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Exercise 14.1

1. Give five examples of data that you can collect from your day-to-day life.

Answer

Five examples from day-to-day life:
(i) Daily expenditures of household.
(ii) Amount of rainfall.
(iii) Bill of electricity.
(iv) Poll or survey results.
(v) Marks obtained by students.

2. Classify the data in Q.1 above as primary or secondary data.

Answer

Primary Data: (i) (iii) and (v)
Secondary Data: (ii) and (iv)

Exercise 14.2

1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Answer

The frequency means the number of students having same blood group. We will represent the data in table:

Most common Blood Group (Highest frequency): O
Rarest Blood Group (Lowest frequency): AB

2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5    3    10    20    25    11    13    7    12    31
19    10    12    17    18    11    32    17    16    2
7    9    7    8    3    5    12    15    18    3
12    14    2    9    6    15    15    7    6    12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?

Answer

The given data is very large. So, we construct a group frequency of class size 5. Therefore, class interval will be 0-5, 5-10, 10-15, 15-20 and so on. The data is represented in the table as:

The classes in the table are not overlapping. Also, 36 out of 40 engineers have their house below 20 km of distance.

3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1     98.6     99.2     90.3     86.5     95.3     92.9      96.3      94.2      95.1
89.2     92.3     97.1     93.5     92.7     95.1     97.2      93.3      95.2      97.3
96.2     92.1     84.9     90.2     95.7     98.3     97.3      96.1      92.1      89
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?

Answer

(i) The given data is very large. So, we construct a group frequency of class size 2. Therefore, class interval will be 84-86, 86-88, 88-90, 90-92 and so on. The data is represented in the table as:

(ii) The humidity is very high in the data which is observed during rainy season. So, it must be rainy season.
(iii) Range of data = Maximum value of data – Minimum = 99.2 − 84.9 = 14.3

4. The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161     150     154     165     168     161     154     162     150     151
162     164     171     165     158     154     156     172     160     170
153     159     161     170     162     165     166     168     165     164
154     152     153     156     158     162     160     161     173     166
161     159     162     167     168     159     158     153     154     159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table?

Answer

(i) The data with class interval 160-165, 165-170 and so on is represented in the table as:

(ii) From the given data, it can be concluded that 35 students i.e. more than 50% are shorter than 165 cm.

5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03     0.08     0.08     0.09     0.04     0.17
0.16     0.05     0.02     0.06     0.18     0.20
0.11     0.08     0.12     0.13     0.22     0.07
0.08     0.01     0.10     0.06     0.09     0.18
0.11     0.07     0.05     0.07     0.01     0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Answer

(i) The data with class interval 0.00 – 0.04, 0.04 – 0.08 and so on is represented in the table as:

(ii) 2 + 4 + 2 = 8 days have the concentration of sulphur dioxide more than 0.11 parts per million.

Page No. 246

6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0     1     2     2     1     2     3     1     3     0
1     3     1     1     2     2     0     1     2     1
3     0     0     1     1     2     3     2     2     0
Prepare a frequency distribution table for the data given above.

Answer

The frequency distribution table for the data given above can be prepared as follow:

7. The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?

Answer

(i)The frequency is given as follow:

(ii) The digit having the least frequency occurs the least and the digit with highest frequency occurs the most. 0 has frequency 2 and thus occurs least frequently while 3 and 9 have frequency 8 and thus occur most frequently.

8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1     6     2     3     5    12     5     8     4     8
10   3     4     12   2     8     15    1     17   6
3     2     8     5     9     6      8     7     14   12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?

Answer

(i) The distribution table for the given data, taking class width 5 and one of the class intervals as 5-10 is as follows:

(ii) We observed from the given table that 2 children television for 15 or more hours a week.

9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6     3.0    3.7     3.2     2.2     4.1     3.5     4.5
3.5     2.3    3.2     3.4     3.8     3.2     4.6     3.7
2.5     4.4    3.4     3.3     2.9     3.0     4.3     2.8
3.5     3.2    3.9     3.2     3.2     3.1     3.7     3.4
4.6     3.8    3.2     2.6     3.5     4.2     2.9     3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.

Answer

A grouped frequency distribution table using class intervals of size 0.5 starting from the interval 2 – 2.5 is constructed.

Exercise 14.4

1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.

Answer

Mean = Sum of all the observations/Total number of observations
= (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8

For Median, we will arrange the given data in ascending order,
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations (n) = 10
Number of observations are even so we will calculate median as,

2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.

Answer

Mean = Sum of all the observations/Total number of observations
= (41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/15 = 822/15 = 54.8

For Median, we will arrange the given data in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations (n) = 15
Number of observations are odd so we will calculate median as,

Exercise 13.1

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m² costs Rs 20.

Answer

Length of plastic box (l) = 1.5 m
Width of plastic box (b) = 1.25 m
Depth of plastic box (h) = 0.65 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×0.65] + (1.5 × 1.25) m² 
= (3.575 + 1.875) m² 
= 5.45 m² 
The sheet required required to make the box is 5.45 m² 

(ii) Cost of 1 m² of sheet = Rs 20
∴ Cost of 5.45 m² of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m².

Answer


length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
 = 2(5+4)×3 + (5×4) m² 
= (54 + 20) m² 
= 74 m² Cost of white washing = ₹7.50 per m²
Total cost = ₹ (74×7.50) = ₹ 555

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m² is ₹15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ₹15000
Rate per m² = ₹10
Area of four walls = 2(l + b) h m² = (250×h) m²
A/q,
(250×h)×10 = ₹15000
⇒ 2500×h = ₹15000 
⇒ h = 15000/2500 m
⇒ h = 6 m 
Thus the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Answer

Volume of paint = 9.375 m² = 93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm² 
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2×468.75 cm² = 937.5 cm²
Number of bricks can be painted = 93750/937.5 = 100

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?

Answer

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm²
= 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5) cm²
= 2(125+80+100) cm²
= (2×305) cm² = 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²



6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?

Answer

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm²
= 2(750 + 625 + 750) cm²
= 4250 cm²

(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm² , find the cost of cardboard required for supplying 250 boxes of each kind.

Answer

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm²
= 2(500 + 100 + 125) cm²
= 1450 cm²

Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm²
= 2(180 + 60 + 75) cm²
= 630 cm²

Total surface area of 250 boxes of each type = 250(1450 + 630) cm²
= 250×2080 cm² = 520000 cm²
Extra area required = 5/100(1450 + 630) × 250 cm² = 26000 cm²

Total Cardboard required = 520000 + 26000 cm² = 546000 cm²
Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3m?

Answer

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb = [2(4+3)×2.5 + 4×3] m²
= (35+12) m²
= 47 m²

Exercise 13.2

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Answer

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m²
= (2 × 22 × 0.1 × 1.7) m²
=7.48 m²

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

 

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²
 = 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh + 2πRh + 2π(R^{2 –} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 – 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm² = 2038.08 cm²

Page No: 217

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Answer

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m² = 3.168 m²
Area of the playground = (500 × 3.168) m² = 1584 m²

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m².

Answer

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = ₹12.50 per m²
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m²
=5.5 m²
Total cost of painting = (5.5 × 12.5) = ₹68.75

6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Answer

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹40 per m².

Answer

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = ₹40 per m²
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m²
= 110 m²

(ii) Cost of plastering = ₹(110 × 40) = ₹4400

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m² = 4.4 m²

9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

Answer

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
= (2 × 22/7 × 2.1 × 4.5) m²
= 59.4 m²

(ii) Total surface area of the tank = 2πr(r + h) m²
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
= 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 – 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35)cm²
= 2200 cm²

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr²
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
= (198 + 198/7) cm²
= 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
= 7920 cm²

Exercise 13.3

Exercise 13.4

1. Find the surface area of a sphere of radius:
(i) 10.5 cm 
(ii) 5.6 cm 
(iii) 14 cm

Answer

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
= (4 × 22/7 × 5.6 × 5.6) cm²
= 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
= (4 × 22/7 × 14 × 14) cm²
= 2464 cm²

2. Find the surface area of a sphere of diameter:
(i) 14 cm 
(ii) 21 cm 
(iii) 3.5 m

Answer

(i) r = 14/2 cm = 7cm
Surface area = 4πr²
= (4 × 22/7 × 7 × 7)cm²
=616cm²

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr² 
= (4 × 22/7 × 1.75 × 1.75) m²
= 38.5 m²

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer

r = 10 cm  
Total surface area of hemisphere = 3πr²
= (3 × 3.14 × 10 ×10) cm²
= 942 cm²

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
= r²/R²
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Answer

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²
Rate of tin – plating is = ₹16 per 100 cm²
Therefor, cost of 1 cm² = ₹16/100 
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72

6. Find the radius of a sphere whose surface area is 154 cm².

Answer

Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = 154/(4 × 22/7)
⇒ r² = 49/4
⇒ r = 7/2 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. 

Answer

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
= (5 + 0.25) cm  = 5.25 cm
Outer curved surface = 2πR²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.

Exercise 13.5

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

Answer

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
= (4 × 2.5 × 1.5)  cm³ = 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15)  cm³ = 180 cm³

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m³ = 1000 l)

Answer

Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m³
=(6×5×4.5)m3=135 m³
Therefore , the tank can hold = 135 × 1000 litres  [Since 1m³=1000litres]
= 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth) 
= 380/(10×8) m
= 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³.

Answer

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
= (8×6×3) m³
= 144 m³
Rate of digging = ₹30 per m³
Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Answer

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m³ = 1000 litres

∴ 50000 litres = 50000/1000 m³ = 50 m³
Breadth = Volume of cuboid/(Length×Depth)
= 50/(2.5×10) m
= 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

Answer

Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m³
= (20×15×6) m³
= 1800 m³
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
= (4000×150)/1000
= 600 m³
Number of days the water will last = Capacity of tank Total water required per day
=(1800/600) = 3
The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

Answer

Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m³ = 10000 m³
Dimension of crates = 1.5m × 1.25m × 0.5m 
Volume of 1 crates = (1.5 × 1.25 × 0.5) m³ = 0.9375 m³
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer

Edge of the cube = 12 cm. 
Volume of the cube = (edge)³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³ 
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm³ = 216 cm³
Side of the smaller cube = a
a³ = 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)²
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
= 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer

Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute 
Volume of water flowing into the sea in a minute = lbh m³
= (100/3 × 40 × 3) m³
= 4000 m³

Exercise 13.6

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)

Answer

Let the base radius of the cylindrical vessel be ‘r’ cm.
∴ Circumference = 2πr
⇒ 2πr = 132     [∵ Circumference = 132 cm]
⇒ 2× 22/7 × r = 132 cm
R=(132×7)/(2×22) cm = 21 cm
∵ Height of the vessel = 25 cm
∴ Volume = πr² x h    [∵ Volume of a cylinder = πr²h]
= (22/7) (21)² × 25 cm³
= (22/7) × 21 × 21 × 25 cm³
= 22 × 3 × 21 × 25 cm³
= 34650 cm³ 
∵ Capacity of the vessel = Volume of the vessel
∴ Capacity of cylindrical vessel = 34650 cm³ 
Since 1000 cm³ = 1 litre
⇒ 34650 cm³ = (34650/1000) litres = 34.65 l

2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer

Here, Inner diameter of the cylindrical pipe = 24 cm
⇒ Inner radius of the pipe (r) = (24/2) cm = 12 cm
Outer diameter of the pipe = 28 cm
⇒ Outer radius of the pipe (R) = (28/2)cm = 14 cm
Length of the pipe (h) = 35 cm
∵ Inner volume of the pipe = πr²h
Outer volume of the pipe = πr²h
∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume
= πR²h – πr²h
= πh(R²-r²)
= πh(R+r)(R-r) [ ∵ a² – b² = (a+b)(a-b)]
= 22/7 x 35 x (14+12) x (14-12) cm³
= 22 x 5 x 26 x 2 cm³ 

Mass of the wood in the pipe = [Mass of wood in 1 m³ of wood] x [Volume of wood in the pipe]
= [0.6g] x [22 x 5 x 26 x 2] cm³
= (6/10)x 22 x 10 x 26 g = 6 x 22 x 26 g
= 3432 g = (3432/1000)= 3.432 kg   [∵ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.

3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer

For rectangular pack: Length (l) = 5 cm
Breadth (b) = 4 cm Height (h) = 15 cm
∴ Volume = l x b x h = 5 x 4 x 15 cm³
= 300 cm³
⇒ Capacity of the rectangular pack = 300 cm³     …(1)
For cylindrical pack: Base diameter = 7 cm
⇒ Radius of the base (r) = (7/2)cm
Height (h) = 10 cm
∴ Volume = πr²h = (22/7) x (7/2)² x 10 cm³
= (22/7) x (7/2) x (7/2) x 10 cm³
= 11 x 7 x 5 cm³ = 385 cm³
⇒ Volume of the cylindrical pack = 385 cm³   …(2)
From (1) and (2),
we have 385 cm³ – 300 cm³ = 85 cm³
⇒ The cylindrical pack has the greater capacity by 85 cm³.

4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find: (i) radius of its base (ii) its volume. (Use π = 3.14)

Answer

(i) Since lateral surface of the cylinder = 2 πrh
But lateral surface of the cylinder = 94.2 cm²
 2πrh = 94.2
2×3.14×r×5 =942/10
⇒{(10×314)/100} × r = 942/10
⇒ r=(942/10)×{100/(10×314)}cm
⇒ r =471/157 cm 
Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr²h
⇒ Volume of the given cylinder = 3.14 x (3)² x 5 cm³
=314×100×3×3×5cm³
=(157×3×3)/10
= 1413/10 = 141.3 cm³

Thus, the required volume = 141.3 cm³

5. It costs ₹ 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m²; find: (i) inner curved surface of the vessel (ii) radius of the base (iii) capacity of the vessel.

Answer

(i) To find inner curved surface
Total cost of painting = ₹ 2200
Rate of painting = ₹ 20 per m²
∴  Area =cost/rate = 2200/20 = 110 m²   
⇒ Inner curved surface of the vessel = 110 m²

(ii) To find radius of the base Let the base radius of the cylindrical vessel.
∵ Curved surface of a cylinder = 2 πrh
∴  2πrh = 110
⇒ 2× 22/7 ×r×10 = 110 [∵Height=10 m ]
⇒ r= (110×7)/(2×22×10)m = 7/4 m
= 1.75 m
⇒ The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel
Since, volume of a cylinder = πr²h
 ∴ Volume (capacity) of the vessel =22/7 × (7/2)² × 10 m³
= 22/7 × 7/4 × 7/4 × 10 m³ 
= (11×7×5)/4 m³ = 385/4 m3 = 96.25 m³

Since, 1 m³ = 1000000 cm³ = 1000 l = 1 kl
∴ 96.5 m³ = 96.5 kl
Thus, the required volume = 96.25 kl

6. The capacity of closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Answer

Capacity of the cylindrical vessel = 15.4 l
= 15.4×1000 cm³ 

⇒ Volume of the vessel = (15.4/1000)m³
Height of the vessel = 1m Let ‘r’ metres be the radius of the base of the vessel
∴ Volume = πr²h

⇒ πr²h= 15.4/1000

Thus, the required sheet = 0.4708 m²

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer

Since, 10 mm = 1cm
∴ 1 mm = (1/10) cm
For graphite cylinder

Thus, the required volume of the graphite = 0.11 cm³ 
For the pencil Diameter of the pencil = 7 mm = (7/10)cm
∴ Radius of the pencil (R) = (7/20) cm

Height of the pencil (h) = 14 cm
Volume of the pencil = πr²h

Volume of the wood Volume of the wood = [Volume of the pencil] – [Volume of the graphite]
= 5.39 cm³ – 0.11 cm³ = 5.28 cm³ 
Thus, the required volume of the wood is 5.28 cm³.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer

The bowl is cylindrical.
Diameter of the base = 7 cm
⇒ Radius of the base (r) = (7/3) cm
Height (h) = 4 cm
Volume of soup = πr²h

= 38500 / 100 liters             

Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

Exercise 13.8

1. Find the volume of a sphere whose radius is
(i) 7 cm        (ii) 0.63 m

Answer

(i) Radius of the sphere(r) = 7 cm
Therefore, Volume of the sphere = 4/3 πr³
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm³
                                                     = 4312/3 cm³

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr³
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³
                                   = 1.05 m³

2. Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm                     (ii) 0.21 m

Answer

(i) Diameter of the spherical ball = 28 cm
Radius = 28/2 cm = 14 cm
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (4/3 × 22/7 × 14 × 14 × 14) cm³
                              = 34496/3 cm³

(ii) Diameter of the spherical ball = 0.21 m
Radius (r) = 0.21/2 m = 0.105 m
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (43×227×0.105×0.105×0.105) m³
                              = 0.004851 m³

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr³
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm³
                               = 38.808 cm³
Density of the metal is 8.9g per cm3
Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon? 

Answer

Let the diameter of the moon be r
Radius of the moon = r/2
A/q,
Diameter of the earth = 4r
Radius(r) = 4r/2 = 2r
Volume of the moon = v = 4/3 π(r/2)³
                                  = 4/3 πr³ × 1/8
⇒ 8v = 4/3 πr³ — (i)
Volume of the earth = r³ = 4/3 π(2r)³
                                = 4/3 πr³× 8
⇒ V/8 = 4/3 πr³ — (ii)
From (i) and (ii), we have
8v = V/8
⇒ v = 1/64 V
Thus, the volume of the moon is 1/64 of the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Answer

Diameter of a hemispherical bowl = 10.5 cm
Radius(r) = (10.5/2) cm = 5.25cm
Volume of the bowl = 2/3 πr³
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm³
                                = 303.1875 cm³
Litres of milk bowl can hold = (303.1875/1000) litres
                                               = 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer

Internal radius = r = 1m
External radius = R = (1 + 0.1) cm = 1.01 cm
Volume of iron used = External volume – Internal volume
                                  = 2/3 πR³ – 2/3 πr³
                                  = 2/3 π(R³ – r³)
                                  = 2/3 × 22/7 × [(1.01)³−(1)³] m³
                                  = 44/21 × (1.030301 – 1) m³
                                  = (44/21 × 0.030301) m³
                                  = 0.06348 m³(approx)

7. Find the volume of a sphere whose surface area is 154 cm².

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr³
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm³
             = 539/3 cm³

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,      (ii) volume of the air inside the dome.

Answer

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
                                                        =  (498.96/2.00) m² = 249.48 m² 

 (ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × 22/7 × r² = 249.48
⇒ r² = (249.48×7)/(2×22) = 39.69
⇒ r²=  39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr³
                                                                    = (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m³
                                                       = 523.9 m³ (approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,      (ii) ratio of S and S′. 

Answer

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr³ — (i)
Volume of the new sphere of radius r′ = 4/3 πr’³ — (ii)
A/q,
4/3 πr’³= 27 × 4/3 πr³
⇒ r’³ = 27r³
⇒ r’³ = (3r)³
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr²/4πr′² = r²/(3r)²
                              = r²/9r² = 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Answer

Diameter of the spherical capsule = 3.5 mm
Radius(r) = 3.52mm
                = 1.75mm
Medicine needed for its filling = Volume of spherical capsule
                                                  = 4/3 πr³
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm³
                                                  = 22.46 mm³ (approx.)

 Question 1.
A wooden bookshelf has external dimensions as follows : Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm² and the rate of pointing is 10 paise per cm², find the total expenses required for palishing and painting the surface of the bookshelf.

Solution:
Here, l = 85 cm b = 25 cm and h = 110 cm
Area of the bookshelf of outer surface = 2 lb + 2bh + hl
= [2(85 x 25)+ 2(110 x 25)+ 85×110] cm2 = (4250 + 5500 + 9350) cm² = 19100 cm²
Cost of polishing of the outer surface of bookshelf
= 19100 x 20100 = ₹ 3820
Thickness of the plank = 5 cm
Internal height of bookshelf = (110 – 2 x 5) = 100 cm
Internal depth of bookshelf = (25 – 5) = 20 cm
Internal breadth of bookshelf = 85 – 2 x 5 = 75 cm
Hence, area of the internal surface of bookshelf
= 2(75 x 20)+ 2(100x 20)+ 75×100
= 3000 + 4000 + 7500 = 14500 cm2
So, cost of painting of internal surface of bookshelf
= 14500 x 10100 = ₹1450
Hence, total costing of polishing and painting = 3820 + 1450= ₹ 5270

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution:
It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint.
Surface area to be silver paint

 Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let d be the diameter of the sphere

Exercise 12.1

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Answer

Length of the side of equilateral triangle = a
Perimeter of the signal board = 3a = 180 cm
∴ 3a = 180 cm ⇒ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
Using heron’s formula,
Area of the signal board = √s (s-a) (s-b) (s-c)
                                       = √(3a/2) (3a/2 – a) (3a/2 – a) (3a/2 – a)
                                       = √3a/2 × a/2 × a/2 × a/2
                                       = √3a⁴/16
                                       = √3a²/4
                                       = √3/4 × 60 × 60 = 900√3 cm²

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer

The sides of the triangle are 122 m, 22 m and 120 m.
Perimeter of the triangle is 122 + 22 + 120 = 264m
Semi perimeter of triangle (s) = 264/2 = 132 m
Using heron’s formula,
Area of the advertisement = √s (s-a) (s-b) (s-c)
                                       = √132(132 – 122) (132 – 22) (132 – 120) m²
                                       = √132 × 10 × 110 × 12 m²
                                       = 1320 m²
Rent of advertising per year = ₹ 5000 per m²
Rent of one wall for 3 months = ₹ (1320 × 5000 × 3)/12 = ₹ 1650000

Page No: 203

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer

Sides of the triangular wall are 15 m, 11 m and 6 m.
Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m
Using heron’s formula,
Area of the message = √s (s-a) (s-b) (s-c)
                                       = √16(16 – 15) (16 – 11) (16 – 6) m²
                                       = √16 × 1 × 5 × 10 m² = √800 m²
                                       = 20√2 m²

4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer

Two sides of the triangle = 18cm and 10cm
Perimeter of the triangle = 42cm
Third side of triangle = 42 – (18+10) cm = 14cm
Semi perimeter of triangle = 42/2 = 21cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √21(21 – 18) (21 – 10) (21 – 14) cm²
                                       = √21 × 3 × 11 × 7 m²
                                       = 21√11 cm²

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Answer

Ratio of the sides of the triangle = 12 : 17 : 25
Let the common ratio be x then sides are 12x, 17x and 25x
Perimeter of the triangle = 540cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
⇒ x = 10
Sides of triangle are,
12x = 12 × 10 = 120cm
17x = 17 × 10 = 170cm
25x = 25 × 10 = 250cm
Semi perimeter of triangle(s) = 540/2 = 270cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √270(270 – 120) (270 – 170) (270 – 250)cm²
                                       = √270 × 150 × 100 × 20 cm²
                                       = 9000 cm²

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer

Length of the equal sides = 12cm
Perimeter of the triangle = 30cm
Length of the third side = 30 – (12+12) cm = 6cm
Semi perimeter of the triangle(s) = 30/2 cm = 15cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √15(15 – 12) (15 – 12) (15 – 6)cm²
                                       = √15 × 3 × 3 × 9 cm²
                                       = 9√15 cm²

Exercise 11.1

1. Construct an angle of 90° at the initial point of a given ray and justify the construction.

Answer

Steps of construction:

Step 1: A ray YZ is drawn.
Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.
Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray XY making an angle 90° with YZ is formed.

Justification for construction:
We constructed ∠BYZ = 60° and also ∠AYB = 60°.
Thus, ∠AYZ = 120°.
Also, bisector of ∠AYB is constructed such that:
∠AYB = ∠XYA + ∠XYB
⇒ ∠XYB = 1/2∠AYB
⇒ ∠XYB = 1/2×60°
⇒ ∠XYB = 30°
Now,
∠XYZ = ∠BYZ + ∠XYB = 60° + 30° = 90°

2. Construct an angle of 45° at the initial point of a given ray and justify the construction.

Answer

Steps of construction:

Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.
Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray making an angle 90° with YZ is formed.
Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of ∠XOY is drawn.

Justification for construction:
By construction,
∠XOY = 90°
We constructed the bisector of ∠XOY as DOY.
Thus,
∠DOY = 1/2 ∠XOY
∠DOY = 1/2×90° = 45°

3. Construct the angles of the following measurements:
(i) 30°        (ii) 22.5°        (iii) 15°

Answer

(i) 30°

Steps of constructions:
Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.
Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.

Thus, ∠XOY is the required angle making 30° with OY.

(ii) 22.5°

 

Steps of constructions:
Step 1: An angle ∠XOY = 90° is drawn.
Step 2: Bisector of ∠XOY is drawn such that ∠BOY = 45° is constructed.
Step 3: Again, ∠BOY is bisected such that ∠AOY is formed.
Thus, ∠AOY is the required angle making 22.5° with OY.

(iii) 15°

Steps of constructions:
Step 1: An angle ∠AOY = 60° is drawn.
Step 2: Bisector of ∠AOY is drawn such that ∠BOY = 30° is constructed.
Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of ∠BOY is drawn.
Thus, ∠XOY is the required angle making 15° with OY.

4. Construct the following angles and verify by measuring them by a protractor:
(i) 75°         (ii) 105°        (iii) 135°

Answer

(i) 75°

Step 1: A ray OY is drawn.
Step 2: An arc BAE is drawn with O as a centre.
Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.
Step 4: With A and B as centres, arcs are made to intersect at X and ∠XOY = 90° is made.
Step 5: With A and C as centres, arcs are made to intersect at D
Step 6: OD is joined and and ∠DOY = 75° is constructed.
Thus, ∠DOY is the required angle making 75° with OY.

(ii) 105°

 

Step 1: A ray OY is drawn.
Step 2: An arc ABC is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With B and C as centres, arcs are made to intersect at X
Step 6: OX is joined and and ∠XOY = 105° is constructed.
Thus, ∠XOY is the required angle making 105° with OY.

Steps of constructions:Step 1: A ray DY is drawn.
Step 2: An arc ACD is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With F and D as centres, arcs are made to intersect at X or bisector of ∠EOD is constructed.
Step 6: OX is joined and and ∠XOY = 135° is constructed.
Thus, ∠XOY is the required angle making 135° with DY.

5. Construct an equilateral triangle, given its side and justify the construction.

Answer

Steps of constructions:
Step 1: A line segment AB=4 cm is drawn.
Step 2: With A and B as centres, two arcs are made.
Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60° each.
Step 5: Lines from A and B are extended to meet each other at C.
Thus, ABC is the required triangle formed.

Justification:
By construction,
AB = 4 cm, ∠A = 60° and ∠B = 60°
We know that,
∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle)
⇒ 60° + 60° + ∠C = 180°
⇒ 120° + ∠C = 180°
⇒ ∠C = 60°
BC = CA = 4 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 4 cm
∠A = ∠B = ∠C = 60°

Exercise 10.1

1. Fill in the blanks:
(i) The centre of a circle lies in ____________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)

(iii) The longest chord of a circle is a _____________ of the circle.

(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.

Answer

(i)   The centre of a circle lies in interior of the circle. (exterior/interior)
(ii)  A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

2. Write True or False: Give reasons for your answers.

(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.

Answer

(i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can’t say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.

Exercise 10.2

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer

A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.

Given,
AB = CD (Equal chords)
To prove,
∠AOB = ∠COD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
∴ ΔAOB ≅ ΔCOD (SSS congruence condition)
Thus, ∠AOB = ∠COD by CPCT.
Equal chords of congruent circles subtend equal angles at their centres.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer

Given,
∠AOB = ∠COD (Equal angles)
To prove,
AB = CD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
∠AOB = ∠COD (Given)
OB = OD (Radii)
∴ ΔAOB ≅ ΔCOD (SAS congruence condition)
Thus, AB = CD by CPCT.
If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Exercise 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Answer

One point P is common.

One point P is common.

Two points P and Q are common.

No point is common.

2. Suppose you are given a circle. Give a construction to find its centre.

Answer

Steps of construction:
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.

Given,
Two circles which intersect each other at P and Q.
To prove,
OO’ is perpendicular bisector of PQ.
Proof,
In ΔPOO’ and ΔQOO’,
OP = OQ (Radii)
OO’ = OO’ (Common)
O’P = OQ (Radii)
∴ ΔPOO’ ≅ ΔQOO’ (SSS congruence condition)
Thus,
∠POO’ = ∠QOO’ — (i)
In ΔPOR and ΔQOR,
OP = OQ (Radii)
∠POR = ∠QOR (from i)
OR = OR (Common)
∴ ΔPOR ≅ ΔQOR (SAS congruence condition)
Thus,
∠PRO = ∠QRO
also,
∠PRO + ∠QRO = 180°
⇒ ∠PRO = ∠QRO = 180°/2 = 90°
Hence,
OO’ is perpendicular bisector of PQ.

Exercise 10.4

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer

OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.

In ΔPOR,
OP² = OR² + PR²
⇒ 5² = (4-x)² + PR²
⇒ 25 = 16 + x² – 8x +  PR²
⇒ PR² = 9 – x² + 8x — (i)


In ΔPRS,
PS² = PR² + RS²
⇒ 3² = PR² + x²
⇒  PR² = 9 – x² — (ii)

Equating (i) and (ii),
9 – x² + 8x = 9 – x²
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR² = 9 – 0²
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND — (i)
and MB = CN — (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT — (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii)  and (iii) we get,
MB – ME = CN – EN
⇒ EB = CE

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ

4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer

OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD — (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC — (ii)
From (i) and (ii),
AM – BM = MD – MC
⇒ AB = CD

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Answer

 Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA² = OM² + AM²
⇒ 5² = x² + y² — (i)
Applying Pythagoras theorem in ΔAMB,
AB² = BM² + AM²
⇒ 6² = (5-x)² + y² — (ii)
Subtracting (i) from (ii), we get
36 – 25 = (5-x)² – x² –
⇒ 11 = 25 – 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y² + 49/25 = 25
⇒ y² = 25 – 49/25
⇒ y² = (625 – 49)/25
⇒ y² = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.

6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB² = BD² + AD²
⇒ AD² = AB² – BD² 
⇒ AD² = a² – (a/2)²
⇒ AD² = 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.

 Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles with centres O and O’ which intersect each other at C and D.

To prove: ∠OCO’ = ∠ODO’
Construction: Join OC, OD, O’C and O’D
Proof: In ∆ OCO’and ∆ODO’, we have
OC = OD (Radii of the same circle)
O’C = O’D (Radii of the same circle)
OO’ = OO’ (Common)
∴ By SSS criterion, we get
∆ OCO’ ≅ ∆ ODO’
Hence, ∠OCO’ = ∠ODO’ (By CPCT)

 Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
Let O be the centre of the given circle and let its radius be cm.
Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm
∴ OM = (6 – a) cm
Join OA and OC.
Then, OA = OC = b c m
Since, the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm
In ∆OAN and ∆OCM, we get
OA² = ON² + AN²
OC² = OM² + CM²
⇒ b² = a² + (2.5)²
and, b² = (6-a)² + (5.5)² …(i)
So, a² + (2.5)² = (6 – a)² + (5.5)²
⇒ a² + 6.25= 36-12a + a² + 30.25
⇒ 12a = 60
⇒ a = 5
On putting a = 5 in Eq. (i), we get
b² = (5)² + (2.5)²
= 25 + 6.25 = 31.25
So, r = 31.25−−−−√ = 5.6cm (Approx.)

 Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?
Solution:
Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of circle.

Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.
∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.
PM = MQ = 12 PQ = 62 = 3 cm
and RN = NS = 12 RS = 82 = 4 cm
In ∆OPM, we have
OP² = OM² + PM²
⇒ a² =4² + 3² = 16 + 9 = 25
⇒ a = 5
In ∆ORN, we have
⇒ OR² = ON² + RN²
⇒ a² = ON² + (4)²
⇒ 25 = ON² + 16
⇒ ON² = 9
⇒ ON = 3cm
Hence, the distance of the chord PS from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ In ∆BDC, we get
∠ADC = ∠DBC + ∠DCB …(i)
Since, angle at the centre is twice at a point on the remaining part of circle.
∴ ∠DCE = 12 ∠DOE
⇒ ∠DCB = 12 ∠DOE (∵ ∠DCE = ∠DCB)
∠ADC = 12 ∠AOC
∴ 12 ∠AOC = ∠ABC + 12 ∠DOE (∵ ∠DBC = ∠ABC)
∴ ∠ABC = 12 (∠AOC – ∠DOE)
Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

 Question 5.
Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
Given: PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.
To prove: A circle drawn on PQ as diameter will pass through O.

Construction: Through O, draw MN || PS and EF || PQ.
Proof : ∵ PQ = SR ⇒ 12 PQ = 12 SR
So, PN = SM
Similarly, PE = ON
So, PN = ON = NQ
Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

 Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°
(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)
∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)
So, ∠ADE + ∠ABC = 180°
(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)
From Eqs. (i) and (ii), we get
∠AED + ∠ABC = ∠ADE + ∠ABC
⇒ ∠AED = ∠ADE
∴ In ∆AED We have
∠AED = ∠ADE
So, AD = AE
(∵ Sides opposite to equal angles of a triangle are equal)

 Question 7.
AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution:
(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get
PA = PC ( ∵ P is the mid-point of AC)
∠APB = ∠CPD (Vertically opposite angles)
and PB = PD (∵ P is the mid-point of BD)
∴ By SAS criterion
∆CPD ≅ ∆APB
∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.
Similarly, AC is a diameter.
(ii) Now, BD and AC bisect each other.
So, ABCD is a parallelogram.
Also, AC = BD
∴ ABCD is a rectangle.

Ex 10.6 Class 9 Maths Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – 12 A, 90° – 12 B and 90° – 12 C.
Solution:
∵ ∠EDF = ∠EDA + ∠ADF
∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.
∴ ∠EDA = ∠EBA
and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

Ex 10.6 Class 9 Maths Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.
∴ ∠BPA = ∠BQA
(∵ Angle subtended by equal chords are equal)
⇒ BP = BQ

Ex 10.6 Class 9 Maths Question 10.
In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.
Solution:
(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.
Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)
and ∠BCM = ∠BAM (Angles in same segment)
But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)
∴ ∠MBC = ∠BCM
So, MB = MC (Sides opposite to equal angles are equal)
So, M must lie on the perpendicular bisector of BC
(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.
Join AM.

Since, M lies on perpendicular bisector of BC.
∴ BM = CM
∠MBC = ∠MCB
But ∠MBC = ∠MAC (Angles in same segment)
and ∠MCB = ∠BAM (Angles in same segment)
So, from Eq. (i),
∠BAM = ∠CAM
AM is the bisector of A.
Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.

Exercise 9.1

1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Answer

(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and  trapezium SMNR lie on the same base SR but not between the same parallel lines.

Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer

Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).

Answer 

Given,
E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH — (i)
also, area of GHF = 1/2 area of HFCD — (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)

3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).

Answer

ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer 

(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) — (i)
Thus,
AD || BC ⇒ AG || BH — (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) — (i)
Thus,
AB || CD ⇒ AE || DF — (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS) 

 

∴ ar(PQRS) = ar(ABRS) — (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) — (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)

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6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Answer

The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS — (i) 
Area of ΔPAQ = 1/2 area of PQRS — (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS — (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.