, The population of Sundar Nagar in 2001
= 2, 35,471 + 72,958 = 3, 08,429
Question: 4
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final days were respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solution:
Total number of tickets sold on all four days is the sum of the tickets sold on the first, second, third and final days.
Therefore, total number of tickets sold on all four days is given by:
= 1094 + 1812 + 2050 + 2751 = 7707
Question: 5
The town newspaper is published every day. One copy has 12 pages. Everyday 11,980 copies are printed. How many pages are in all printed every day? Every month?
Solution:
As given in the question,
Number of pages in 1 copy of newspaper = 12
Therefore, Number of pages in 11,980 copies of newspaper
= 11,980 × 12 = 1, 43,760
Thus, 1, 43,760 pages are printed every day.
Now, number of pages printed every day = 1, 43,760
Therefore, Number of pages printed in a month = 1, 43,760 × 30 = 43, 12,800
Thus, 43, 12,800 pages are printed in a month.
Question: 6
A machine, on an average, manufactures 2825 screws a day. How many screws did it produce in the month of January 2006?
Solution:
As given in the question,
Number of screws produced by a machine in a day = 2,825
Therefore, Number of screws produced by the same machine in the month of January 2006 = 2, 825 × 31 = 87,575
Thus, machine-produced 87,575 screws in the month of January 2006.
Question: 7
A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Solution:
Runs scored by cricket player in test matches = 6,978
Therefore, Remaining runs required to complete 10,000 runs
= 10,000 – 6,978 = 3,022
Thus, the player needs to score 3,022 more runs to complete 10,000 runs.
Question: 8
Ravish has Rs. 78,592 with him. He placed an order for purchasing 39 radio sets at Rs. 1234 each. How much money will remain with him after the purchase?
Solution:
Ravish’s initial money = Rs.78, 592
He purchased 39 radio sets at Rs.1, 234 each.
Therefore, Money spent by him on purchasing 39 radio sets
= 1,234 × 39 = Rs. 48,126
Therefore, Remaining money with Ravish after the purchase = Initial money – Money spent on purchasing 39 radio sets = Rs. 78,592 – Rs. 48,126 = Rs. 30,466
Thus, 230,466 are left with him after the purchase.
Question: 9
In an election, the successful candidate registered 5, 77,570 votes and his nearest rival secured 3, 48,685 votes. By what margin did the successful candidate win the election?
Solution:
Margin of victory in the election for the successful candidate = Number of votes registered by the winner – Number of votes secured by nearest rival candidate
Votes registered by the winner = 5, 77,570
Votes secured by the rival = 3, 48,685
Therefore, Margin of victory for the successful candidate
= 5, 77, 570 – 3, 48, 685 = 2, 28, 885
Question: 10
To stitch a shirt 2m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Solution:
As given in the question, Total length of available cloth = 40 m = 4,000 cm (1 m = 100 cm)
As given in the question, Length of cloth required to stitch a shirt
= 215 cm = 200 + 15 = 215 cm
Therefore, The number of shirts that can be stitched from the 40-metre cloth
= 4,000 / 215 = 18.60
As the number of shirts has to be a whole number, we consider the whole part only. That is, 18 such shirts can be stitched.
Therefore, Cloth required for stitching 18 shirts = 215 x 18 = 3870 cm. Therefore, Remaining cloth = 4,000 — 3870 = 130 cm = 1.3 m
Question: 11
A vessel has 4 litre and 650 ml of curd. In how many glasses, each of 25 ml capacity, can it be distributed?
Solution:
The number of glasses in which curd can be distributed = Total amount of curd/Capacity of each glass.
Total amount of curd in the vessel = 4,650 ml = 4,000 + 650 = 4,650 ml
( 1 L = 1,000 ml )
Capacity of each glass = 25 ml
Therefore, Number of glasses in which curd can be distributed = 4,650/25 = 186
Question: 12
Medicine in packed in boxes, each such boxes weighing 4kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 Kg?
Solution:
Sol :
As given in the question,
Total capacity of a van carrying boxes of medicines = 800 kg = 8, 00,000 g (1 kg = 1,000 g)
As given in the question, Weight of each packed box
= 4,500 g = 4,000 + 500 = 4,500 g
Therefore, Total number of boxes that can be loaded in the van
= 8, 00,000 / 4,500 = 177.77
The obtained number of boxes is not a whole number.
Therefore, Weight of 177 boxes = 177 × 4,500 = 7, 96,500 g (under permissible limit)
Therefore, Weight of 178 boxes = 178 × 4,500 = 8, 01,000 g (beyond permissible limit)
Therefore, we can’t load 178 boxes; hence, we can say that 177 boxes can be loaded in the van.
Question: 13
The Distance between the school and the house of a student is 1 Km 875 m. Every day she walks both ways between her school and home. Find the total distance covered by her in a week?
Solution:
Therefore, Distance between the school and the house of a student
= 1,875 m = 1,000 + 875 = 1,875 m (1 km = 1,000 m)
As given in the question, Distance covered by a student in a day
= 2 × 1,875 = 3,750 m
Total distance covered by her in a week = 7 × 3,750 = 26,250 m = 26.25 km
Exercise 1.6
Question: 1
Round off each of the following numbers to nearest tens :
i) 84
ii) 98
iii) 984
iv) 808
v) 998
vi) 12,096
vii) 10,908
viii) 28,925
Solution:
i) 80
ii) 100
iii) 980
iv) 810
v) 1,000
vi) 12,100
vii) 10,910
viii) 28,930
Question: 2
Round off each of the following numbers to nearest hundreds :
i) 3,985
ii) 7289
iii) 8074
iv) 14,627
v) 28,826
vi) 4,20,387
vii) 43,68,973
viii) 7,42,898
Solution:
i) 4,000
ii) 7,300
iii) 8,100
iv) 14,600
v) 28,800
vi) 4,20,400
vii) 43,69,000
viii) 7,42,900
Question: 3
Round off each of the numbers to nearest thousands :
i) 2401
ii) 9600
iii) 4278
iv) 7832
v) 9567
vi) 26,019
vii) 20,963
viii) 4,36,952
Solution:
i) 2000
ii) 10000
iii) 4000
iv) 8000
v) 10000
vi) 26000
vii) 21000
viii) 4,37,000
Question: 4
Round off each of the following numbers to nearest tens, hundreds and thousands.
i) 964
ii) 1049
iii) 45,634
iv) 79,085
Solution:
Tens :
i) 970
ii) 1050
iii) 45,630
iv) 79,090
Hundreds :
i)1000
ii) 1000
iii) 45,600
iv) 79,100
Thousands :
i) 1000
ii) 1000
iii) 46000
iv) 79000
Question: 5
Round off the following measures to the nearest hundreds :
i) Rs 666
ii) Rs 850
iii) Rs 3,428
iv) Rs 9,080
v) 1265 km
vi) 417 m
vii) 550 cm
viii) 2486 m
ix) 360 gm
x) 940 kg
xi) 273 l
xii) 820 mg
Solution:
i) Rs. 700
ii) Rs. 900
iii) Rs. 3,500
iv) Rs.9100
v) 1300 km
vi) 400 m
vii) 600 cm
viii) 2500 m
ix) 400 gm
x) 900 kg
xi) 300 l
xii) 800 mg
Question: 6
List all numbers which are rounded off to the nearest ten as 370.
A coin is tossed 1000 times with the following frequencies:
Head: 445, Tail: 555
When a coin is tossed at random, what is the probability of getting?
(i). a head?
(ii). a tail?
Solution:
Total number of times a coin is tossed = 1000
Number of times a head comes up = 445
Number of times a tail comes up = 555
Question: 2
A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?
Solution:
No. of socks in the box = 4
Let B and W denote black and white socks respectively. Then we have:
S = {B,B,W,W}
If a white sock is picked out, then the total no. of socks left in the box = 3
No. of white socks left = 2 – 1 = 1
Question: 3
Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:
If same pair of coins is tossed at random, find the probability of getting:
The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2019, 2540, 2820
Find the mean enrollment of the school for this period.
Solution:
The mean enrolment = Sum of the enrolments in each year ÷ Total number of years
The mean enrolment = (1555 + 1670 + 1750 + 2019 + 2540 + 2820) ÷ 6
= 12354 ÷ 6
= 2059.
Thus, the mean enrolment of the school for the given period is 2059.
Question: 5
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
Day
Mon
Tue
Wed
Thu
Fri
Sat
Sun
Rainfall (in mm)
0.0
12.2
2.1
0.0
20.5
5.3
1.0
(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.
Solution:
(i) The range of the rainfall = Maximum rainfall – Minimum rainfall
= 20.5 – 0.0
= 20.5 mm.
(ii) The mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0) ÷ 7
= 41.1 ÷ 7
= 5.87 mm.
(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.
Question: 6
If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.
Solution:
The mean height = Sum of the heights ÷ Total number of persons
= (140 + 150 + 152 + 158 + 161) ÷ 5
= 761 ÷ 5
= 152.2 cm.
Question: 7
Find the mean of 994, 996, 998, 1002 and 1000.
Solution:
Mean = Sum of the observations ÷ Total number of observations
Mean = (994 + 996 + 998 + 1002 + 1000) ÷ 5
= 4990 ÷ 5
= 998.
Question: 8
Find the mean of first five natural numbers.
Solution:
The first five natural numbers are 1, 2, 3, 4 and 5.
Question: 9
Find the mean of all factors of 10.
Solution:
Question: 10
Find the mean of first 10 even natural numbers.
Solution:
Question: 11
Find the mean of x, x + 2, x + 4, x + 6, x + 8
Solution:
Mean = Sum of observations ÷ Number of observations
→ Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) ÷ 5
→ Mean = (5x + 20) ÷ 5
→ Mean = 5(x + 4)5
→ Mean = x + 4
Question: 12
Find the mean of first five multiples of 3.
Solution:
The first five multiples of 3 are 3, 6, 9, 12 and 15.
Question: 13
Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6 Find the mean
Solution:
We Have
Question: 14
The percentage of marks obtained by students of a class in mathematics are:
The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 3, 5, 1, 1, 5 Find the mean number of children per family.
Solution:
= 3.
Thus, on an average there are 3 children per family in the locality.
Question: 16
The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
We have:
n = The number of observations = 100, Mean = 40
→ Sum of the observations = 40 x 100
Thus, the incorrect sum of the observations = 40 x 100 = 4000.
Now,
The correct sum of the observations = Incorrect sum of the observations – Incorrect observation + Correct observation
→ The correct sum of the observations = 4000 – 83 + 53
→ The correct sum of the observations = 4000 – 30 = 3970
Question: 17
The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.
Solution:
We have:
So, sum of the five numbers = 5 x 27 = 135.
Now,
So, sum of the four numbers = 4 x 25 = 100.
Therefore, the excluded number = Sum of the five number – Sum of the four numbers
→ The excluded number = 135 – 100 = 35.
Question: 18
The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Solution:
We have:
Let the weight of the seventh student be x kg.
Thus, the weight of the seventh student is 61 kg.
Question: 19
The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?
Solution:
Let x1, x2, x3…x8 be the eight numbers whose mean is 15 kg. Then,
x1 + x2 + x3 + …+ x8 = 15 × 8
→x1 + x2 + x3 +…+ x8 = 120.
Let the new numbers be 2x1, 2x2, 2x3 …2x8. Let M be the arithmetic mean of the new numbers.
Then,
Question: 20
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Solution:
Let x1, x2, x3, x4 and x5 be five numbers whose mean is 18. Then,
18 = Sum of five numbers ÷ 5
∴ Sum of five numbers = 18 × 5 = 90
Now, if one number is excluded, then their mean is 16.
So,
16 = Sum of four numbers ÷ 4
∴ Sum of four numbers = 16 × 4 = 64.
The excluded number = Sum of five observations – Sum of four observations
∴ The excluded number = 90 – 64
∴ The excluded number = 26.
Question: 21
The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Solution:
n = Number of observations = 200
→ Sum of the observations = 50 x 200 = 10,000.
Thus, the incorrect sum of the observations = 50 x 200
Now,
The correct sum of the observations = Incorrect sum of the observations – Incorrect observations + Correct observations
→ Correct sum of the observations = 10,000 – (92 + 8) + (192 + 88)
→ Correct sum of the observations = 10,000 – 100 + 280
→ Correct sum of the observations = 9900 + 280
→ Correct sum of the observations = 10,180.
Question: 22
The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.
Solution:
We have:
Mean = Sum of five numbers ÷ 5
→ Sum of the five numbers = 27 × 5 = 135.
Now, New mean = 25
25 = Sum of six numbers ÷ 6
→ Sum of the six numbers = 25 × 6 = 150.
The included number = Sum of the six numbers – Sum of the five numbers
→ The included number = 150 – 135
→ The included number = 15.
Question: 23
The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.
Solution:
Let x1, x2, x3…x75 be 75 numbers with their mean equal to 35. Then,
x1 + x2 + x3 +…+x75 = 35 × 75
→ x1 + x2 + x3 +…+ x75 = 2625
The new numbers are 4 x 1, 4 x 2, 4 x 3…4 x 75
Let M be the arithmetic mean of the new numbers. Then,
→ M = 140.
Exercise 23.2
Question: 1
A die was thrown 20 times and the following scores were recorded:
Prepare the frequency table and find the mean wage.
Solution:
Exercise 23.3
Question: 1
Data – 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
Question: 2
Data – 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution:
Question: 3
Data – 31, 38, 27, 28, 36, 25, 35, 40
Solution:
Question: 4
Data – 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
Question: 5
Data – 41, 43,127, 99, 71, 92, 71, 58, 57
Solution:
Question: 6
Data – 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
Question: 7
Data – 12, 17, 3, 14, 5, 8, 7, 15
Solution:
Question: 8
Data – 92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
Question: 9
Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.
Solution:
Question: 10
Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution:
Question: 11
Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?
Solution:
Question: 12
The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Solution:
Question: 13
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Exercise 23.4
Question: 1
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
By using the empirical relation also find the mean.
Solution:
Arranging the data in ascending order such that same numbers are put together, we get:
12, 12, 13, 13, 14, 14, 14, 16, 19
Here, n = 9.
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.
Now,
Mode = 3 Median – 2 Mean
→ 14 = 3 x 14 – 2 Mean
→ 2 Mean = 42 – 14 = 28
→ Mean = 28 ÷ 2 = 14.
Question: 2
Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34
Solution:
Arranging the data in ascending order such that same numbers are put together, we get:
32, 32, 34, 35, 35, 38, 42
Here, n = 7
Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.
Question: 3
Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Solution:
Arranging the data in ascending order such that same values are put together, we get:
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
Here, 2, 3 and 4 occur three times each. Therefore, 2, 3 and 4 are the three modes.
Question: 4
The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10
Find the mean, mode and median of this data.
Solution:
Arranging the data in ascending order such that same values are put together, we get:
6, 8, 10, 10, 15, 15, 50, 80, 100, 120
Here, n = 11
Here, 10 occur three times. Therefore, 10 is the mode of the given data.
Now,
Mode = 3 Median – 2 Mean
→ 10 = 3 x 15 – 2 Mean
→ 2 Mean = 45 – 10 = 35
→ Mean = 35 ÷ 2 = 17.5
Question: 5
Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Solution:
Arranging the data in ascending order such that same values are put together, we get:
10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18
Here, clearly, 14 occurs the most number of times.
Find the circumference of a circle whose radius is
(i) 14 cm (ii) 10 m (iii) 4 km
Solution:
Question: 2
Find the circumference of a circle whose diameter is
(i) 7 cm (ii) 4.2 cm (iii) 11.2 km
Solution:
Question: 3
Find the radius of a circle whose circumference is
(i) 52.8 cm (ii) 42 cm (iii) 6.6 km
Solution:
Question: 4
Find the diameter of a circle whose circumference is
(i) 12.56 cm (ii) 88 m (iii) 11.0 km
Solution:
Question: 5
The ratio of the radii of two circles is 3 : 2. What is the ratio of their circumferences?
Solution:
We have, the ratio of the radii = 3 : 2
So, let the radii of the two circles be 3r and 2r respectively.
Question: 6
A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.
Solution:
Question: 7
A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. It is re-bent to form a circular ring. What is the diameter of the ring?
Solution:
Question: 8
The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.
Solution:
It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Now, the diameter of the wheel = 63 cm
∴ Circumference of the wheel = πd = 227 x 63 = 198 cm.
Thus, the cycle covers 198 cm in one revolution.
∴ The distance covered by the cycle in 1000 revolutions = (198 x 1000) = 198000 cm = 1980 m.
Question: 9
The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.
Solution:
Question: 10
The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.
Solution:
Question: 11
How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr?
Solution:
Question: 12
The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.
Solution:
Question: 13
A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m. Find the radius of the circle.
Solution:
Question: 14
A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.
Solution:
Question: 15
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Solution:
Question: 16
A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.
Solution:
Question: 17
The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?
Solution:
Question: 18
A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.
Solution:
Question: 19
A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm. then find the width of the parapet.
Solution:
Question: 20
An ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?
Solution:
Exercise 21.2
Question: 1
Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
Question: 2
Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
Question: 3
The area of a circle is 154 cm2. Find the radius of the circle.
Solution:
Question: 4
Find the radius of a circle, if its area is
(i) 4 it cm2
(ii) 55.44 m2
(iii) 1.54 km2
Solution:
Question: 5
The circumference of a circle is 3.14 m, find its area.
Solution:
Question: 6
If the area of a circle is 50.24 m2, find its circumference.
Solution:
Question: 7
A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22 / 7).
Solution:
Question: 8
A steel wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.
Solution:
Question: 9
A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.
Solution:
Question: 10
Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is πh(2r + h).
Solution:
Question: 11
The perimeter of a circle is 4πr cm. What is the area of the circle?
Solution:
Question: 12
A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.
Solution:
Question: 13
The radius of a circle is 14 cm. Find the radius of the circle whose area is double of the area of the circle.
Solution:
Question: 14
The radius of one circular field is 20 m and that of another is 48 m. find the radius of the third circular field whose area is equal to the sum of the areas of two fields.
Solution:
Question: 15
The radius of one circular field is 5 m and that of the other is 13 m. Find the radius of the circular field whose area is the difference of the areas of first and second field.
Solution:
Question: 16
Two circles are drawn inside a big circle with diameters 2/3rd and 1/3rd of the diameter of the big circle as shown in Figure. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm.
Solution:
Question: 17
In Figure, the radius of quarter circular plot taken is 2 m and radius of the flower bed is 2 m. Find the area of the remaining field.
Solution:
Question: 18
Four equal circles, each of radius 5 cm, touch each other as shown in Figure. Find the area included between them. (Take π = 3.14).
Solution:
Question: 19
The area of circle is 100 times the area of another circle. What is the ratio of their circumferences?
Find the area, in square meters, of a rectangle whose
(i) Length = 5.5 m, breadth = 2.4 m
(ii) Length = 180 cm, breadth = 150 cm
Solution:
We have,
(i) Length = 5.5 m, Breadth = 2.4 m Therefore, Area of rectangle = Length x Breadth = 5.5 m x 2.4 m = 13.2 m2
(ii) Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m [ Since 100 cm = 1 m] Therefore, Area of rectangle = Length x Breadth = 1.8 m x 1.5 m = 2.7 m2
Question: 2
Find the area, in square centimeters, of a square whose side is
(i) 2.6 cm
(ii) 1.2 dm
Solution:
We have,
(i) Side of the square = 2.6 cm
Therefore, area of the square = (Side)2 = (2.6 cm)2= 6.76 cm2
(ii) Side of the square = 1.2 dm = 1.2 x 10 cm = 12 cm
Therefore, area of the square = (Side)2 = (12 cm)2= 144 cm2 [ Since 1 dm = 10 cm]
Question: 3
Find in square metres, the area of a square of side 16.5 dam.
Solution:
We have,
Side of the square = 16.5
dam = 16.5 x 10 m = 165 m
Area of the square = (Side)2 = (165 m)2 = 27225 m2
[Since 1 dam/dm (decameter) = 10 m]
Question: 4
Find the area of a rectangular field in acres whose sides are:
(1) 200 m and 125 m
(ii) 75 m 5 dm and 120 m
Solution:
We have,
(i) Length of the rectangular field = 200 m
Breadth of the rectangular field = 125 m
Therefore, Area of the rectangular field = Length x Breadth = 200 m x 125 m
= 25000 m2 = 250 acres [Since 100 m2 = 1 are]
(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m
= 75.5 m [Since 1 dm = 10 cm = OA m]
Breadth of the rectangular field = 120 m
Therefore, Area of the rectangular field = Length x Breadth
= 75.5 m x 120 m = 9060 m2 = 90.6 acres [Since 100 m2 = 1 are]
Question: 5
Find the area of a rectangular field in hectares whose sides are:
(i) 125 m and 400 m
(ii) 75 m 5 dm and 120 m
Solution:
We have,
(i) Length of the rectangular field = 125 m
Breadth of the rectangular field = 400 m
Therefore, Area of the rectangular field = Length x Breadth
= 125 m x 400 m = 50000 m2 = 5 hectares [Since 10000 m2 = 1 hectare]
(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m
= 75.5 m [Since 1 dm = 10 cm = 0.1 m]
Breadth of the rectangular field = 120 m
Therefore, Area of the rectangular field = Length x Breadth
= 75.5 m x 120 m = 9060 m2 = 0.906 hectares [Since 10000 m2 = 1 hectare]
Question: 6
A door of dimensions 3 m x 2m is on the wall of dimension 10 m x 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.
Solution:
We have,
Length of the door = 3 m
Breadth of the door = 2 m
Side of the wall = 10 m
Area of the wall = Side x Side = 10 m x 10 m
= 100 m2
Area of the door = Length x Breadth = 3 m x 2 m = 6 m
Thus, required area of the wall for painting = Area of the wall – Area of the door
= (100 – 6) m2= 94 m2
Rate of painting per square metre = Rs. 2.50
Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235
Question: 7
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side? Also, find which side encloses more area?
Solution:
We have,
Perimeter of the rectangle = 2(Length + Breadth)
= 2(40 cm + 22 cm) = 124 cm
It is given that the wire which was in the shape of a rectangle is now bent into a square.
Therefore, the perimeter of the square = Perimeter of the rectangle
→ Perimeter of the square = 124 cm
4 x side = 124 cm
Side = 124/4 = 31 cm
Now, Area of the rectangle = 40 cm x 22 cm = 880 cm2
Area of the square = (Side)2 = (31 cm)2 = 961 cm2.
Therefore, the square-shaped wire encloses more area.
Question: 8
How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?
Solution:
We have,
Length of the glass pane = 25 cm
Breadth of the glass pane = 16 cm
Area of one glass pane = 25 cm x 16 cm
= 400 cm2 = 0.04 m2
[Since 1 m2 = 10000 cm2 ]
Thus, Area of 12 such panes = 12 x 0.04 = 0.48 m2
Question: 9
A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m x 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.
Solution:
We have,
Area of the wall = 3 m x 4 m = 12 m2
Area of one marble tile = 10 cm x 12 cm
= 120 c m2 = 0.012 m2 [Since 1 m2 = 10000 c m2 ]
Thus, Number of tiles = Area of wall
Area of one tile=12 m2 = 0.012 m2=1000
Cost of one tile = Rs. 2
Total cost = Number of tiles x Cost of one tile
= Rs. (1000 x 2)
= Rs. 2000
Question: 10
A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?
Solution:
We have,
Length of the table top = 9 dm 5 cm = (9 x 10 + 5) cm = 95 cm [ Since 1 dm = 10 cm]
Breadth of the table top = 6 dm 5 cm = (6 x 10 + 5) cm = 65 cm
Area of the table top = Length x Breadth = (95 cm x 65 cm) = 6175 c m2
Rate of polishing per square centimetre = 20 paise = Rs. 0.20
Total cost = Rs. (6175 x 0.20) = Rs. 1235
Question: 11
A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile.
Solution:
We have,
Length of the floor of the room = 9.68 m
Breadth of the floor of the room = 6.2 m
Area of the floor = 9.68 m x 6.2 m = 60.016 m2
Length of the tile = 22 cm
Breadth of the tile = 10 cm
Area of one tile = 22 cm x 10 cm = 220 c m2 = 0.022 m2 [Since 1 m2 = 10000 c m2]
Thus, Number of tiles = 60.016 m2/0.022 m2=2728
Cost of one tile = Rs. 2.50
Total cost = Number of tiles x Cost of one tile = Rs. (2728 x 2.50) = Rs. 6820
Question: 12
One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.
Solution:
We have,
Side of the square field = 179 m
Area of the field = (Side) 2 = (179 m) 2 = 32041 m2
Rate of raising a lawn on the field per square metre = Rs. 1.50 Thus,
Total cost of raising a lawn on the field = Rs. (32041 x 1.50) = Rs. 48061.50
Question: 13
A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec?
Solution:
We have,
Length of the rectangular field = 290 m
Breadth of the rectangular field = 210 m
Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m
Distance covered by the girl = 2 x Perimeter of the rectangular field = 2 x 1000 = 2000 m
The girl walks at the rate of 1.5 m/sec. Or, Rate = 1.5 x 60 m/min = 90 m/min
Thus, required time to cover a distance of 2000 m = 2000 m/90 m/min
= 2229min
Hence, the girl will take 2229 min to go two times around the field.
Question: 14
A corridor of a school is 8 m long and 6 m wide. It is to be covered with canvas sheets. If the available canvas sheets have the size 2 m x 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet.
Solution:
We have,
Length of the corridor = 8 m
Breadth of the corridor = 6 m
Area of the corridor of a school = Length x Breadth = (8 m x 6 m) = 48 m2
Length of the canvas sheet = 2 m
Breadth of the canvas sheet = 1 m
Area of one canvas sheet = Length x Breadth = (2 m x 1 m) = 2 m2
Thus, Number of canvas sheets = 48 m2 /2m2=24
Cost of one canvas sheet = Rs. 8
Total cost of the canvas sheets = Rs. (24 x 8) = Rs. 192
Question: 15
The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second?
Solution:
We have,
Length of a playground = 62 m 60 cm = 62.6 m [ Since 10 cm = 0.1 m]
Breadth of a playground = 25 m 40 cm = 25.4 m
Area of a playground = Length x Breadth= 62.6 m x 25.4 m = 1590.04 m2
Rate of turfing = Rs. 2.50/ m2 Total cost of turfing = Rs. (1590.04 x 2.50) = Rs. 3975.10
Again, Perimeter of a rectangular field = 2(Length + Breadth) = 2(62.6 + 25.4) = 176 m
Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field
= 3 x 176 m = 528 m
The man walks at the rate of 2 m/sec. Or, Rate = 2 x 60 m/min = 120 m/min
Thus, required time to cover a distance of 528 m = 528 m120 m/min=4.4 min
A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs 750 per thousand.
Solution:
We have,
Length of the lane = 180 m
Breadth of the lane = 5 m
Area of a lane = Length x Breadth = 180 m x 5 m = 900 m2
Length of the brick = 20 cm
Breadth of the brick = 15 cm
Area of a brick = Length x Breadth = 20 cm x 15 cm
= 300 cm2 = 0.03 m2 [Since 1 m2 = 10000 cm2]
Required number of bricks = 900 m2/0.03 m2 = 30000
Cost of 1000 bricks = Rs. 750
Total cost of 30,000 bricks = Rs. 750×30,000/1000 = Rs. 22,500
Question: 17
How many envelopes can be made out of a sheet of paper 125 cm by 85 cm; supposing one envelope requires a piece of paper of size 17 cm by 5 cm?
Solution:
We have,
Length of the sheet of paper = 125 cm
Breadth of the sheet of paper = 85 cm
Area of a sheet of paper = Length x Breadth = 125 cm x 85 cm = 10,625 cm2
Length of sheet required for an envelope = 17 cm
Breadth of sheet required for an envelope = 5 cm
Area of the sheet required for one envelope = Length x Breadth
= 17 cm x 5 cm = 85 c m2
Thus, required number of envelopes = 10,625 cm2/85 c m2 = 125
Question: 18
The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.
Solution:
We have,
Length of the diaper = 50 cm
Breadth of the diaper = 17 cm
Area of cloth to make 1 diaper = Length x Breadth = 50 cm x 17 cm = 850 cm2
Thus, Area of 25 such diapers = (25 x 850) c m2 = 21,250 cm2
Area of total cloth = Area of 25 diapers = 21,250 cm2
It is given that width of a cloth = 170 cm
Length of the cloth = Area of cloth
Width of a cloth = 21,250 cm 2170 cm = 125 cm
Hence, length of the cloth will be 125 cm.
Question: 19
The carpet for a room 6.6 m by 5.6 m costs Rs 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metre.
Solution:
We have,
Length of a room = 6.6 m
Breadth of a room = 5.6 m
Area of a room = Length x Breadth = 6.6 m x 5.6 m = 36.96 m2
Width of a carpet = 70 cm = 0.7 m [Since 1 m = 100 cm]
Length of a carpet = Area of a room
Width of a carpet = 36.96 m 20.7 m = 52.8 m
Cost of 52.8 m long roll of carpet = Rs. 3960
Therefore, Cost of 1 m long roll of carpet = Rs. 396052.8 = Rs. 75
Question: 20
A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white T4L washing the walls at Rs 3.80 per square metre.
Solution:
We have,
Length of a room = 9 m
Breadth of a room = 8 m
Height of a room = 6.5 m
Area of 4 walls = 2(1 + b)h = 2(9 m + 8 m) x 6.5 m = 2 x 17 m x 6.5 m = 221 m2
Length of a door = 2 m
Breadth of a door = 1.5 m
Area of a door = Length x Breadth = 2 m x 1.5 m = 3 m2
Length of a window = 1.5 m
Breadth of a window = 1 m
Since, area of one window = Length x Breadth = 1.5 m x 1 m = 1.5 m2
Thus, Area of 3 such windows = 3 x 1.5 m2 = 4.5 m2
Area to be white-washed = Area of 4 walls – (Area of one door + Area of 3 windows)
Area to be white-washed = [221 – (3 + 4.5)] m2 = (221 – 7.5) m2 = 213.5 m2
Cost of white-washing for 1 m2 area = Rs. 3.80
Cost of white-washing for 213.5 m2 area = Rs. (213.5 x 3.80) = Rs. 811.30
Question: 21
A hall 36 m long and 24 m broad allowing 80 m2 for doors and windows, the cost of papering the walls at Rs 8.40 per m2 is Rs 9408. Find the height of the hall.
Solution:
We have,
Length of the hall = 36 m
Breadth of the hall = 24 m
Let h be the height of the hall.
Now, in papering the wall, we need to paper the four walls excluding the floor and roof of the hall. So, the area of the wall which is to be papered = Area of 4 walls
= 2h(I + b)
= 2h (36 + 24)
= 120h m2
Now, area left for the door and the windows = 80 m2
So, the area which is actually papered = (120h – 80) m2
Again, The cost of papering the walls at Rs 8.40 per m2 = Rs. 9408
→ (120h – 80) m2 x Rs. 8.40 per m2= Rs. 9408
→ (120h – 80) m2 = Rs. 9408/Rs. 8.40
→ (120h – 80) m2 = 1120 m2
→ 120h m2 = (1120 + 80) m2
→ 120h m2= 1200 m2
h = 1200 m2 120 m = 10 m
Hence, the height of the wall would be 10 m.
Exercise 20.2
Question: 1
A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25 per square metre.
Solution:
We have,
Length AB = 40 m and breadth BC = 25 m
Area of lawn ABCD = 40 m x 25 m = 1000 m2
Length PQ = (40 + 2 + 2 ) m = 44 m
Breadth QR = ( 25 + 2 + 2 ) m = 29 m
Area of PQRS = 44 m x 29 m = 1276 m2
Now, Area of the path = Area of PQRS – Area of the lawn ABCD
= 1276 m2 – 1000 m2 = 276 m2
Rate of leveling the path = Rs. 8.25 per m2
Cost of leveling the path = Rs.( 8.25 x 276) = Rs. 2277
Question: 2
One metre wide path is built inside a square park of side 30 m along its sides. The remaining part of the park is covered by grass. If the total cost of covering by grass is Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
We have,
Side of square garden (a) = 30 m
Area of the square garden including the path = a2= (30)2 = 900 m2
From the figure, it can be observed that the side of the square garden, when the path is not included, is 28 m.
Area of the square garden not including the path = (28)2 = 784 m2
Total cost of covering the park with grass = Area of the park covering with green grass x Rate per square metre
1176 = 784 x Rate per square metre
Rate per square metre at which the park is covered with grass = Rs. (1176 ÷ 784 ) = Rs. 1.50
Question: 3
Through a rectangular field of sides 90 m x 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the field. if the width of the roads is 3 m, find the total area covered by the two roads.
Solution:
Length of the rectangular sheet = 90 m
Breadth of the rectangular sheet = 60 cm
Area of the rectangular field = 90 m x 60 m = 5400 m2
Area of the road PQRS = 90 m x 3 m = 270 m2
Area of the road ABCD = 60 m x 3 m = 180 m2
Clearly, area of KLMN is common to the two roads.
Thus, area of KLMN = 3 m x 3 m = 9 m2
Hence, Area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)
= (270 + 180) m2 – 9 m2 = 441 m2
Question: 4
from a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.
Solution:
Length of the rectangular sheet = 100 cm
Breadth of the rectangular sheet = 80 cm
Area of the rectangular sheet of tin = 100 cm x 80 cm
= 8000 c m2
Side of the square at the corner of the sheet = 10 cm
Area of one square at the corner of the sheet = (10 cm)2 = 100 cm2
Area of 4 squares at the corner of the sheet = 4 x 100 cm2 = 400 cm2
Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4 squares
Area of the remaining sheet of tin = (8000 – 400) cm2 = 7600 cm2
Question: 5
A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
We have, Length of the cardboard = 8 cm and breadth of the cardboard = 5 cm
Area of the cardboard including the margin = 8 cm x 5 cm = 40 c m2
From the figure, it can be observed that,
New length of the painting when the margin is not included = 8 cm – (1.5 cm + 1.5 cm)
= (8 – 3) cm = 5 cm
New breadth of the painting when the margin is not included = 5 cm – (1.5 cm + 1.5 cm)
= (5 – 3) cm = 2 cm
Area of the painting not including the margin = 5 cm x 2 cm = 10 cm2
Hence, Area of the margin = Area of the cardboard including the margin – Area of the painting
= (40 – 10) cm2 = 30 cm2
Question: 6
Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to make a garden 10 m long and 4 m broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining part of the field, he wants to apply manures. Find the cost of applying the manures at the rate of Rs 300 per area.
Solution:
Length of the rectangular field = 80 m
Breadth of the rectangular field = 60 m
Area of the rectangular field = 80 m x 60 = 4800 m2
Again, Area of the garden = 10 m x 4 m = 40 m2
Area of one flower bed = 4 m x 1.5 m = 6 m2
Thus, Area of two flower beds = 2 x 6 m2 = 12 m2
Remaining area of the field for applying manure = Area of the rectangular field – (Area of the garden + Area of the two flower beds)
Remaining area of the field for applying manure = 4800 m2 – (40 + 12) m2
= (4800 – 52 ) m2 = 4748 m2
Since 100 m2 = 1 acre → 4748 m2 = 47.48 acres
So, cost of applying manure at the rate of Rs. 300 per are will be Rs. (300 x 47.48) = Rs. 14244
Question: 7
Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip 30 cm wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.
Solution:
We have ,
Side of the flower bed = 2 m 80 cm = 2.80 m [since 100 cm = 1 m ]
Area of the square flower bed = (Side)2 = (2.80 m )2 = 7.84 m2
Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm
= (2.80 + 0.3 + 0.3) m
= 3.4 m
Area of the enlarged flower bed with the digging strip = (Side) 2 = (3.4)2 = 11.56 m2
Thus, Increase in the area of the flower bed = 11.56 m2 – 7.84 m2 = 3.72 m2
Question: 8
A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m2 , find the width of the varandah.
Solution:
Let the width of the verandah be x m.
Length of the room AB = 5 m and BC = 4 m
Area of the room = 5 m x 4 m = 20 m2
Length of the verandah PQ = (5 + x + x) = (5 + 2x) m
Breadth of the verandah QR = ( 4 + x + x) = (4 + 2x) m
Area of verandah PQRS = (5 + 2x) x (4 + 2x) = (4×2 + 18x + 20 ) m2
Area of verandah = Area of PQRS – Area of ABCD
→ 22 = 4x2 + 18x + 20 – 20
22 = 4x2 + 18x
11 = 2x2 + 9x
2x2 + 9x – 11 = 0
2x2 + 11x – 2x – 11 =0
x(2x+11)-1(2x+11)=0
(x- 1)(2x+11)= 0
When x – 1 = 0, x = 1
When 2x + 11 = 0, x = -11/2
The width cannot be a negative value. So, width of the verandah = x = 1 m.
Question: 9
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m2 , find the area of the lawn.
Solution:
Let ABCD be the square lawn and PQRS be the outer boundary of the square path.
Let side of the lawn AB be x m.
Area of the square lawn = x2
Length PQ = (x m + 2 m + 2 m) = (x + 4) m
Area of PQRS = (x + 4)2 = (x2 + 8x + 16) m2
Now, Area of the path = Area of PQRS – Area of the square lawn
136 = x2 + 8x + 16 – x2
136 = 8x + 16
136 – 16 = 8x
120 = 8x
x = 120/ 8 = 15
Side of the lawn = 15 m Hence, Area of the lawn = (Side)2 = (15 m)2 = 225 m2
Question: 10
A poster of size 10 cm by 8 cm is pasted on a sheet of cardboard such that there is a margin of width 1.75 cm along each side of the poster. Find (i) the total area of the margin (ii) the cost of the cardboard used at the rate of Re 0.60 per c m2 .
Solution:
We have,
Length of poster = 10 cm and breadth of poster = 8 cm
Area of the poster = Length x Breadth = 10 cm x 8 cm = 80 cm2
From the figure, it can be observed that,
Length of the cardboard when the margin is included = 10 cm + 1.75 cm + 1.75 cm = 13.5 cm
Breadth of the cardboard when the margin is included = 8 cm + 1.75 cm + 1.75 cm = 11.5 cm
Area of the cardboard = Length x Breadth = 13.5 cm x 11.5 cm = 155.25 c m2
Hence,
(i) Area of the margin = Area of cardboard including the margin – Area of the poster
= 155.25 c m2 – 80 c m2
= 75.25 c m2
(ii) Cost of the cardboard = Area of cardboard x Rate of the cardboard Rs 0.60 per cm2
= Rs. (155.25 x 0.60)
= Rs. 93.15
Question: 11
A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The widths of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the area of the remaining portion of the field.
Solution:
Let ABCD be the rectangular field and KLMN and PQRS the two rectangular roads with width 1.8 m and 2.5 m, respectively.
Length of the rectangular field CD = 50 cm and breadth of the rectangular field BC = 40 m
Area of the rectangular field ABCD = 50 m x 40 m = 2000 m2
Area of the road KLMN = 40 m x 2.5 m = 100 m2
Area of the road PQRS = 50 m x 1.8 m = 90 m2
Clearly area of EFGH is common to the two roads.
Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m2
Hence, Area of the roads = Area (KLMN) + Area (PQRS) – Area (EFGH)
= (100 m2 + 90 m2) – 4.5 m2
= 185.5 m2
Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads
= (2000 – 185.5) m2
= 1814.5 m2
Question: 12
There is a rectangular field of size 94 m x 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.
Solution:
Let ABCD be the rectangular field.
Here, Two roads which are parallel to the breadth of the field KLMN and EFGH with width 2 m each. One road which is parallel to the length of the field PQRS with width 2 m.
Length of the rectangular field AB = 94 m and breadth of the rectangular field BC = 32 m
Area of the rectangular field = Length x Breadth = 94 m x 32 m = 3008 m2
Area of the road KLMN = 32 m x 2 m = 64 m2
Area of the road EFGH = 32 m x 2 m = 64 m2
Area of the road PQRS = 94 m x 2 m = 188 m2
Clearly area of TUVI and WXYZ is common to these three roads.
Thus, Area of TUV1 = 2 m x 2 m = 4 m2
Area of WXYZ = 2 m x 2 m = 4 m2
Hence,
(i) Area of the field covered by the three roads: = Area (KLMN) + Area (EFGH) + Area (PQRS) – {Area (TUVI) + Area (WXYZ)}
= [ 64+ 64 + 188 – (4 + 4 )] m2
= 316 m2 – 8 m2
= 308 m2
(ii) Area of the field not covered by the roads: = Area of the rectangular field ABCD – Area of the field covered by the three roads
= 3008 m2 – 308 m2
= 2700 m2
Question: 13
A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find the cost at the rate of Rs 18 per metre.
Solution:
We have,
Length of hall PQ = 22 m and breadth of hall QR = 15.5 m
Area of the school hall PQRS = 22 m x 15.5 m = 341 m2
Length of the carpet AB = 22 m – ( 0.75 m + 0.75 m) = 20.5 m [ Since 100 cm = 1 m]
Breadth of the carpet BC = 15.5 m – ( 0.75 m + 0.75 m) = 14 m
Area of the carpet ABCD = 20.5 m x 14 m = 287 m2
Area of the strip = Area of the school hall PQRS – Area of the carpet ABCD
= 341 m2 – 287 m2 = 54 m2
Again, Area of the 1 m length of carpet = 1 m x 0.82 m = 0.82 m2
Thus, Length of the carpet whose area is 287 m2 = 287 m2 + 0.82 m2 = 350 m
Cost of the 350 m long carpet = Rs. 18 x 350 = Rs. 6300
Question: 14
Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m2 .
Solution:
Let ABCD be the rectangular park then EFGH and IJKL the two rectangular roads with width 5 m.
Length of the rectangular park AD = 70 cm
Breadth of the rectangular park CD = 45 m
Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m2
Area of the road EFGH = 70 m x 5 m = 350 m2
Area of the road JKIL = 45 m x 5 m = 225 m2
Clearly area of MNOP is common to the two roads.
Thus, Area of MNOP = 5 m x 5 m = 25 m2
Hence,
Area of the roads = Area (EFGH) + Area (JKIL) – Area (MNOP)
= (350 + 225) m2– 25 m2 = 550 m2
Again, it is given that the cost of constructing the roads = Rs. 105 per m2
Therefore,
Cost of constructing 550 m2 area of the roads = Rs. (105 x 550)
= Rs. 57750.
Question: 15
The length and breadth of a rectangular park are in the ratio 5: 2. A 2.5 m wide path running all around the outside the park has an area 305 m2 . Find the dimensions of the park.
Solution:
We have,
Area of path = 305 m2
Let the length of the park be 5x m and the breadth of the park be 2x m
Thus,
Area of the rectangular park = (5x) x (2x) = 10x2 m2
Width of the path = 2.5 m
Outer length PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m
Outer breadth QR = 2x + 2.5 m + 2.5 m = (2x + 5) m
Area of PQRS = (5x + 5) m x (2x + 5) m = (10x2 + 25x + 10x + 25) m2= (10x2 + 35x + 25) m2
Area of the path = [(10x2 + 35x + 25) – 10x2] m2
→ 305 = 35x + 25
→ 305 – 25 = 35x
→ 280 = 35x
→ x = 280 + 35 = 8
Therefore,
Length of the park = 5x = 5 x 8 = 40 m
Breadth of the park = 2x = 2 x 8 = 16 m
Question: 16
A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m2 , find the area of the lawn.
Solution:
Let the side of the lawn be x m.
Given that width of the path = 2.5 m
Side of the lawn including the path = (x + 2.5 + 2.5) m = (x + 5 ) m
So, area of lawn = (Area of the lawn including the path) – (Area of the path)
We know that the area of a square = (Side)2
Area of lawn (x2) = (x + 5)2 – 165
→ x2 = (X2 + 10X + 25) – 165
→ 165 = 10x + 25
→ 165 – 25 =10x
→ 140 = 10x
Therefore x = 140 / 10 = 14
Thus the side of the lawn = 14 m
Hence,
The area of the lawn = (14 m) 2 = 196 m2
Exercise 20.3
Question: 1
Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
We have,
Base = 8 cm and altitude = 4.5 cm
Thus, Area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
= 36 cm2
Question: 2
Find the area in square metres of the parallelogram whose base and altitudes are as under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
We have,
(i) Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
Thus, Area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
= 0.96 m2
(ii) Base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m
Thus, Area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m2 [Since 100 cm = 1 m]
Question: 3
Find the altitude of a parallelogram whose area is 54 d m2 and base is 12 dm.
Solution:
We have,
Area of the given parallelogram = 54 d m2
Base of the given parallelogram = 12 dm
Altitude of the given parallelogram = Area/Base = 54/12 dm = 4.5 dm
Question: 4
The area of a rhombus is 28 m2. If its perimeter be 28 m, find its altitude.
Solution:
We have,
Perimeter of a rhombus = 28 m 4(Side) = 28 m [Since perimeter = 4(Side)]
Side = 28 m4=7 m
Now, Area of the rhombus = 28 m2
(Side x Altitude) = 28 m2 (7 m x Altitude) = 28 m2
Altitude = 28 m 27m = 4 m
Question: 5
In Fig., ABCD is a parallelogram, DL ⊥ AB and DM ⊥ BC. If AB = 18 cm, BC =12 cm and DM= 9.3 cm, find DL.
Solution:
We have,
Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm
Area of parallelogram ABCD = Base x Altitude = (12 cm x 9.3 cm) = 111.6 c m2 —–(i)
Now, Taking AB as the base,
We have, Area of the parallelogram ABCD = Base x Altitude
= (18 cm x DL) —–(ii)
From (i) and (ii), we have 18 cm x DL = 111.6 c m2
DL = 111.6 cm 218 cm = 6.2 cm
Question: 6
The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.
Solution:
We have,
ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude AE = 16 cm. The shorter side is BC and the corresponding altitude is CF = 24 cm.
Area of a parallelogram = base x height.
We have two altitudes and two corresponding bases.
So,
½ x BC x CF = ½ x AB x AE
= BC x CF = AB x AE
= BC x 24 = 54 x 16
= BC = (54×16)/24 = 36 cm
Hence, the length of the shorter side BC = AD = 36 cm.
Question: 7
In Fig. 21, ABCD is a parallelogram, DL ⊥ AB. If AB = 20 cm, AD = 13 cm and area of the parallelogram is 100 c m2, find AL.
Solution:
We have,
ABCD is a parallelogram with base AB = 20 cm and corresponding altitude DL.
It is given that the area of the parallelogram ABCD = 100 c m2
Now, Area of a parallelogram = Base x Height
100 c m2 = AB x DL
100 c m2 = 20 cm x DL
DL = 100 c m2= 5 cm
Again by Pythagoras theorem, we have,
(AD)2 = (AL)2 + (DL)2
= (13)2 = (AL)2+ (5)2
(AL)2= (13)2 – (5)2
= 169 – 25 = 144
(AL)2 = (12)2
AL = 12 cm
Hence, length of AL is 12 cm.
Question: 8
In Fig. 21, if AB = 35 cm, AD= 20 cm and area of the parallelogram is 560 cm2, find LB.
Solution:
We have,
ABCD is a parallelogram with base AB = 35 cm and corresponding altitude DL.
The adjacent side of the parallelogram AD = 20 cm.
It is given that the area of the parallelogram ABCD = 560 c m2
Now, Area of the parallelogram = Base x Height
560 cm2 = AB x DL 560 c m2 = 35 cm x DL
DL = 560 cm/235 cm= 16 cm
Again by Pythagoras theorem, we have, (AD)2 = (AL)2 + (DL)2
(20)2= (AL)2 + (16)2
(AL)2 = (20)2 – (16)2
= 400 – 256
= 144
(AL)2 = (12)2
= AL = 12 cm
From the figure, AB = AL + LB 35 cm
= 12 cm + LB
LB = 35 cm – 12 cm = 23 cm
Hence, length of LB is 23 cm.
Question: 9
The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides.
Solution:
We have,
ABCD is a parallelogram with side AB = 10 m and corresponding altitude AE = 4 m.
The adjacent side AD = 8 m and the corresponding altitude is CF.
Area of a parallelogram = Base x Height
We have two altitudes and two corresponding bases.
So, AD x CF = AB x AE = 8 m x CF = 10 m x 4 m
= CF = (10 x 4)8 = 5 m
Hence, the distance between the shorter sides is 5 m.
Question: 10
The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm2, find the base and height.
Solution:
Let the height of the parallelogram be x cm.
Then the base of the parallelogram is 2x cm.
It is given that the area of the parallelogram = 512 cm2
So, Area of a parallelogram = Base x Height
512 c m2 = (2x) (x)
512 c m2 = 2x2
X2 = 512 cm2/2 = 256 c m2
X2 = (16 cm)2
X = 16 cm
Hence, base = 2x = 2 x 16 = 32 cm and height = x = 16 cm.
Question: 11
Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.
Solution:
Let ABCD be the rhombus where diagonals intersect at 0.
Then AB = 15 cm and AC = 24 cm.
The diagonals of a rhombus bisect each other at right angles.
Therefore, triangle A0B is a right-angled triangle, right angled at O such that
OA = ½(AC) = 12 cm and AB = 15 cm.
By Pythagoras theorem, we have,
(AB)2 = (OA)2 + (OB)2
(15)2 = (12)2 + (OB)2
(OB)2= (15)2– (12)2
(OB)2 = 225 – 144 = 81
(OB)2 = (9)2
OB = 9 cm
BD = 2 x OB = 2 x 9 cm = 18 cm
Hence, Area of the rhombus ABCD = (½ x AC x BD)
=( 1/2 x 24 x 18 )
= 216 cm2
Question: 12
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Let ABCD be the rhombus whose diagonals intersect at 0.
Then AB = 20 cm and AC = 24 cm.
The diagonals of a rhombus bisect each other at right angles.
Therefore Triangle AOB is a right-angled triangle, right angled at O
Such that;
OA = ½ AC =12 cm and AB =20 cm
By Pythagoras theorem, we have,
(AB)2 = (OA)2+ (OB) 2
(20)2 = (12)2 + (OB) 2
(OB) 2= (20) 2– (12) 2
(OB)2 = 400 – 144
= 256
(OB) 2 = (16) 2
= OB = 16 cm
BD =2 x OB = 2 x 16 cm = 32 cm
Hence, Area of the rhombus ABCD = ½ x AC x BD
= 1/2 x 24 x 32
= 384 c m2
Question: 13
The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?
Solution:
We have,
Area of the rhombus = Area of the square of side 4 m
= 1/2 x AC x 130 = 4 m2
= 1/2 x AC x 2 m =16 m2
= AC = 16 m
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
= AO = 1/2 (AC) = 8 m and BO = 1/2 (BD) = 1 m
By Pythagoras theorem, we have:
AO2 + BO2 = AB2
AB2 = (8 m) 2 + (1 m) 2 = 64 m2 + 1 m2 = 65 m2
Side of a rhombus = AB = √65 m.
Let DX be the altitude.
Area of the rhombus = AB x DX 16 m2
= √65m × DX
DX = 16/(√65) m
Hence, the altitude of the rhombus will be 16/√65 m.
Question: 14
Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.
Solution:
We have,
ABCD is a parallelogram with longer side AB = 25 cm and altitude AE = 10 cm.
As ABCD is a parallelogram. Hence AB = CD (opposite sides of parallelogram are equal).
The shorter side is AD = 20 cm and the corresponding altitude is CF.
Area of a parallelogram = Base x Height
We have two altitudes and two corresponding bases.
So, = AD x CF = CD x AE
= 20 x CF = 25 x 10
CF = 12.5 cm
Hence, the altitude corresponding to the other pair of the side AD is 12.5 cm.
Question: 15
The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.
Solution:
We have,
ABCD is a parallelogram with side AB = CD = 10 cm (Opposite sides of parallelogram are equal) and corresponding altitude AM = 12 cm. The other side is AD and the corresponding altitude is CN = 8 cm.
Area of a parallelogram = Base x Height
We have two altitudes and two corresponding bases.
So,
= AD x CN = CD x AM
= AD x 8 = 10 x 12
= AD = (10×12)/8=15 cm
Hence, the length of the other pair of the parallel side = 15 cm.
Question: 16
A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm2.
Solution:
We have,
Attitude of a tile = 3 cm
Base of a tile = 5 cm
Area of one tile = Attitude x Base = 5 cm x 3 cm = 15 c m2
Area of 280 tiles = 280 x 15 c m2 = 4200 c m2
Rate of polishing the tiles at 50 paise per c m2 = Rs. 0.5 per c m2
Thus, Total cost of polishing the design = Rs. (4200 x 0.5) = Rs. 2100
Exercise 20.4
Question: 1
Find the area in square centimeters of a triangle whose base and altitude are as under :
(i) base =18 cm, altitude = 3.5 cm
(ii) base = 8 dm, altitude =15 cm
Solution:
We know that the area of a triangle = 1/2 (Base x Height)
(i) Here, base = 18 cm and height = 3.5 cm
Area of the triangle = 1/2 x 18 x 3.5
= 31.5 cm2
(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm] and height = 3.5 cm
Area of the triangle = 1/2 x 80 x 15
= 600 c m2
Question: 2
Find the altitude of a triangle whose area is 42 cm2 and base is 12 cm.
Solution:
We have,
Attitude of a triangle = (2 x Area)/Base
Here, base = 12 cm and area = 42 cm2
Attitude = (2 x 42)/12 = 7 cm
Question: 3
The area of a triangle is 50 cm2. If the altitude is 8 cm, what is its base?
Solution:
We have,
Base of a triangle = (2 x Area)/ Altitude
Here, altitude = 8 cm and area = 50 cm2
Altitude = (2 x 50)/ 8 = 12.5 cm
Question: 4
Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 141 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m2
Question: 5
The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, we have,
Base = 15 cm and altitude = 7 cm
Thus, area of a triangle = 1/2 (Base x Altitude)
= 1/2 (15 x 7)
= 52.5 cm2
It is given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
Question: 6
A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?
Solution:
We have,
Length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth = 48 m x 20 m = 960 m2
Area of one right triangular flower bed = ½ (12 m x 5m) = 30 m2
Therefore,
Required number of right triangular flower beds = 960 m2/30 m2= 32
Question: 7
In Figure, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ⊥AC, BM ⊥ AC, DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
We have,
AC = 84 cm, DL = 16.5 cm and BM = 12 cm
Area of triangle ADC = 1/2 (AC x DL) = 1/2 (84 cm x 16.5 cm) = 693 cm2
Area of triangle ABC = 1/2 (AC x BM) = 1/2 (84 cm x 12 cm) = 504 cm2
Hence, Area of quadrilateral ABCD = Area of ADC + Area of ABC = (693 + 504) cm2 = 1197 cm2
Question: 8
Find the area of the quadrilateral ABCD given in Figure. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.
Solution:
We have,
Diagonal AC = 48 cm and diagonal BD = 32 m
Area of a quadrilateral = 1/2 (Product of diagonals)
= 1/2(AC x BD) = 1/2 (48 x 32) m2
= (24 x 32) m2 = 768 m2
Question: 9
In Fig below, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF⊥ AD and EF= 14 cm. Calculate the area of the shaded region.
Solution:
We have,
Area of the rectangle = AB x BC = 32 m x 18 m = 576 m2
Area of the triangle = 1/2 (AD x FE)
= 1/2 (BC x FE) [Since AD = BC]
= 1/2 (18 m x 14 m)
= 9 m x 14 m
= 126 m2
Area of the shaded region = Area of the rectangle – Area of the triangle
= (576 – 126) m2
= 450 m2
Question: 10
In Fig. below, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q, R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the shaded region.
Solution:
Join points PR and SQ. These two lines bisect each other at point 0.
Here, AB = DC = SQ = 40 cm and AD = BC = RP = 25 cm
Also OP = OR = RP/2 = 25/2 = 12.5 cm
From the figure we observed that,
Area of Triangle SPQ = Area of Triangle SRQ
Hence, area of the shaded region = 2 x (Area of SPQ)
= 2 x (1/2 (SQ x OP))
= 2 x (1/2 (40 cm x 12.5 cm))
= 500 cm2
Question: 11
Calculate the area of the quadrilateral ABCD as shown in Figure, given that BD = 42 cm, AC = 28 cm, OD = 12 cm and AC ⊥ BO.
Solution:
We have,
BD = 42 cm, AC = 28 cm, OD= 12 cm
Area of Triangle ABC = 1/2 (AC x OB)
= 1/2 (AC x (BD – OD))
= 1/2 (28 cm x (42 cm – 12 cm))
= 1/2 (28 cm x 30 cm)
= 14 cm x 30 cm
= 420 cm2
Area of Triangle ADC = 1/2 (AC x OD) = 1/2 (28 cm x 12 cm)
= 14 cm x 12 cm
= 168 cm2
Hence, Area of the quadrilateral ABCD = Area of ABC + Area of ADC
= (420 + 168) cm2
= 588 cm2
Question: 12
Find the area of a figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.
Solution:
Let x cm be one of the equal sides of an isosceles triangle.
Given that the perimeter of the isosceles triangle = 18 cm
Then, x + x + 8 =18
2x = (18 – 8) cm = 10 cm
x = 5 cm
Area of the figure formed = Area of the square + Area of the isosceles triangle
Question: 13
Find the area of Figure, in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle — sum of the areas of five triangles
Solution:
We have,
(i) P is the midpoint of AD.
Thus AP = PD = 25 cm and AB = CD = 20 cm
From the figure, we observed that,
Area of Triangle APB = Area of Triangle PDC
Area of Triangle APB = 1/2 (AB x AP ) = 1/2 (20 cm x 25 cm) = 250 cm2
Area of Triangle PDC = Area of Triangle APB = 250 c m2
Area of Triangle RPQ = 1/2 (Base x Height) = 1/2 (25 cm x 10 cm) = 125 cm2
Hence, Sum of the three triangles = (250 + 250 + 125) cm2 = 625 cm2
(ii) Area of the rectangle ABCD = 50 cm x 20 cm = 1000 cm2
Thus, Area of the rectangle – Sum of the areas of three triangles
= (1000 – 625 ) cm2 = 375 cm2
Question: 14
Calculate the area of quadrilateral field ABCD as shown in Figure, by dividing it into a rectangle and a triangle.
Solution:
We have,
Join CE , which intersect AD at point E.
Here, AE = ED = BC = 25 m and EC = AB = 30 m
Area of the rectangle ABCE = AB x BC = 30 m x 25 m = 750 m2
Area of Triangle CED = 1/2 (EC x ED) = 1/2 ( 30 m x 25 m) = 375 m2
Hence, Area of the quadrilateral ABCD = (750 + 375) m2 = 1125 m2
Question: 15
Calculate the area of the pentagon ABCDE, where AB = AE and with dimensions as shown in Figure.
Solution:
Join BE.
Area of the rectangle BCDE = CD x DE
= 10 cm x 12 cm
= 120 c m2
Area of Triangle ABE = 1/2 (BE x height of the triangle)
= 1/2 (10 cm x (20 – 12) cm)
= 1/2 (10 cm x 8 cm)
= 40 cm2
Hence, Area of the pentagon ABCDE = (120 + 40) cm2 = 160 cm2
Question: 16
The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height.
Solution:
Let altitude of the triangular field be h m
Then base of the triangular field is 3h m.
Area of the triangular field = 1/2 (h x 3h )=3h2/2 m2 —–(i)
The rate of cultivating the field is Rs 24.60 per hectare.
Hence, Height of the triangular field = 300 m and base of the triangular field = 3 x 300 m = 900 m
Question: 17
A wall is 4.5 m long and 3 m high. It has two equal windows, each having form and dimensions as shown in Figure. below. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m2.
Solution:
We have,
Length of a wall = 4.5 m
Breadth of the wall = 3 m
Area of the wall = Length x Breadth
= 4.5 m x 3 m = 13.5 m2
From the figure we observed that,
Area of the window = Area of the rectangle + Area of the triangle
= (0.8 m x 0.5 m) + (12 x 0.8 m x 0.2 m) [Since 1 m = 100 cm]
= 0.4 m2 + 0.08 m2
= 0.48 m2
Area of two windows = 2 x 0.48 = 0.96 m2
Area of the remaining wall (leaving windows) = (13.5 – 0.96) m2
= 12.54 m2
Cost of painting the wall per m2 = Rs. 15
Hence, the cost of painting on the wall = Rs. (15 x 12.54)
= Rs. 188.1
(In the book, the answer is given for one window, but we have 2 windows.)
State the number of lines of symmetry for the following figures:
(i) An equilateral triangle
(ii) An isosceles triangle
(iii) A scalene triangle
(iv) A rectangle
(v) A rhombus
(vi) A square
(vii) A parallelogram
(viii) A quadrilateral
(ix) A regular pentagon
(x) A regular hexagon
(xi) A circle
(xii) A semi-circle
Solution:
(i) An equilateral triangle has 3 lines of symmetry.
(ii) An isosceles triangle has 1 line of symmetry.
(iii) A scalene triangle has no line of symmetry.
(iv) A rectangle has 2 lines of symmetry.
(v) A rhombus has 2 lines of symmetry.
(vi) A square has 4 lines of symmetry.
(vii) A parallelogram has no line of symmetry.
(viii) A quadrilateral has no line of symmetry.
(ix) A regular pentagon has 5 lines of symmetry.
(x) A regular hexagon has 6 lines of symmetry.
(xi) A circle has an infinite number of lines of symmetry all along the diameters.
(xii) A semicircle has only one line of symmetry.
Question: 2
What other name can you give to the line of symmetry of
(i) An isosceles triangle?
(ii) A circle?
Solution:
(i) An isosceles triangle has only 1 line of symmetry.
This line of symmetry is also known as the altitude of an isosceles triangle.
(ii) A circle has infinite lines of symmetry all along its diameters.
Question: 3
Identify three examples of shapes with no line of symmetry.
Solution:
A scalene triangle, a parallelogram and a trapezium do not have any line of symmetry.
Question: 4
Identify multiple lines of symmetry, if any, in each of the following figures:
Solution:
(A) The given figure has 3 lines of symmetry. Therefore it has multiple lines of symmetry.
(B) The given figure has 2 lines of symmetry. Therefore it has multiple lines of symmetry.
(C) The given figure has 3 lines of symmetry. Therefore it has multiple lines of symmetry.
(D) The given figure has 2 lines of symmetry. Therefore it has multiple lines of symmetry.
(E) The given figure has 4 lines of symmetry. Therefore it has multiple lines of symmetry.
(F) The given figure has only 1 line of symmetry.
(G) The given figure has 4 lines of symmetry. Therefore it has multiple lines of symmetry.
(H) The given figure has 6 lines of symmetry. Therefore it has multiple lines of symmetry.
Exercise 18.2
Question: 1
In the following figures, the mirror line (i.e. the line of symmetry) is given as dotted line. Complete each figure performing reflection in the dotted (mirror) line. Also, try to recall name of the complete figure.
Solution:
(a)it will be a rectangle.
(b) It will be a triangle.
(c) It will be a rhombus.
(d) It will be a circle.
(e) It will be a pentagon.
(f) It will be an octagon.
Question: 2
Each of the following figures shows paper cuttings with punched holes. Copy these figures on a plane sheet and mark the axis of symmetry so that if the paper is folded along it, then the wholes on one side of it coincide with the holes on the other side.
Solution:
The lines of symmetry in the given figures are as follows:
Question: 3
In the following figures if the dotted lines represent the lines of symmetry, find the other hole (s).
Solution:
The other holes in the figure are as follows:
Exercise 18.3
Question: 1
Give the order of rotational symmetry for each of the following figures when rotated about the marked point (x):
Solution:
(i) The given figure has its rotational symmetry as 4.
(ii) The given figure has its rotational symmetry as 3.
(iii) The given figure has its rotational symmetry as 3.
(iv) The given figure has its rotational symmetry as 4.
(v) The given figure has its rotational symmetry as 2.
(vi) The given figure has its rotational symmetry as 4.
(vii) The given figure has its rotational symmetry as 5.
(viii) The given figure has its rotational symmetry as 6.
(ix) The given figure has its rotational symmetry as 3.
Question: 2
Name any two figures that have both line symmetry and rotational symmetry.
Solution:
An equilateral triangle and a square have both lines of symmetry and rotational symmetry.
Question: 3
Give an example of a figure that has a line of symmetry but does not have rotational symmetry.
Solution:
A semicircle and an isosceles triangle have a line of symmetry but do not have rotational symmetry.
Question: 4
Give an example of a geometrical figure which has neither a line of symmetry nor a rotational symmetry.
Solution:
A scalene triangle has neither a line of symmetry nor a rotational symmetry.
Question: 5
Give an example of a letter of the English alphabet which has
(i) No line of symmetry
(ii) Rotational symmetry of order 2.
Solution:
(i) The letter of the English alphabet which has no line of symmetry is Z.
(ii) The letter of the English alphabet which has rotational symmetry of order 2 is N.
Question: 6
What is the line of symmetry of a semi-circle? Does it have rotational symmetry?
Solution:
A semicircle (half of a circle) has only one line of symmetry. In the figure, there is one line of symmetry. The figure is symmetric along the perpendicular bisector I of the diameter XY.
A semi-circle does not have any rotational symmetry.
Question: 7
Draw, whenever possible, a rough sketch of
(i) a triangle with both line and rotational symmetries.
(ii) a triangle with only line symmetry and no rotational symmetry.
(iii) a quadrilateral with a rotational symmetry but not a line of symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry.
Solution:
(i) An equilateral triangle has 3 lines of symmetry and a rotational symmetry of order 3.
(ii) An isosceles triangle has only 1 line of symmetry and no rotational symmetry.
(iii) A parallelogram is a quadrilateral which has no line of symmetry but a rotational symmetry of order 2.
(iv) A kite is a quadrilateral which has only one line of symmetry and no rotational symmetry.
Draw an ∠BAC of measure 50° such that AB = 5 cm and AC = 7 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD
Solution:
Steps of construction:
Draw angle BAC = 50° such that AB = 5 cm and AC = 7 cm.
Cut an arc through C at an angle of 50°
Draw a straight line passing through C and the arc. This line will be parallel to AB since ∠CAB =∠RCA=50°
Alternate angles are equal; therefore the line is parallel to AB.
Again through B, cut an arc at an angle of 50° and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
∠SBA =∠BAC = 50°, since they are alternate angles. Therefore BD parallel to AC
Also we can measure BD = 7 cm and CD = 5 cm.
Question: 2
Draw a line PQ. Draw another line parallel to PQ at a distance of 3 cm from it.
Solution:
Steps of construction:
Draw a line PQ.
Take any two points A and B on the line.
Construct ∠PBF = 90° and ∠QAE = 90°
With A as centre and radius 3 cm cut AE at C.
With B as centre and radius 3 cm cut BF at D.
Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
Question: 3
Take any three non-collinear points A, B, C and draw ∠ABC. Through each vertex of the triangle, draw a line parallel to the opposite side.
Solution:
Steps of construction:
Mark three non collinear points A, B and C such that none of them lie on the same line.
Join AB, BC and CA to form triangle ABC.
Parallel line to AC
With A as centre, draw an arc cutting AC and AB at T and U, respectively.
With centre B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.
With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.
Join BY and produce in both directions to obtain the line parallel to AC.
Parallel line to AB
With B as centre, draw an arc cutting BC and BA at W and V, respectively.
With centre C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.
With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.
Join CQ and produce in both directions to obtain the line parallel to AB.
Parallel line to BC
With B as centre, draw an arc cutting BC and BA at W and V, respectively (already drawn).
With centre A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.
With centre R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.
Join AS and produce in both directions to obtain the line parallel to BC.
Question: 4
Draw two parallel lines at a distance of 5kms apart.
Solution:
Steps of construction:
Draw a line PQ.
Take any two points A and B on the line.
Construct ∠PBF = 90° and ∠QAE = 90°
With A as centre and radius 5 cm cut AE at C.
With B as centre and radius 5 cm cut BF at D.
Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
Exercise 17.2
Question: 1
Draw △ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.
Solution:
Steps of construction:
Draw a line segment AB of length 5.5 cm.
From B, cut an arc of radius 6 cm.
With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
Join AC and BC to obtain the desired triangle.
With centre B and radius more than half of BC, draw two arcs on both sides of BC.
With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
Join XY to get the perpendicular bisector of BC.
Question: 2
Draw ∆PQR in which PQ = 3 cm, QR. 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q
Solution:
Steps of construction:
Draw a line segment PQ of length 3 cm.
With Q as centre and radius 4 cm, draw an arc.
With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
Join PR and OR to obtain the required triangle.
From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
From M and N, cut arcs of equal radius intersecting at point S.
Join QS and extend to produce the angle bisector of angle PQR.
Verify that angle PQS and angle SQR are equal to 45° each.
Question: 3
Draw an equilateral triangle one of whose sides is of length 7 cm.
Solution:
Steps of construction:
Draw a line segment AB of length 7 cm.
With centre A, draw an arc of radius 7 cm.
With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
Join AC and BC to get the required triangle.
Question: 4
Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.
Solution:
Steps of construction:
Draw a line segment PR of length 7 cm.
With centre P, draw an arc of radius 5 cm.
With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
Join PQ and QR to obtain the required triangle.
From P, draw arcs with radius more than half of PR on either sides.
With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
MN is the required perpendicular bisector of the largest side.
Question: 5
Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of ∠A and (ii) perpendicular AL from A on BC. Measure LAD.
Solution:
Steps of construction:
Draw a line segment BC of length 7 cm.
With centre B, draw an arc of radius 6 cm.
With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
Join AC and BC to get the required triangle.
Angle bisector steps:
From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.
From E and F, cut arcs of equal radius intersecting at point H.
Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.
Perpendicular from Point A to line BC steps:
From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).
From P and Q, cut arcs of equal radius intersecting at M.
Join AM cutting BC at L.
AL is the perpendicular to the line BC.
Angle LAD is 15°.
Question: 6
Draw △DEF such that DE= DF= 4 cm and EF = 6 cm. Measure ∠E and ∠F.
Solution:
Steps of construction:
Draw a line segment EF of length 6 cm.
With E as centre, draw an arc of radius 4 cm.
With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
Join DE and DF to get the desired triangle DEF.
By measuring we get, ∠E= ∠F= 40°..
Question: 7
Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.
Solution:
Steps of construction:
We first draw a triangle ABC with each side = 6 cm.
Steps to bisect line AB:
Draw an arc from A on either side of line AB.
With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.
Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.
Parallel line to BC:
With B as centre, draw an arc cutting BC and BA at M and N, respectively.
With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
Join XD and extend it to intersect AC at E.
DE is the required parallel line.
Exercise 17.3
Question: 1
Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠Q = 70°.
Solution:
Steps of construction:
Draw a line segment AB of length 3 cm.
Draw ∠XBA=70°.
Cut an arc on BX at a distance of 5 cm at C.
Join AC to get the required triangle.
Question: 2
Draw △ABC in which ∠A=70°., AB = 4 cm and AC= 6 cm. Measure BC.
Solution:
Steps of construction:
Draw a line segment AC of length 6 cm.
Draw ∠XAC=70°.
Cut an arc on AX at a distance of 4 cm at B.
Join BC to get the desired triangle.
We see that BC = 6 cm.
Question: 3
Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.
Solution:
Steps of construction:
Draw a line segment PQ of length 3 cm.
Draw ∠QPX=45°.
Cut an arc on PX at a distance of 3 cm at R.
Join QR to get the required triangle.
Question: 4
Draw △ABC in which ∠A = 120°, AB = AC = 3 cm. Measure ∠B and ∠C.
Solution:
Steps of construction:
Draw a line segment AC of length 3 cm.
Draw ∠XAC = 120°.
Cut an arc on AX at a distance of 3 cm at B.
Join BC to get the required triangle.
By measuring, we get ∠B = ∠C = 30°.
Question: 5
Draw △ABC in which ∠C = 90° and AC = BC = 4 cm.
Solution:
Steps of construction:
Draw a line segment BC of length 4 cm.
At C, draw ∠BCY=90°.
Cut an arc on CY at a distance of 4 cm at A.
Join AB. ABC is the required triangle.
Question: 6
Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45°. Also, draw a perpendicular from A on BC.
Solution:
Steps of construction:
Draw a line segment AB of length 3 cm.
Draw an angle of 45° and cut an arc at this angle at a radius of 4 cm at C.
Join AC to get the required triangle.
With A as centre, draw intersecting arcs at M and N.
With centre M and radius more than half of MN, cut an arc on the opposite side of A.
With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
Join AE, it meets BC at D, then AE is the required perpendicular.
Question: 7
Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60°. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.
Solution:
Steps of construction:
Draw a line segment BC = 4 cm.
Draw ∠CBX = 60°.
Draw an arc on BX at a radius of 3 cm cutting BX at A.
Join AC to get the required triangle.
Angle bisector for angle A:
With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
From P and Q cut arcs of same radius intersecting at R.
Join AR to get the angle bisector of angle A.
Angle bisector for angle C:
With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
From M and N, cut arcs of the same radius intersecting at T
Join CT to get the angle bisector of angle C.
Mark the point of intersection of CT and AR as 0.
Angle ∠COA = 120°.
Exercise 17.4
Question: 1
Construct ∆ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.
Solution:
Steps of construction:
Draw a line segment BC of length 4 cm.
Draw ∠CBX such that ∠CBX=50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY=70°.
Let CY and BX intersect at A.
ABC is the required triangle.
Question: 2
Draw ∆ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.
∠ABC + ∠BCA + ∠CAB = 180°
∠BCA = 180° − ∠CAB − ∠ABC
∠BCA = 180°− 100° = 80°
Solution:
Steps of construction:
Draw a line segment BC of length 8 cm.
Draw ∠CBX such that ∠CBX = 50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°.
Let CY and BX intersect at A.
Question: 3
Draw ∆ABC in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.
Solution:
Steps of construction:
Draw a line segment QR = 4.5 cm.
Draw ∠RQX = 80° and ∠QRY = 55°.
Let QX and RY intersect at P so that PQR is the required triangle.
With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
Join MN
MN is the required perpendicular bisector of QR.
Question: 4
Construct ∆ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°
Solution:
Steps of construction:
Draw a line segment AB = 6.4 cm.
Draw ∠BAX = 45°.
Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°.
Let AX and BY intersect at C.
ABC is the required triangle.
Question: 5
Draw ∆ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.
∠A + ∠B + ∠C = 180°
Therefore ∠C = 180°− 60°− 90°= 30°
Solution:
Steps of construction:
Draw a line segment AC = 6 cm.
Draw ∠ACX = 30°.
Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°.
Join CX and AY. Let these intersect at B.
ABC is the required triangle where angle ∠ABC = 60°.
Exercise 17.5
Question: 1
Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.
Solution:
Steps of construction:
Draw a line segment QR = 4 cm.
Draw ∠QRX of measure 90°.
With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
Join PQ to obtain the desired triangle PQR.
PQR is the required triangle.
Question: 2
Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.
Solution:
Steps of construction:
Draw a line segment QR = 2.5 cm.
Draw ∠QRX of measure 90°.
With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
Join PQ to obtain the desired triangle PQR.
PQR is the required triangle.
Question: 3
Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30°
Solution:
Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let cC = 30°.
Explain the concept of congruence of figures with the help of certain examples.
Solution:
Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.
Consider Ball A and Ball B. These two balls are congruent.
Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars.
Question: 2
Fill in the blanks:
(i) Two line segments are congruent if ___
(ii) Two angles are congruent if ___
(iii) Two square are congruent if ___
(iv) Two rectangles are congruent if ___
(v) Two circles are congruent if ___
Solution:
(i) They have the same length, since they can superpose on each other.
(ii) Their measures are the same. On superposition, we can see that the angles are equal.
(iii) Their sides are equal. All the sides of a square are equal and if two squares have equal sides, then all their sides are of the same length. Also angles of a square are 90° which is also the same for both the squares.
(iv) Their lengths are equal and their breadths are also equal. The opposite sides of a rectangle are equal. So if two rectangles have lengths of the same size and breadths of the same size, then they are congruent to each other.
(v) Their radii are of the same length. Then the circles will have the same diameter and thus will be congruent to each other.
Question: 3
In Figure, ∠POQ ≅ ∠ROS, can we say that ∠POR ≅ ∠QOS
Solution:
We have,
∠POQ ≅ ∠ROS (1) Also, ∠ROQ ≅ ∠ROQ Therefore adding ∠ROQ to both sides of (1), Weget, ∠POQ + ∠ROQ ≅ ∠ROQ + ∠ROS Therefore, ∠PQR = ∠QOS
Question: 4
In figure, a = b = c, name the angle which is congruent to ∠AOC
Solution:
We have,
∠ AOB = ∠ BOC = ∠ COD
Therefore, ∠ AOB = ∠ COD
Also, ∠ AOB + ∠ BOC = ∠ BOC + ∠ COD
∠ AOC = ∠ BOD
Hence, ∠ BOD is congruent to ∠ AOC
Question: 5
Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.
Solution:
Two right angles are congruent to each other because they both measure 90 degrees.
We know that two angles are congruent if they have the same measure.
Question: 6
In figure, ∠AOC ≅ ∠PYR and ∠BOC ≅ ∠QYR. Name the angle which is congruent to ∠AOB.
Solution:
∠AOC ≅ ∠PYR…. (i) Also, ∠BOC ≅ ∠QYR Now, ∠AOC = ∠AOB + ∠BOC ∠PYR = ∠PYQ +∠QYR By putting the value of ∠AOC and ∠PYR in equation (i), we get, ∠AOB + ∠BOC ≅ ∠PYQ + ∠QYR ∠AOB ≅ ∠PYQ (∠BOC ≅ ∠QYR) Hence, ∠AOB ≅ ∠PYQ
Question: 7
Which of the following statements are true and which are false;
(i) All squares are congruent.
(ii) If two squares have equal areas, they are congruent.
(iii) If two rectangles have equal areas, they are congruent.
(iv) If two triangles have equal areas, they are congruent.
Solution:
(i) False.
All the sides of a square are of equal length.
However, different squares can have sides of different lengths. Hence all squares are not congruent.
(ii) True.
Area of a square = side x side
Therefore, two squares that have the same area will have sides of the same lengths. Hence they will be congruent.
(iii) False Area of a rectangle = length x breadth
Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.
Example: Suppose rectangle 1 has sides 8 m and 8 m and area 64 meter square. Rectangle 2 has sides 16 m and 4 m and area 64 meter square. Then rectangle 1 and 2 are not congruent.
(iv) False
Area of a triangle = 12 x base x height
Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.
Exercise 16.2
Question: 1
In the following pairs of triangle (Figures), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.
Solution:
1) In Δ ABC and Δ DEF
AB = DE = 4.5 cm (Side)
BC = EF = 6 cm (Side) and
AC = DF = 4 cm (Side)
Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF
2)
In Δ ACB and Δ ADB
AC = AD (Side)
BC = BD (Side) and
AB = AB (Side)
Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB
3) In Δ ABD and Δ FEC,
AB = FE (Side)
AD = FC (Side)
BD = CE (Side)
Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC
4) In Δ ABO and Δ DOC,
AB = DC (Side)
AO = OC (Side)
BO = OD (Side)
Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC
Question: 2
In figure, AD = DC and AB = BC
(i) Is ΔABD ≅ ΔCBD?
(ii) State the three parts of matching pairs you have used to answer (i).
Solution:
Yes ΔABD = ΔCBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:
AD = DC
AB = BC and
DB = BD
Question: 3
In Figure, AB = DC and BC = AD.
(i) Is ΔABC ≅ ΔCDA?
(ii) What congruence condition have you used?
(iii) You have used some fact, not given in the question, what is that?
Solution:
We have AB = DC
BC = AD
and AC = AC
Therefore by SSS ΔABC ≅ ΔCDA
We have used Side congruence condition with one side common in both the triangles.
Yes, have used the fact that AC = CA.
Question: 4
In ΔPQR ≅ ΔEFD,
(i) Which side of ΔPQR equals ED?
(ii) Which angle of ΔPQR equals angle E?
Solution:
ΔPQR ≅ ΔEFD
(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.
(ii) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.
Question: 5
Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?
It ∠B = 50°, what is the measure of ∠R?
Solution:
We have AB = AC in isosceles ΔABC
And PQ = PR in isosceles ΔPQR.
Also, we are given that AB = PQ and QR = BC.
Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, ΔABC ≅ ΔPQR
Now
∠ABC = ∠PQR (Since triangles are congruent)However, ΔPQR is isosceles.
Therefore, ∠PRQ = ∠PQR = ∠ABC = 50°
Question: 6
ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.
Solution:
∠BAD = ∠CAD (c.p.c.t)
∠BAD + ∠CAD = 40°/ 2 ∠BAD = 40°
∠BAD = 40°/2 =20°
∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle,
∠ABC = ∠BCA ∠ABC +∠ABC + 40°= 180°
2 ∠ABC = 180° – 40° = 140° ∠ABC = 140°/2 = 70°
∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD ∠DBC + ∠DBC + 100o = 180°
Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Figure. Which of the following statements is true?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔADB
(iii) ΔABC ≅ ΔBAD
Solution:
In ΔABC and ΔBAD we have,
AC = BD (given)
BC = AD (given)
and AB = BA (common)
Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD
There option (iii) is true.
Question: 8
In Figure, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts you use to arrive at your answer.
Solution:
We have AB = AC.
Also since D is the midpoint of BC, BD = DC
Also, AD = DA
Therefore by SSS condition,
ΔADB ≅ ΔADC
We have used AB, AC : BD, DC AND AD, DA
Question: 9
In figure, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.
Solution:
Yes, ΔABC ≅ ΔACB by SSS condition.
Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB
Question: 10
Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?
Solution:
Yes,
Given,
Δ ABC and Δ DBC have side BC common, AB = BD and AC = CD
By SSS criterion of congruency, ΔABC ≅ ΔDBC
No, ∠ABD and ∠ACD are not equal because AB ≠ AC
Exercise 16.3
Question: 1
By applying SAS congruence condition, state which of the following pairs of triangle are congruent. State the result in symbolic form
Solution:
(i)
We have OA = OC and OB = OD and
∠AOB = ∠COD which are vertically opposite angles. Therefore by SAS condition, ΔAOC ≅ ΔBOD
(ii)
We have BD = DC
∠ADB = ∠ADC = 90° and
Therefore, by SAS condition, ΔADB ≅ ΔADC.
(iii)
We have AB = DC
∠ABD = ∠CDB and
Therefore, by SAS condition, ΔABD ≅ ΔCBD
(iv)
We have BC = QR
ABC = PQR = 90°
And AB = PQ
Therefore, by SAS condition, ΔABC≅ ΔPQR.
Question: 2
State the condition by which the following pairs of triangles are congruent.
Solution:
(i)
AB = AD
BC = CD and AC = CA
Therefore by SSS condition, ΔABC≅ ΔADC
(ii)
AC = BD
AD = BC and AB = BA
Therefore, by SSS condition, ΔABD ≅ ΔADC
(iii)
AB = AD
∠BAC = ∠DAC and
Therefore by SAS condition, ΔBAC ≅ ΔBAC
(iv)
AD = BC
∠DAC = ∠BCA and
Therefore, by SAS condition, ΔABC ≅ ΔADC
Question: 3
In figure, line segments AB and CD bisect each other at O. Which of the following statements is true?
(i) ΔAOC ≅ ΔDOB
(ii) ΔAOC ≅ ΔBOD
(iii) ΔAOC ≅ ΔODB
State the three pairs of matching parts, you have used to arrive at the answer.
Solution:
We have,
And, CO = OD
Also, AOC = BOD
Therefore, by SAS condition, ΔAOC ≅ ΔBOD
Question: 4
Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?
Solution:
We have AO = OB and CO = OD since AB and CD bisect each other at 0.
Also ∠AOC = ∠BOD since they are opposite angles on the same vertex.
Therefore by SAS congruence condition, ΔAOC ≅ ΔBOD
Question: 5
ΔABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) We have AB = AC (Given)
∠BAD = ∠CAD (AD bisects ∠BAC)
Therefore by SAS condition of congruence, ΔABD ≅ ΔACD
(ii) We have used AB, AC; ∠BAD = ∠CAD; AD, DA.
(iii) Now, ΔABD≅ΔACD therefore by c.p.c.t BD = DC.
Question: 6
In Figure, AB = AD and ∠BAC = ∠DAC.
(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.
(ii) Complete each of the following, so as to make it true:
(a) ∠ABC =
(b) ∠ACD =
(c) Line segment AC bisects ___ and ___
Solution:
i) AB = AD (given)
∠BAC = ∠DAC (given)
AC = CA (common)
Therefore by SAS condition of congruency, ΔABC ≅ ΔADC
ii) ∠ABC = ∠ADC (c.p.c.t)
∠ACD = ∠ACB (c.p.c.t)
Question: 7
In figure, AB || DC and AB = DC.
(i) Is ΔACD ≅ ΔCAB?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Which angle is equal to ∠CAD ?
(iv) Does it follow from (iii) that AD || BC?
Solution:
(i) Yes by SAS condition of congruency, ΔDCA ≅ ΔBAC
(ii) We have used AB = DC, AC = CA and ∠DCA = ∠BAC.
(iii) ∠CAD = ∠ACB since the two triangles are congruent.
(iv) Yes this follows from AD // BC as alternate angles are equal. lf alternate angles are equal the lines are parallel
Exercise 16.4
Question: 1
Which of the following pairs of triangle are congruent by ASA condition?
Solution:
i)
We have,
Since ∠ABO = ∠CDO = 45° and both are alternate angles, AB // DC, ∠BAO = ∠DCO (alternate angle, AB // CD and AC is a transversal line)
∠ABO = ∠CDO = 45° (given in the figure) Also, AB = DC (Given in the figure)
Therefore, by ASA ΔAOB ≅ ΔDOC
ii)
In ABC,
Now AB =AC (Given)
∠ABD = ∠ACD = 40° (Angles opposite to equal sides)
BC = QR but none of the angles of ΔABC and ΔPQR are equal.
Therefore, ΔABC and Cong ΔPRQ
Question: 2
In figure, AD bisects A and AD and AD ⊥ BC.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
Solution:
(i) Yes, ΔADB≅ΔADC, by ASA criterion of congruency.
(ii) We have used ∠BAD = ∠CAD ∠ADB = ∠ADC = 90°
Since, AD ⊥ BC and AD = DA
(iii) Yes, BD = DC since, ΔADB ≅ ΔADC
Question: 3
Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.
Solution:
We have drawn
Δ ABC with ∠ABC = 65° and ∠ACB = 70°
We now construct ΔPQR ≅ ΔABC has ∠PQR = 65° and ∠PRQ = 70°
Also we construct ΔPQR such that BC = QR
Therefore by ASA the two triangles are congruent
Question: 4
In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC
(i) Is ΔABC ≅ ΔACB
(ii) State the three pairs of matching parts you have used to answer (i).
(iii) Is it true to say that AB = AC?
Solution:
(i) Yes ΔABC ≅ ΔACB
(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.
Also BC = CB
(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.
Question: 5
In Figure, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD
Solution:
As per the given conditions, ∠CAD = ∠BAD and ∠CDA = ∠BDA (because AX bisects ∠BAC)
AD = DA (common)
Therefore, by ASA, ΔACD ≅ ΔABD
Question: 6
In Figure, AO = OB and ∠A = ∠B.
(i) Is ΔAOC ≅ ΔBOD
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ∠ACO = ∠BDO?
Solution:
We have
∠OAC = ∠OBD,
AO = OB
Also, ∠AOC = ∠BOD (Opposite angles on same vertex)
Therefore, by ASA ΔAOC ≅ ΔBOD
Exercise 16.5
Question: 1
In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form.
Solution:
i)
∠ADC = ∠BCA = 90°
AD = BC and hyp AB = hyp AB
Therefore, by RHS ΔADB ≅ ΔACB
ii)
AD = AD (Common)
hyp AC = hyp AB (Given)
∠ADB + ∠ADC = 180° (Linear pair)
∠ADB + 90° = 180°
∠ADB = 180° – 90° = 90°
∠ADB = ∠ADC = 90°
Therefore, by RHS Δ ADB = Δ ADC
iii)
hyp AO = hyp DOBO = CO ∠B = ∠C = 90°
Therefore, by RHS, ΔAOB≅ΔDOC
iv)
Hyp A = Hyp CABC = DC ∠ABC = ∠ADC = 90°
Therefore, by RHS, ΔABC ≅ ΔADC
v)
BD = DB Hyp AB = Hyp BC, as per the given figure,
∠BDA + ∠BDC = 180°
∠BDA + 90° = 180°
∠BDA= 180°- 90° = 90°
∠BDA = ∠BDC = 90°
Therefore, by RHS, ΔABD ≅ ΔCBD
Question: 2
Δ ABC is isosceles with AB = AC. AD is the altitude from A on BC.
i) Is ΔABD ≅ ΔACD?
(ii) State the pairs of matching parts you have used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.
(ii) We have used Hyp AB = Hyp AC
AD = DA
∠ADB = ∠ADC = 90° (AD ⊥ BC at point D)
(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.
Question: 3
ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?
Solution:
We have AB = AC …… (i)
AD = DA (common) ……(ii)
And, ∠ADC = ∠ADB (AD ⊥ BC at point D) ……(iii)
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.
Therefore, BD = CD.
And ∠ABD = ∠ACD (c.p.c.t)
Question: 4
Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.
Solution:
Consider
Δ ABC with ∠B as right angle.
We now construct another triangle on base BC, such that ∠C is a right angle and AB = DC
Also, BC = CB
Therefore, BC = CB
Therefore by RHS, ΔABC ≅ ΔDCB
Question: 5
In figure, BD and CE are altitudes of Δ ABC and BD = CE.
(i) Is ΔBCD ≅ ΔCBE?
(ii) State the three pairs or matching parts you have used to answer (i)
Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name
(i) The side opposite to ∠B
(ii) The angle opposite to side AB
(iii) The vertex opposite to side BC
(iv) The side opposite to vertex B.
Solution:
(i) AC
(ii) ∠B
(iii) A
(iv) AC
Question: 2
Take three collinear points A, B and C on a page of your note book. Join AB. BC and CA. Is the figure a triangle? If not, why?
Solution:
No, the figure is not a triangle. By definition a triangle is a plane figure formed by three non-parallel line segments
Question: 3
Distinguish between a triangle and its triangular region.
Solution:
A triangle is a plane figure formed by three non-parallel line segments, whereas, its triangular region includes the interior of the triangle along with the triangle itself.
Question: 4
D is a point on side BC of a ∆CAD is joined. Name all the triangles that you can observe in the figure. How many are they?
Solution:
We can observe the following three triangles in the given figure
(i) ∆ABC
(ii) ∆ACD
(iii) ∆ADB
Question: 5
A, B. C and D are four points, and no three points are collinear. AC and BD intersed at O. There are eight triangles that you can observe. Name all the triangles
Solution:
(i) ∆ABC
(ii) ∆ABD
(iii) ∆ABO
(iv) ∆BCD
(v) ∆DCO
(vi) ∆AOD
(vii) ∆ACD
(viii) ∆BCD
Question: 6
What is the difference between a triangle and triangular region?
Solution:
Plane of figure formed by three non-parallel line segments is called a triangle where as triangular region is the interior of triangle ABC together with the triangle ABC itself is called the triangular region ABC
Question: 7
Explain the following terms:
(i) Triangle
(a) Parts or elements of a triangle
(iii) Scalene triangle
(iv) Isosceles triangle
(v) Equilateral triangle
(vi) Acute triangle
(vii) Right triangle
(viii) Obtuse triangle
(ix) Interior of a triangle
(x) Exterior of a triangle
Solution:
(i) A triangle is a plane figure formed by three non-parallel line segments.
(ii) The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.
(iii) A scalene triangle is a triangle in which no two sides are equal.
(iv) An isosceles triangle is a triangle in which two sides are equal. Isosceles triangle
(v) An equilateral triangle is a triangle in which all three sides are equal. Equilateral triangle
(vi) An acute triangle is a triangle in which all the angles are acute (less than 90°).
(vii) A right angled triangle is a triangle in which one angle is right angled, i.e. 90°.
(viii) An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).
(ix) The interior of a triangle is made up of all such points that are enclosed within the triangle.
(x) The exterior of a triangle is made up of all such points that are not enclosed within the triangle.
Question: 8
In Figure, the length (in cm) of each side has been indicated along the side. State for each triangle angle whether it is scalene, isosceles or equilateral:
Solution:
(i) This triangle is a scalene triangle because no two sides are equal.
(ii) This triangle is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.
(iii) This triangle is an equilateral triangle because all its three sides are equal.
(iv) This triangle is a scalene triangle because no two sides are equal.
(v) This triangle is an isosceles triangle because two of its sides are equal.
Question: 9
There are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.
Solution:
(i) This is a right triangle because one of its angles is 90°.
(ii) This is an obtuse triangle because one of its angles is 120°, which is greater than 90°.
(iii) This is an acute triangle because all its angles are acute angles (less than 90°).
(iv) This is a right triangle because one of its angles is 90°.
(v) This is an obtuse triangle because one of its angles is 110°, which is greater than 90°.
Question: 10
Fill in the blanks with the correct word/symbol to make it a true statement:
(i) A triangle has _____ sides.
(ii) A triangle has ______ vertices.
(iii) A triangle has _____ angles.
(iv) A triangle has ________ parts.
(v) A triangle whose no two sides are equal is known as ________
(v0 A triangle whose two sides are equal is known as ________
(vii) A triangle whose all the sides are equal is known as ________
(viii) A triangle whose one angle is a right angle is known as _________
(ix) A triangle whose all the angles are of measure less than 90′ is known as _________
(x) A triangle whose one angle is more than 90′ is known as _________
Solution:
(i) three
(ii) three
(iii) three
(iv) six (three sides + three angles)
(v) a scalene triangle
(vi) an isosceles triangle
(vii) an equilateral triangle
(viii) a right triangle
(ix) an acute triangle
(x) an obtuse triangle
Question: 11
In each of the following, state if the statement is true (T) or false (F):
(i) A triangle has three sides.
(ii) A triangle may have four vertices.
(iii) Any three line-segments make up a triangle.
(iv) The interior of a triangle includes its vertices.
(v) The triangular region includes the vertices of the corresponding triangle.
(vi) The vertices of a triangle are three collinear points.
(vii) An equilateral triangle is isosceles also.
(viii) Every right triangle is scalene.
(ix) Each acute triangle is equilateral.
(x) No isosceles triangle is obtuse.
Solution:
(i) True.
(ii) False. A triangle has three vertices.
(iii) False. Any three non-parallel line segments can make up a triangle.
(iv) False. The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.
(v) True. The triangular region includes the interior region and the triangle itself.
(vi) False. The vertices of a triangle are three non-collinear points.
(vii) True. In an equilateral triangle, any two sides are equal.
(viii) False. A right triangle can also be an isosceles triangle.
(ix) False. Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.
(x) False. An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle
Exercise 15.2
Question: 1
Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.
Solution:
Let the third angle be x
Sum of all the angles of a triangle = 180°
105° + 30° + x = 180°
135° + x = 180°
x = 180° – 135°
x = 45°
Therefore the third angle is 45°
Question: 2
One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?
Solution:
Let the second and third angle be x
Sum of all the angles of a triangle = 180°
130° + x + x = 180°
130° + 2x = 180°
2x = 180°– 130°
2x = 50°
x = 50/2
x = 25°
Therefore the two other angles are 25° each
Question: 3
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
Solution:
Let the each angle be x
Sum of all the angles of a triangle =180°
x + x + x = 180°
3x = 180°
x = 180/3
x = 60°
Therefore angle is 60° each
Question: 4
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Solution:
If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’
Sum of all the angles of a triangle=180°
x + 2x + 3x = 180°
6x = 180°
x = 180/6
x = 30°
2x = 30° × 2 = 60°
3x = 30° × 3 = 90°
Therefore the first angle is 30°, second angle is 60° and third angle is 90°
Question: 5
The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.
Solution:
Question: 6
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.
Solution:
Let the first angle be x
Second angle be x + 10°
Third angle be x + 10° + 10°
Sum of all the angles of a triangle = 180°
x + x + 10° + x + 10° +10° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
First angle is 50°
Second angle x + 10° = 50 + 10 = 60°
Third angle x + 10° +10° = 50 + 10 + 10 = 70°
Question: 7
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle
Solution:
Let the first and second angle be x
The third angle is greater than the first and second by 30° = x + 30°
The first and the second angles are equal
Sum of all the angles of a triangle = 180°
x + x + x + 30° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
Third angle = x + 30° = 50° + 30° = 80°
The first and the second angle is 50° and the third angle is 80°
Question: 8
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
One angle of a triangle is equal to the sum of the other two
x = y + z
Let the measure of angles be x, y, z
x + y + z = 180°
x + x = 180°
2x = 180°
x = 180/2
x = 90°
If one angle is 90° then the given triangle is a right angled triangle
Question: 9
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Each angle of a triangle is less than the sum of the other two
Measure of angles be x, y and z
x > y + z
y < x + z
z < x + y
Therefore triangle is an acute triangle
Question: 10
In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°
Solution:
(i) 63°, 37°, 80° = 180°
Angles form a triangle
(ii) 45°, 61°, 73° is not equal to 180°
Therefore not a triangle
(iii) 59°, 72°, 61° is not equal to 180°
Therefore not a triangle
(iv) 45°, 45°, 90° = 180°
Angles form a triangle
(v) 30°, 20°, 125° is not equal to 180°
Therefore not a triangle
Question: 11
The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle
Solution:
Given that
Angles of a triangle are in the ratio: 3: 4: 5
Measure of the angles be 3x, 4x, 5x
Sum of the angles of a triangle =180°
3x + 4x + 5x = 180°
12x = 180°
x = 180/12
x = 15°
Smallest angle = 3x
=3 × 15°
= 45°
Question: 12
Two acute angles of a right triangle are equal. Find the two angles.
Solution:
Given acute angles of a right angled triangle are equal
Right triangle: whose one of the angle is a right angle
Measured angle be x, x, 90°
x + x + 180°= 180°
2x = 90°
x = 90/2
x = 45°
The two angles are 45° and 45°
Question: 13
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
Solution:
Angle of a triangle is greater than the sum of the other two
Measure of the angles be x, y, z
x > y + z or
y > x + z or
z > x + y
x or y or z > 90° which is obtuse
Therefore triangle is an obtuse angle
Question: 14
AC, AD and AE are joined. Find
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
Solution:
We know that sum of the angles of a triangle is 180°
Find x, y, z (whichever is required) from the figures given below:
Solution:
(i) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (corresponding angles)
x = 40°
∠AED = ∠ACB (corresponding angles)
y = 30°
We know that the sum of all the three angles of a triangle is equal to 180°
x + y + z = 180° (Angles of ∆ADE)
Which means: 40° + 30° + z = 180°
z = 180° – 70°
z = 110°
Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.
(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.
(∆ADC is a right triangle)
We also know that the sum of all the angles of a triangle is equal to 180°.
Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)
135° + y = 180°
y = 180° – 135°.
y = 45°.
We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.
(Sum of the angles of ∆ABC)
40° + y + (x + 45°) = 180°
40° + 45° + x + 45° = 180° (y = 45°)
x = 180° –130°
x = 50°
Therefore, we can say that the required angles are 45° and 50°.
(iii) We know that the sum of all the angles of a triangle is equal to 180°.
Therefore, for △ABD:
∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)
50° + x + 50° = 180°
100° + x = 180°
x = 180° – 100°
x = 80°
For ∆ABC:
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° – 130°
z = 50°
Using the same argument for ∆ADC:
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)
y +z + 30° =180°
y + 50° + 30° = 180° (z = 50°)
y = 180° – 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°.
(iv) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (Corresponding angles)
y = 50°
Also, ∠AED = ∠ACB (Corresponding angles)
z = 40°
We know that the sum of all the three angles of a triangle is equal to 180°.
Which means: x + 50° + 40° = 180° (Angles of ∆ADE)
x = 180° – 90°
x = 90°
Therefore, we can conclude that the required angles are 50°, 40° and 90°.
Question: 16
If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles
Solution:
We know that one of the angles of the given triangle is 60°. (Given)
We also know that the other two angles of the triangle are in the ratio 1: 2.
Let one of the other two angles be x.
Therefore, the second one will be 2x.
We know that the sum of all the three angles of a triangle is equal to 180°.
60° + x + 2x = 180°
3x = 180° – 60°
3x = 120°
x = 120/3
x = 40°
2x = 2 × 40
2x = 80°
Hence, we can conclude that the required angles are 40° and 80°.
Question: 17
It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.
Solution:
We know that one of the angles of the given triangle is 100°.
We also know that the other two angles are in the ratio 2: 3.
Let one of the other two angles be 2x.
Therefore, the second angle will be 3x.
We know that the sum of all three angles of a triangle is 180°.
100° + 2x + 3x = 180°
5x = 180° – 100°
5x = 80°
x = 80/5
2x = 2 ×16
2x = 32°
3x = 3×16
3x = 48°
Thus, the required angles are 32° and 48°.
Question: 18
In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.
Solution:
We know that for the given triangle, 3∠A = 6∠C
∠A = 2∠C — (i)
We also know that for the same triangle, 4∠B = 6∠C
∠B = (6/4)∠C — (ii)
We know that the sum of all three angles of a triangle is 180°.
Therefore, we can say that:
∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)
On putting the values of ∠A and ∠B in equation (iii), we get:
2∠C + (6/4)∠C +∠C = 180°
(18/4) ∠C = 180°
∠C = 40°
From equation (i), we have:
∠A = 2∠C = 2 × 40
∠A = 80°
From equation (ii), we have:
∠B = (6/4)∠C = (6/4) × 40°
∠B = 60°
∠A = 80°, ∠B = 60°, ∠C = 40°
Therefore, the three angles of the given triangle are 80°, 60°, and 40°.
Question: 19
Is it possible to have a triangle, in which
(i) Two of the angles are right?
(ii) Two of the angles are obtuse?
(iii) Two of the angles are acute?
(iv) Each angle is less than 60°?
(v) Each angle is greater than 60°?
(vi) Each angle is equal to 60°
Solution:
Give reasons in support of your answer in each case.
(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C.
As per the given information,
∠A < 60° … (i)
∠B< 60° … (ii)
∠C < 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C < 60°+ 60°+ 60°
∠A + ∠B + ∠C < 180°
We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.
(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A > 60° … (i)
∠B > 60° … (ii)
∠C > 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C > 60°+ 60°+ 60°
∠A + ∠B + ∠C > 180°
We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.
(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A = 60° … (i)
∠B = 60° …(ii)
∠C = 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C = 60°+ 60°+ 60°
∠A + ∠B + ∠C =180°
We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.
Question: 20
In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B
Solution:
Consider ∆ABD
∠BAD = 100/2 (AD bisects ∠A)
∠BAD = 50°
∠ADB = 90° (AD perpendicular to BC)
We know that the sum of all three angles of a triangle is 180°.
Thus,
∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)
Or,
∠ABD + 50° + 90° = 180°
∠ABD =180° – 140°
∠ABD = 40°
Question: 21
In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC
Solution:
We know that the sum of all three angles of a triangle is equal to 180°.
Therefore, for the given △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
50° + 70° + ∠C = 180°
∠C= 180° –120°
∠C = 60°
∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )
∠ACD = ∠BCD = 60/2= 30°
Using the same logic for the given ∆ACD, we can say that:
∠DAC + ∠ACD + ∠ADC = 180°
50° + 30° + ∠ADC = 180°
∠ADC = 180°– 80°
∠ADC = 100°
If we use the same logic for the given ∆BCD, we can say that
∠DBC + ∠BCD + ∠BDC = 180°
70° + 30° + ∠BDC = 180°
∠BDC = 180° – 100°
∠BDC = 80°
Thus,
For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°
∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°
Question: 22
In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C, meet at O. Find
(i) ∠C
(ii) ∠BOC
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
60° + 80° + ∠C= 180°.
∠C = 180° – 140°
∠C = 140°.
For △OBC,
∠OBC = ∠B2 = 80/2 (OB bisects ∠B)
∠OBC = 40°
∠OCB =∠C2 = 40/2 (OC bisects ∠C)
∠OCB = 20°
If we apply the above logic to this triangle, we can say that:
∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)
20° + 40° + ∠BOC = 180°
∠BOC = 180° – 60°
∠BOC = 120°
Question: 23
The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠A + ∠B + ∠C = 180°
∠A + 90° + ∠C = 180°
∠A + ∠C = 180° – 90°
∠A + ∠C = 90°
For ∆OAC:
∠OAC = ∠A2 (OA bisects LA)
∠OCA = ∠C2 (OC bisects LC)
On applying the above logic to △OAC, we get:
∠AOC + ∠OAC + ∠OCA = 180° (Sum of angles of ∆AOC)
∠AOC + ∠A2 + ∠C2 = 180°
∠AOC + ∠A + ∠C2 = 180°
∠AOC + 90/2 = 180°
∠AOC = 180° – 45°
∠AOC = 135°
Question: 24
In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.
Solution:
In the given triangle,
∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)
We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we can say that:
Question: 25
In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for ∆ABC, we can say that:
Question: 26
Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.
Solution:
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
∠CAB = ∠DBA (Alternate interior angles)
∠DBA = 35°
We also know that the sum of all three angles of a triangle is 180°.
In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P
Solution:
In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.
∠QCA = ∠CQP (Alternate interior angles)
Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,
∠ABC = ∠PRQ (alternate interior angles).
We know that the sum of all three angles of a triangle is 180°.
(ii) the interior opposite angles to exterior ∠CBX
Also, name the interior opposite angles to an exterior angle at A.
Solution:
(i) ∠ABC
(ii) ∠BAC and ∠ACB
Also the interior angles opposite to exterior are ∠ABC and ∠ACB
Question: 2
In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?
Solution:
In ∆ABC, ∠A = 50° and ∠B = 55°
Because of the angle sum property of the triangle, we can say that
∠A + ∠B + ∠C = 180°
50°+ 55°+ ∠C = 180°
Or
∠C = 75°
∠ACB = 75°
∠ACX = 180°− ∠ACB = 180°−75° = 105°
Question: 3
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.
Solution:
We know that the sum of interior opposite angles is equal to the exterior angle.
Hence, for the given triangle, we can say that:
∠ABC+ ∠BAC = ∠BCO
55° + ∠BAC = 95°
Or,
∠BAC= 95°– 95°
= ∠BAC = 40°
We also know that the sum of all angles of a triangle is 180°.
Hence, for the given △ABC, we can say that:
∠ABC + ∠BAC + ∠BCA = 180°
55° + 40° + ∠BCA = 180°
Or,
∠BCA = 180° –95°
= ∠BCA = 85°
Question: 4
One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?
Solution:
Let us assume that A and B are the two interior opposite angles.
We know that ∠A is equal to ∠B.
We also know that the sum of interior opposite angles is equal to the exterior angle.
Hence, we can say that:
∠A + ∠B = 80°
Or,
∠A +∠A = 80° (∠A= ∠B)
2∠A = 80°
∠A = 40/2 =40°
∠A= ∠B = 40°
Thus, each of the required angles is of 40°.
Question: 5
The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In the given figure, ∠ABE and ∠ABC form a linear pair.
∠ABE + ∠ABC =180°
∠ABC = 180°– 136°
∠ABC = 44°
We can also see that ∠ACD and ∠ACB form a linear pair.
∠ACD + ∠ACB = 180°
∠AUB = 180°– 104°
∠ACB = 76°
We know that the sum of interior opposite angles is equal to the exterior angle.
Therefore, we can say that:
∠BAC + ∠ABC = 104°
∠BAC = 104°– 44° = 60°
Thus,
∠ACE = 76° and ∠BAC = 60°
Question: 6
In Fig, the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC
Solution:
In a ∆ABC, ∠BAC and ∠EAF are vertically opposite angles.
Hence, we can say that:
∠BAC = ∠EAF = 45°
Considering the exterior angle property, we can say that:
∠BAC + ∠ABC = ∠ACD = 105°
∠ABC = 105°– 45° = 60°
Because of the angle sum property of the triangle, we can say that:
∠ABC + ∠ACS +∠BAC = 180°
∠ACB = 75°
Therefore, the angles are 45°, 65° and 75°.
Question: 7
In Figure, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.
Solution:
In the given triangle, the angles are in the ratio 3: 2: 1.
Let the angles of the triangle be 3x, 2x and x.
Because of the angle sum property of the triangle, we can say that:
3x + 2x + x = 180°
6x = 180°
Or,
x = 30° … (i)
Also, ∠ACB + ∠ACE + ∠ECD = 180°
x + 90° + ∠ECD = 180° (∠ACE = 90°)
∠ECD = 60° [From (i)]
Question: 8
A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?
Solution:
Here,
Internal angle at A + External angle at A = 180°
Internal angle at A + 103° =180°
Internal angle at A = 77°
Internal angle at B + External angle at B = 180°
Internal angle at B + 74° = 180°
Internal angle at B = 106°
Sum of internal angles at A and B = 77° + 106° =183°
It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.
Question: 9
In Figure, AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆FCB, we can say that:
∠FCB + ∠CBF + ∠BFC = 180°
50° + ∠CBF + 90°= 180°
Or,
∠CBF = 180° – 50°– 90° = 40° … (i)
Using the above rule for ∆ABD, we can say that:
∠ABD + ∠BDA + ∠BAD = 180°
∠BAD = 180° – 90°– 40° = 50° [from (i)]
Question: 10
In Figure, measures of some angles are indicated. Find the value of x.
Solution:
Here,
∠AED + 120° = 180° (Linear pair)
∠AED = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for △ADE, we can say that:
∠ADE + ∠AED + ∠DAE = 180°
60°+ ∠ADE + 30° =180°
Or,
∠ADE = 180°– 60°– 30° = 90°
From the given figure, we can also say that:
∠FDC + 90° = 180° (Linear pair)
∠FDC = 180°– 90° = 90°
Using the above rule for △CDF, we can say that:
∠CDF + ∠DCF + ∠DFC = 180°
90° + ∠DCF + 60° =180°
∠DCF = 180°−60°− 90°= 30°
Also,
∠DCF + x = 180° (Linear pair)
30° + x = 180°
Or,
x = 180°– 30° = 150°
Question: 11
In Figure, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130°, find
(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC
Solution:
(i) Here,
∠BAF + ∠FAD = 180° (Linear pair)
∠FAD = 180°- ∠BAF = 180°– 90° = 90°
Also,
∠AFE = ∠ADF + ∠FAD (Exterior angle property)
∠ADF + 90° = 130°
∠ADF = 130°− 90° = 40°
(ii) We know that the sum of all the angles of a triangle is 180°.
ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°. Find ∠ACB.
Solution:
Here,
∠CAX = ∠DAX (AX bisects ∠CAD)
∠CAX =70°
∠CAX +∠DAX + ∠CAB =180°
70°+ 70° + ∠CAB =180°
∠CAB =180° –140°
∠CAB =40°
∠ACB + ∠CBA + ∠CAB = 180° (Sum of the angles of ∆ABC)
∠ACB + ∠ACB+ 40° = 180° (∠C = ∠B)
2∠ACB = 180°– 40°
∠ACB = 140/2
∠ACB = 70°
Question: 13
The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30° and ∠ACD = 115°, find ∠ALC
Solution:
∠ACD and ∠ACL make a linear pair.
∠ACD+ ∠ACB = 180°
115° + ∠ACB =180°
∠ACB = 180°– 115°
∠ACB = 65°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
30° + ∠BAC + 65° = 180°
Or,
∠BAC = 85°
∠LAC = ∠BAC/2 = 85/2
Using the above rule for ∆ALC, we can say that:
∠ALC + ∠LAC + ∠ACL = 180°
Question: 14
D is a point on the side BC of ∆ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find
(i) ∠AQD
(ii) ∠APD
Solution:
∠ABD and ∠QBD form a linear pair.
∠ABC + ∠QBC =180°
60° + ∠QBC = 180°
∠QBC = 120°
∠PDC = ∠BDQ (Vertically opposite angles)
∠BDQ = 75°
In ∆QBD:
∠QBD + ∠QDB + ∠BDQ = 180° (Sum of angles of ∆QBD)
120°+ 15° + ∠BQD = 180°
∠BQD = 180°– 135°
∠BQD = 45°
∠AQD = ∠BQD = 45°
In ∆AQP:
∠QAP + ∠AQP + ∠APQ = 180° (Sum of angles of ∆AQP)
80° + 45° + ∠APQ = 180°
∠APQ = 55°
∠APD = ∠APQ
Question: 15
Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y
Solution:
The interior angles of a triangle are the three angle elements inside the triangle.
The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Using these definitions, we will obtain the values of x and y.
(I) From the given figure, we can see that:
∠ACB + x = 180° (Linear pair)
75°+ x = 180°
Or,
x = 105°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠BAC+ ∠ABC +∠ACB = 180°
40°+ y +75° = 180°
Or,
y = 65°
(ii) x + 80°= 180° (Linear pair)
= x = 100°
In ∆ABC:
x+ y+ 30° = 180° (Angle sum property)
100° + 30° + y = 180°
= y = 50°
(iii) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ACD, we can say that:
30° + 100° + y = 180°
Or,
y = 50°
∠ACB + 100° = 180°
∠ACB = 80° … (i)
Using the above rule for ∆ACD, we can say that:
x + 45° + 80° = 180°
= x = 55°
(iv) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆DBC, we can say that:
30° + 50° + ∠DBC = 180°
∠DBC = 100°
x + ∠DBC = 180° (Linear pair)
x = 80°
And,
y = 30° + 80° = 110° (Exterior angle property)
Question: 16
Compute the value of x in each of the following figures
Solution:
(i) From the given figure, we can say that:
∠ACD + ∠ACB = 180° (Linear pair)
Or,
∠ACB = 180°– 112° = 68°
We can also say that:
∠BAE + ∠BAC = 180° (Linear pair)
Or,
∠BAC = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠BAC + ∠ACB = 180°
x = 180°– 60°– 68° = 52°
= x = 52°
(ii) From the given figure, we can say that:
∠ABC + 120° = 180° (Linear pair)
∠ABC = 60°
We can also say that:
∠ACB+ 110° = 180° (Linear pair)
∠ACB = 70°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠ABC + ∠ACB = 180°
x = 50°
(iii) From the given figure, we can see that:
∠BAD = ∠ADC = 52° (Alternate angles)
We know that the sum of all the angles of a triangle is 180°.
Therefore, for ∆DEC:
x + 40°+ 52° = 180°
= x = 88°
(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.
Thus,
35° + 45° + 50° + reflex ∠ADC = 360°
Or,
reflex ∠ADC = 230°
230° + x = 360° (A complete angle)
= x = 130°
Exercise 15.4
Question: 1
In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:
(i) 5, 7, 9
(ii) 2, 10.15
(iii) 3, 4, 5
(iv) 2, 5, 7
(v) 5, 8, 20
Solution:
(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side. Here, 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5
(ii) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.
(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side. Here, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3
(iv) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 2 + 5 = 7
(v) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 5 + 8 < 20
Question: 2
In Fig, P is the point on the side BC. Complete each of the following statements using symbol ‘=’,’ > ‘or ‘< ‘so as to make it true:
(i) AP… AB+ BP
(ii) AP… AC + PC
(iii) AP…. 1/2(AB + AC + BC)
Solution:
(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.
(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.
(iii) AP < 12(AB + AC + BC) In triangles ABP and ACP, we can see that:
AP < AB + BP…(i) (Because the sum of any two sides of a triangle is greater than the third side)
AP < AC + PC…(ii) (Because the sum of any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we have:
AP + AP < AB + BP + AC + PC
2AP < AB + AC + BC (BC = BP + PC)
AP < (AB – FAC + BC)
Question: 3
P is a point in the interior of △ABC as shown in Fig. State which of the following statements are true (T) or false (F):
(i) AP + PB < AB
(ii) AP + PC > AC
(iii) BP + PC = BC
Solution:
(i) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
(ii) True
We know that the sum of any two sides of a triangle is greater than the third side: it is true for the given triangle.
(iii) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
Question: 4
O is a point in the exterior of △ABC. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that
OA + OB + OC > 1/2(AB + BC +CA)
Solution:
Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:
OA + OB > AB — (i)
OB + OC > BC — (ii)
OA + OC > CA — (iii)
On adding equations (i), (ii) and (iii) we get:
OA + OB + OB + OC + OA + OC > AB + BC + CA
2(OA + OB + OC) > AB + BC + CA
OA + OB + OC > (AB + BC + CA)/2
Question: 5
In ∆ABC, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.
Solution:
Because the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.
Exercise 15.5
Question: 1
State Pythagoras theorem and its converse.
Solution:
The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.
Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.
Question: 2
In right ∆ABC, the lengths of the legs are given. Find the length of the hypotenuse
(i) a = 6 cm, b = 8 cm
(ii) a = 8 cm, b = 15 cm
(iii) a = 3 cm, b = 4 cm
(iv) a = 2 cm, b =1.5 cm
Solution:
According to the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
(i) c2 = a2 + b2
c2 = 62 + 82
c2 = 36 + 64 = 100
c = 10 cm
(ii) c2 = a2 + b2
c2 = 82 + 152
c2 = 64 + 225 = 289
c = 17cm
(iii) c2 = a2 + b2
c2 = 32 + 42
c2 = 9 + 16 = 25
c = 5 cm
(iv) c2 = a2 + b2
c2 = 22 + 1.52
c2 = 4 + 2.25 = 6.25
c = 2.5 cm
Question: 3
The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side.
Solution:
Let the hypotenuse be ” c ” and the other two sides be ” b ” and ” c”.
Using the Pythagoras theorem, we can say that:
c2 = a2 + b2
2.52 = 1.52 + b2
b2 = 6.25 −2.25 = 4
c = 2 cm
Hence, the length of the other side is 2 cm.
Question: 4
A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.
Solution:
Let the hypotenuse be h.
Using the Pythagoras theorem, we get:
3.72 = 1.22 + h2
h2 = 13.69 – 1.44 = 12.25
h = 3.5 m
Hence, the height of the wall is 3.5 m.
Question: 5
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.
Solution:
In the given triangle, the largest side is 6 cm.
We know that in a right angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side.
Therefore,
32 + 42 = 9 + 16 = 25
But,
62 = 36
32 + 42 not equal to 62
Hence, the given triangle is not a right angled triangle.
Question: 6
The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c= 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
Solution:
(i) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 25 cm.
c2 = 625
We have:
a2+ b2 = 72 + 242 = 49 + 576 = 625 = c2
Thus, the given triangle is a right triangle.
(ii) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 18 cm.
c2 = 324
We have:
a2+ b2 = 92+162 = 81 + 256 = 337 not equal to c2
Thus, the given triangle is not a right triangle.
Question: 7
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.
(Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 m and 12 m)
Solution:
The distance between the tops of the poles is the distance between points A and B.
We can see from the given figure that points A, B and C form a right triangle, with AB as the hypotenuse.
On using the Pythagoras Theorem in ∆ABC, we get:
(11−6)2 + 122 = AB2
AB2 = 25 + 144
AB2 = 169
AB = 13
Hence, the distance between the tops of the poles is 13 m.
Question: 8
A man goes 15 m due west and then 8 m due north. How far is he from the starting point?
Solution:
Let O be the starting point and P be the final point.
By using the Pythagoras theorem, we can find the distance OP.
OP2 = 152 + 82
OP2 = 225 + 64
OP2 = 289
OP = 17
Hence, the required distance is 17 m.
Question: 9
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
Solution:
Given Let the length of the ladder be L m.
By using the Pythagoras theorem, we can find the length of the ladder.
62 + 82 = L2
L2 = 36 + 64 =100
L = 10
Thus, the length of the ladder is 10 m.
When the ladder is shifted:
Let the height of the ladder after it is shifted be H m.
By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.
82 + H2 = 102
H2 = 100 – 64 = 36
H = 6
Thus, the height of the ladder is 6 m.
Question: 10
A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?
Solution:
Let the distance of the lower end of the ladder from the wall be x m.
On using the Pythagoras theorem, we get:
x2 + 482 = 502
x2 = 502 − 482 = 2500 – 2304 = 196
H = 14 dm
Hence, the distance of the lower end of the ladder from the wall is 14 dm.
Question: 11
The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg.
Solution:
Let the length of each leg of the given triangle be x units.
Using the Pythagoras theorem, we get:
x2 + x2 = (Hypotenuse)2
x2 + x2 = 50
2x2 = 50
x2 = 25
x = 5
Hence, we can say that the length of each leg is 5 units.
Question: 12
Verity that the following numbers represent Pythagorean triplet:
(i) 12, 35, 37
(ii) 7, 24, 25
(iii) 27, 36, 45
(iv) 15, 36, 39
Solution:
We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.
(i) 372 =1369
122 + 352 = 144 + 1225 = 1369
122 + 352 = 372
Yes, they represent a Pythagorean triplet.
(ii) 252 = 625
72 + 242 = 49 + 576 = 625
72 + 242 = 252
Yes, they represent a Pythagorean triplet.
(iii) 452 = 2025
272 + 362 = 729 + 1296 = 2025
272 + 362 = 452
Yes, they represent a Pythagorean triplet.
(iv) 392 = 1521
152 + 362 = 225 + 1296 = 1521
152 + 362 = 392
Yes, they represent a Pythagorean triplet.
Question: 13
In ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD perpendicular to AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
100° + 35° + ∠ACB = 180°
∠ACB = 180° –135°
∠ACB = 45°
∠C = 45°
If we apply the above rule on ∆BCD, we can say that:
∠BCD + ∠BDC + ∠CBD = 180°
45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)
∠CBD = 180°– 135°
∠CBD = 45°
We know that the sides opposite to equal angles have equal length.
Thus, BD = DC
DC = 2 cm
Question: 14
In a ∆ABC, AD is the altitude from A such that AD = 12 cm. BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.
Solution:
In ∆ADC,
∠ADC = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 162 = AC2
AC2 = 144 + 256 = 400
AC = 20 cm
In ∆ADB,
∠ADB = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 92 = AB2
AB2 = 144 + 81 = 225
AB = 15 cm
In ∆ABC,
BC2 = 252 = 625
AB2 + AC2 = 152 + 202 = 625
AB2 + AC2 = BC2
Because it satisfies the Pythagoras theorem, we can say that ∆ABC is right angled at A.
Question: 15
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 105° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB, which will be approximately 5.5 cm.
AC2 + BC2 = 42 + 32 = 9 +16 = 25
AB2 = 5.52 = 30.25
AB2 not equal to AC2+ BC2
Here,
AB2 > AC2 + BC2
Question: 16
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 80° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name
Therefore in ∆ABC, we have
