Exercise Ex. 19A

Question 1

Fill in the blanks:

 (i) The probability of an impossible event is ……. .

 (ii) The probability of a sure event is ……. .

 (iii) For any event E, P(E) + P(not E)= …… .

(iv) The probability of a possible but not a sure event lies between …… and ……. .

 (v) The sum of probabilities of all the outcomes of an experiment is …….. .Solution 1

(i) The probability of an impossible event is zero.

(ii) The probability of a sure event is one.

(iii) For any event E, P(E) + P(not E)= one .

(iv) The probability of a possible but not a sure event lies between zero and one.

(v) The sum of probabilities of all the outcomes of an experiment is one.Question 2

A coin is tossed once. What is the probability of getting a tail?Solution 2

Question 3

Two coins are tossed simultaneously. Find the probability of getting

(i) exactly 1 head

(ii) at most 1 head

(iii) at least 1 headSolution 3

Question 4

A die is thrown once. Find the probability of getting:

(i)An even number

(ii)A number less than 5

(iii)A number greater than 2

(iv)A number between 3 and 6

(v)A number other than 3

(vi)The number 5Solution 4

In a throw of a dice, all possible outcomes are 1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

(i)Let E be event of getting even number

Then, the favorable outcomes are 2, 4, 6

Number of favorable outcomes= 3

                P(getting a even number)= P(E) =  

(ii)Let R be the number less than 5

Then, the favorable outcomes are 1, 2, 3, 4

Number of favorable outcomes = 4

                     P(getting a number less than 5)= P(R) =  

(iii)Let M be the event of getting a number greater than 2

Then, the favorable outcomes are 3, 4, 5, 6

Number of favorableoutcomes = 4 

P(getting a number greater than 2)= P(M)  

(iv)Let N be the number lying between 3 and 6

Then the favorable outcomes are 4, 5

Number of favorable outcomes = 2

                        P(getting a number 3 and 6)= P(N) =  

(v)Let G be event of getting a number other than 3

Then the favorable outcomes are 1, 2, 4, 5, 6

Number of favorable outcomes = 5

                     P(getting a number other than 5)=P(G) =  

(vi)Let T be event of getting a number 5

Then the favorable outcome is 5

Number of favorable outcomes = 1

                    P(getting a number 5)=P(T)  Question 5

A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.Solution 5

Question 6

Solution 6

Question 7

It is known that a box of 200 electric bulbs contains 16 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is

(i)Defective

(ii)Non – defective?Solution 7

Total number of bulbs = 200

Number of defective bulbs = 16

(i)Let be the event of getting a defective bulb

Total number of defective bulbs = 16

P(getting defective bulbs) = P() = 

(ii)Let be the event of “getting non – defective bulb”

P(getting non defective bulb) = Question 8

If the probability of winning a game is 0.7, what is the probability of losing it?Solution 8

Question 9

 There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is a (i) boy, (ii) girl?Solution 9

Question 10

In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?Solution 10

Question 11

250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?Solution 11

Total number of tickets sold = 250

Number of prizes = 5

Let E be the event getting a prize

Number of favorable outcomes = 5

P(getting a prize) = Question 12

17 cards numbered 1, 2, 3, 4, ….., 17 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card drawn bears (i) an odd number (ii) a number divisible by 5.Solution 12

Question 13

A game of chance consists of spinning an arrow, which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8.Solution 13

Question 14

In a family of 3 children, find the probability of having at least one boy.Solution 14

Question 15

A bag contains 4 white balls, 5 red balls, 2 blacks balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is (i) black, (ii) not green, (iii) red or white, (iv) neither red nor green.Solution 15

Question 16

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting (i) a red king, (ii) a queen or a jack. Solution 16

Question 17

A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.Solution 17

There are 26 red cards containing a 2 queensand 2 more black queens are there in a pack of cards

P(getting a red card or a queen) 

P(getting neither a red card nor a queen) = Question 18

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting (i) a red face card (ii) a black king.Solution 18

Question 19

Two different dice are tossed together. Find the probability that (i) the number on each dice is even, (ii) the sum of the numbers appearing on the two dice is 5.Solution 19

Question 20

Two different dice are rolled simultaneously. Find the probability that the sum of numbers in the two dice is 10.Solution 20

Question 21

When two dice are tossed together, find the probability that the sum of numbers on their tops is less than 7.Solution 21

Question 22

Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is a perfect square.Solution 22

Question 23

Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.Solution 23

Question 24

Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the number on the taken out card is (i) a prime number less than 10 (ii) a number which is a perfect square.Solution 24

Question 25

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3,…. , 12 as shown in the figure. What is the probability that it will point to

(i)6?

(ii)An even number?

(iii)A prime number?

(iv)A number which is a multiple of 5?

Solution 25

Spinning arrow may come to rest at one of the 12 numbers

total number of outcomes = 12

(i)Probability that it will point at 6 = 

(ii)Even numbers are 2, 4, 6, 8, 10 and 12. There are 6 numbers.

Probability that it points at even numbers = 

(iii)The prime numbers are 2,3 5, 7 and 11. There are 5 prime numbers.

Probability that it points at prime number = 

(iv)There are 2 numbers divisible by 5. These are 5 and 10.

Probability that a number is a multiple of 5 = Question 26

12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out is good one.Solution 26

Question 27

A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvy will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) she will buy it, (ii) she will not buy it?Solution 27

Question 28

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.Solution 28

Question 29

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and not replaced. Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?Solution 29

Question 30

A bag contains lemon-flavoured candies only. Hema takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?Solution 30

Question 31

There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate card, the cards being identical. Then the puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?Solution 31

Question 32

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing

(i)An ace

(ii)A ‘4’ of spades

(iii)A ‘9’ of a black suit

(iv)A red kingSolution 32

Total number of all possible outcomes = 52

(i)P(getting an ace) = 

(ii)P(getting a ‘4’ of spades) = 

(iii)P(a ‘9’ of a black suit) = 

(iv)P(getting a red king) = Question 33

A card is drawn at random from a well- shuffled deck of 52 cards. Find the probability of getting

(i)A queen

(ii)A diamond

(iii)A king or an ace

(iv)A red aceSolution 33

Total numbers of cards = 52

(i)There are 4 queen cards in a pack of cards

Probability of getting a queen card = 

(ii)There are 13 cards of diamond in a pack of cards

probability ofgetting a diamond card = 

(iii)In a pack of cards there are 4 kings and 4 aces

Number of such cards = 4 + 4 = 8

Probability of getting either a king or an ace = 

(iv)There are two red aces in a pack of cards

probability of getting a red ace = Question 34

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red suit

(ii) a face card

(iii) a red face card

(iv) a queen of black suit

(v) a jack of hearts

(vi) a spadeSolution 34

Question 35

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is

(i)A card of spades or an ace

(ii)A red king

(iii)Either a king or a queen

(iv)Neither a king nor a queenSolution 35

Total number of cards = 52

(i)There are 13 cards of spade (including 1 ace) and 3 more ace cards are there in a pack of cards

P(getting a card of spades or an ace) = 

(ii)There are 2 red kings in a pack of cards

P(getting a red king) = 

(iii)There are 4 kings and 4 queens in a pack of cards

P(getting either a king or a queen) = 

(iv)P(getting neither a king nor a queen) = 

Exercise Ex. 19B

Question 1

A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) divisible by 2 or 3, (ii) a prime number.Solution 1

Question 2

A box contains cards numbered 3, 5, 7, 9, …., 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.Solution 2

Question 3

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) not divisible by 3, (ii) a prime number greater that 7, (iii) not a perfect square number.Solution 3

Question 4

Cards bearing numbers 1, 3, 5, …., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (i) a prime number less than 15, (ii) a number divisible by 3 and 5.Solution 4

Question 5

A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, Find the probability that it bears (i) a one-digit number, (ii) a number divisible by 5, (iii) an odd number less than 30, (iv) a composite number between 50 and 70.Solution 5

Question 6

Cards marked with numbers 1, 3, 5, ……, 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) less than 19, (ii) a prime number less than 20.Solution 6

Question 7

Tickets numbered 2, 3, 4, 5,.. , 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is

(i)An even number

(ii)A number less than 16

(iii)A number which is a perfect square

(iv)A prime number less than 40Solution 7

Total number of tickets = 100

(i)Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100

Total number of even number = 50

P(getting a even number) = 

(ii)Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Total number of numbers less than 16 is 14

P(getting a number less than 16) = 

(iii)Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100

Total number of perfect squares = 9

P(getting a perfect square) = 

(iv)Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Total number of prime numbers =12

P(getting a prime number less 40) = Question 8

A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.Solution 8

Question 9

 A piggy bank contains hundred 50-p coins, seventy Rs. 1 coin, fifty Rs. 2 coins and thirty Rs. 5 coins. If it equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a Rs. 1 coin? (ii) will not be a Rs. 5 coins (iii) will be 50-p or a Rs. 2 coin?Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

A carton consists of 100 shirts of which 88 are good and 8 have minor defects. Rohit, a trader, will only accept the shirts which are good. But, Kamal, another trader, will only reject the shirts which have major defect. One shirt is drawn at random from the carton. What is the probability that it is acceptable to (i) Rohit, (ii) Kamal?Solution 14

Question 15

A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient, (ii) extremely kind or honest. Which of the above values you prefer more?Solution 15

Question 16

Two dice are thrown simultaneously. What is the probability that

(i)5 will not come up on either of them

(ii)5 will not come up on at least one,

(iii)5 will come up at both the diceSolution 16

Two dice are thrown simultaneously

Total number of outcomes = 6 6 = 36

(i)Favourable cases are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) = 25.

Probability that 5 will not come upon either die 

(ii)Favourable cases are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) = 11

Probability that 5 will come at least once 

(iii)5 will come up on both dice in 1 case = (5,5)

probability that 5 will come on both dice = Question 17

Two dice are rolled once. Find the probability of getting such numbers on two dice whose product is a perfect square. Solution 17

Question 18

A letter is chosen at random from the letters of the word ‘ASSOCIATION’. Find the probability that the chosen letter is a (i) vowel (ii) consonant (iii) an S.Solution 18

Question 19

Five cards-the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random. (a) What is the probability that the drawn card is the queen?

(b) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is (i) an ace, (ii) a queen.Solution 19

Question 20

A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen?Solution 20

Question 21

What is the probability that an ordinary year has 53 Mondays?Solution 21

Question 22

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (i) a red card, (ii) a face card, (iii) a card of clubs.Solution 22

Question 23

All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the draw card is (i) a black face card, (ii) a red carSolution 23

Question 24

A game consists of tossing a one-rupee coin three times and noting its outcome each time. Find the probability of getting (i) three heads, (ii) at least 2 tails.Solution 24

Question 25

Find the probability that a leap year selected at random will contain 53 Sundays.Solution 25

Exercise MCQ

Question 1

If P(E) denotes the probability of an event E then

(a) P(E)< 0

(b) P(E) > 1

(c) 0 ≤ P(E) ≤1

(d) -1≤  P(E) ≤ 1Solution 1

Question 2

If the probability of occurrence of an event is p then the probability of non-happening of this event is

Solution 2

Question 3

What is the probability of an impossible event?

Solution 3

Question 4

What is the probability of a sure event?

Solution 4

Question 5

Which of the following cannot be the probability of an event?

Solution 5

Question 6

A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number?

Solution 6

Question 7

The probability that a number selected at random from the numbers 1,2,3,….15 is a multiple of 4, is

Solution 7

Question 8

A box contains cards numbered 6 to 50.A card is drawn at random from the box. The probability that draw card has a number which is a perfect square is

Solution 8

Question 9

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is

Solution 9

Question 10

Cards bearing numbers 2, 3, 4…, 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is

Solution 10

Question 11

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is

Solution 11

Question 12

Which of the following cannot be the probability of an event?

Solution 12

Question 13

If the probability of winning a game is 0.4 then the probability of losing it, is

Solution 13

Question 14

If an event cannot occur then its probability is

Solution 14

Question 15

There are 20 tickets numbered as 1, 2, 3…., 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5?

Solution 15

Question 16

There are 25 tickets numbered as 1, 2, 3, 4…., 25 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 3 or 5?

Solution 16

Question 17

Card, each market with one of the numbers 6, 7, 8, …, 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with number less than 10?

Solution 17

Question 18

A die is thrown once. The probability of getting an even number is

Solution 18

Question 19

The probability of throwing a number greater than 2 with a fair die is

Solution 19

Question 20

A die is thrown once. The probability of getting an odd number greater than 3 is

Solution 20

Question 21

A die is thrown once. The probability of getting a prime number is

Solution 21

Question 22

Two dice are thrown together. The probability of getting the same number on both dice is

Solution 22

Question 23

The probability of getting 2 heads, when two coins are tossed, is

Solution 23

Question 24

Two dice are thrown together. The probability of getting a doublet is

Solution 24

Question 25

Two coins are tossed simultaneously. What is the probability of getting at most one head?

Solution 25

Question 26

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Solution 26

Question 27

In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?

Solution 27

Question 28

In a lottery, there are 6 prizes and 24 blacks. What is the probability of not getting a prize?

Solution 28

Question 29

A box contains 3 blue, 2 white and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?

Solution 29

Question 30

A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?

Solution 30

Question 31

A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?

Solution 31

Question 32

A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?

Solution 32

Question 33

A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?

Solution 33

Question 34

From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability of getting a queen?

Solution 34

Question 35

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a face card?

Solution 35

Question 36

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?

Solution 36

Question 37

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a 6?

Solution 37

Exercise Ex. 15A

Question 1

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14. 5 cm.Solution 1

Question 2

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.Solution 2

Let a = 42 cm, b = 34 cm and c = 20 cm

(i)Area of triangle = 

(ii)Let base = 42 cm and corresponding height = h cm

          Then area of triangle = 

          Hence, the height corresponding to the longest side = 16 cmQuestion 3

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side of the triangle.Solution 3

Let a = 18 cm, b = 24 cm, c = 30 cm

Then,2s = (18 + 24 + 30) cm = 72 cm 

           s = 36 cm

(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(i)Area of triangle = 

(ii)Let base = 18 cm and altitude = x cm

          Then, area of triangle = 

           Hence, altitude corresponding to the smallest side = 24 cmQuestion 4

The sides of a triangle are in the ratio 5 : 12 : 13, and it perimeter is 150 m. find the area of the triangle.Solution 4

On dividing 150 m in the ratio 5 : 12 : 13, we get

Length of one side = 

Length of the second side = 

Length of third side = 

Let a = 25 m, b = 60 m, c = 65 m

(s a) = 50 cm, (s b) = 15 cm, and (s c) = 10 cm

Hence, area of the triangle = 750 m²Question 5

The perimeter of a triangular field is 540 m, and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs. 18. 80 per Solution 5

On dividing 540 m in ratio 25 : 17 : 12, we get

Length of one side = 

Length of second side = 

Length of third side = =120 m

Let a = 250m, b = 170 m and c = 120 m

Then, (s a) = 29 m, (s b) = 100 m, and (s c) = 150m

The cost of ploughing 100 area is = Rs. 18. 80

The cost of ploughing 1 is = 

The cost of ploughing 9000 area = 

                                                      = Rs. 1692

Hence, cost of ploughing = Rs 1692.Question 6

The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.Solution 6

Let the length of one side be x cm

Then the length of other side = {40 (17 + x)} cm = (23 – x) cm

Hypotenuse = 17 cm

Applying Pythagoras theorem, we get

Hence, area of the triangle = 60 cm²Question 7

The difference between the sides at right angles in a right – angled triangle is 7 cm. The area of the triangle is. Find its perimeter.Solution 7

Let the sides containing the right – angle be x cm and (x – 7) cm

One side = 15 cm and other = (15 – 7) cm = 8 cm

perimeter of triangle (15 + 8 + 17) cm = 40 cmQuestion 8

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is , find the perimeter of the triangle.Solution 8

Let the sides containing the right angle be x and (x 2) cm

One side = 8 cm, and other (8 2) cm = 6 cm

                      = 10 cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cmQuestion 9

Each side of an equilateral triangle is 10 cm. Find

(i) the area of a triangle and (ii) the height of the triangle.Solution 9

Question 10

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take Solution 10

Let each side of the equilateral triangle be a cm

Question 11

If the area of an equilateral triangle is find its perimeter.Solution 11

Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cmQuestion 12

If the area of an equilateral triangle is find its height.Solution 12

Let each side of the equilateral triangle be a cm

area of equilateral triangle =

Height of equilateral triangle

Question 13

The base of a right – angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.Solution 13

Base of right angled triangle = 48 cm

Height of the right angled triangle =

Question 14

The hypotenuse of a right – angled triangle measure 6.5 m and its base measures 6 m. Find the length of perpendicular, and hence calculate the area of the triangle.Solution 14

Let the hypotenuse of right – angle triangle = 6.5 m

Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm²Question 15

Find the area of a right – angled triangle, the radius of whose circumcircle measure 8 cm and the altitude drawn to the hypotenuse measure 6 cm.Solution 15

The circumcentre of a right – triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)

= (2 × 8) cm = 16 cm

Base = 16 cm, height = 6 cm

Area of right angled triangle

Hence, area of the triangle= 48 cm²Question 16

Find the length of the hypotenuse of an isosceles right × angled triangle whose area is 200 . Also, find the perimeter. Take .Solution 16

Let each equal side be a cm in length.

Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cmQuestion 17

The base of an isosceles triangle measure 80 cm and its area is . Find the perimeter of the triangle.Solution 17

Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm

= (2 41 + 80) cm

= (82 + 80) cm = 162 cm

Hence, perimeter of the triangle = 162 cmQuestion 18

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measure 12 cm. Find the area of the triangle.Solution 18

Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Squaring both sides,

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

Hence, area of the triangle = 48 cm².Question 19

Find the area and perimeter of an isosceles right – angled triangle, each of whose equal sides measures 10 cm. Take Solution 19

Let ABC is a isosceles triangle. Let AC, BC be the equal sides

Then AC = BC = 10cm. Let AB be the base of ABC right angle at C.

Area of right isosceles triangle ABC

Hence, area = 50 cm² and perimeter = 34.14 cmQuestion 20

In the given figure, ABC is an equilateral triangle the length of whose side is equal to 10 cm, and DBC is right angled triangle at D and BD = 8 cm. Find the area of the shaded region. 

Solution 20

Area of shaded region = Area of ABC – Area of DBC

First we find area of ABC

Second we find area of _{DBC which} is right angled

Area of shaded region = Area of ABC – Area of DBC

= (43.30 – 24) = 19. 30 

Area of shaded region = 19.3 

Exercise Ex. 15B

Question 2

The length of a rectangular park is twice its breadth and its perimeter is 840 m. Find the area of the park.Solution 2

Question 5

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is 3375 m². Find the cost of fencing the lawn at Rs. 65 per metre.Solution 5

Question 6

A room is 16 m long and 13.5 m broad. Find the cost of covering its floor with 0.75-m-wide carpet at Rs. 60 per metre.Solution 6

Question 7

The floor of a rectangular hall is 24 m long and 18 m wide. How many carpets, each of length 2.5 m and breadth 80 cm, will be required to cover the floor of the hall?Solution 7

Area of floor = Length Breadth

Area of carpet = Length Breadth

                    ₌ 

Number of carpets = 

                                                          = 216

Hence the number of carpet pieces required = 216Question 8

A 36 m-long, 15 m-broad verandah is to be paved with stones, each measuring 6 dm by 5 dm. How many stones will be required?Solution 8

Area of verandah = (36 × 15) = 540 

Area of stone = (0.6 × 0.5) [10 dm = 1 m]

Number of stones required = 

Hence, 1800 stones are required to pave the verandah.Question 9

The area of a rectangle is 192 and its perimeter is 56 cm. Find the dimensions of the rectangle.Solution 9

Perimeter of rectangle = 2(l + b)

2(l + b) = 56 Þ l + b = 28 cm

b = (28 l) cm

Area of rectangle = 192

l (28 l) = 192

28l – = 192

– 28l + 192 = 0

– 16l 12l + 192 = 0

l(l 16) 12(l 16) = 0

(l 16) (l 12) = 0

l = 16 or l = 12

Therefore, length = 16 cm and breadth = 12 cmQuestion 10

A rectangular park 35 m long and 18 m wide is to be covered with grass, leaving 2.5 m uncovered all around it. Find the area to be laid with grass.Solution 10

Length of the park = 35 m

Breadth of the park = 18 m

Area of the park = (35 18) ₌ 630 

Length of the park with grass =(35 5) = 30 m

Breadth of the park with grass = (18- 5) m = 13 m

Area of park with grass = (30 13) = 390 

Area of path without grass = Area of the whole park area of park with grass

                                     = 630 – 390 = 240 

Hence, area of the park to be laid with grass = 240 m²Question 11

A rectangular plot measures 125 m by 78 m. It has a gravel path 3 m wide all around on the outside. Find the area of the path and the cost of gravelling it at Rs. 75 per .Solution 11

Length of the plot = 125 m

Breadth of the plot = 78 m

Area of plot ABCD = (125 78) = 9750 

Length of the plot including the path= (125 + 3 + 3) m = 131 m

Breadth of the plot including the path = (78 + 3 + 3) m = 84 m

Area of plot PQRS including the path

                              = (131 84) = 11004 

Area of path = Area of plot PQRS Area of plot ABCD

                  = (11004 9750) 

                  = 1254 

Cost of gravelling = Rs 75 per m²

Cost of gravelling the whole path = Rs. (1254 75) = Rs. 94050

Hence, cost of gravelling the path = Rs 94050Question 12

A footpath of uniform width runs all around the inside of a rectangular field 54m long and 35 m wide. If the area of the path is 420 , find the width of the path.Solution 12

Area of rectangular field including the foot path = (54 35) 

Let the width of the path be x m

Then, area of rectangle plot excluding the path = (54 2x) (35 2x)

Area of path = (54 35) (54 2x) (35 2x)

                      (54 35) (54 2x) (35 2x) = 420

                      1890 1890 + 108x + 70x – 4 = 420

                      178x – 4 = 420

                      4 – 178x + 420 = 0

                      2 – 89x + 210 = 0

                      2 – 84x 5x + 210 = 0

                      2x(x 42) 5(x 42) = 0

                      (x 42) (2x 5) = 0

Question 13

The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911. Find the dimensions of the garden.Solution 13

Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45

Length of park excluding the path = (9x 7) m

Breadth of the park excluding the path = (5x 7) m

Area of the park excluding the path = (9x 7)(5x 7) 

Area of the path = 

(98x 49) = 1911

     98x = 1911 + 49

Length = 9x = 9 20 = 180 m

Breadth = 5x = 5 20 = 100 m

Hence, length = 180 m and breadth = 100 mQuestion 14

A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at Rs.80 per metre.Solution 14

Question 15

A carpet is laid on the floor of a room 8m by 5m. There is a border of constant width all around the carpet. If the area of the border is , find its width.Solution 15

Let the width of the carpet = x meter

Area of floor ABCD = (8 5) 

Area of floor PQRS without border

= (8 2x)(5 2x)

= 40 16x 10x + 

= 40 26x + 

Area of border = Area of floor ABCD Area of floor PQRS

= [40 (40 26x + )] 

=[40 40 + 26x – ] 

= (26x – )

Question 16

A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at Rs.40 per m².Solution 16

Question 17

The dimensions of a room are 14 m x 10 m x 6.5 m. There are two doors and 4 windows in the room. Each door measures 2.5 m x 1.2 m and each window was of the measure 1.5 m x 1 m. Find the cost of painting the four walls of the room at Rs.35 per m².Solution 17

Question 18

The cost of painting the four walls of a room 12 m long at Rs.30 per m² is Rs.7560 and the cost of covering the floor with mat at Rs.25 per m² is Rs.2700. Find the dimensions of the room.Solution 18

Question 19

Find the area and perimeter of a square plot of land whose diagonal is 24 m long. []Solution 19

Question 20

Find the length of the diagonal of a square of area 128 . Also find the perimeter of the square, correct to two decimal places.Solution 20

Area of the square = 

Let diagonal of square be x

Length of diagonal = 16 cm

Side of square = 

Perimeter of square = [4 side] sq. units

=[ 4 11.31] cm = 45.24 cmQuestion 21

The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km per hour?Solution 21

Let d meter be the length of diagonal

Area of square field = 

Time taken to cross the field along the diagonal

Hence, man will take 6 min to cross the field diagonally.Hence, man will take 6 min to cross the field diagonally.Question 22

The cost of harvesting a square field at Rs.900 per hectare is Rs.8100. Find the cost of putting a fence around it at Rs.18 per metre.Solution 22

Question 23

The cost of fencing a square at Rs. 14 per meter is Rs. 28000. Find the cost of mowing the lawn at Rs. 54 per 100 .Solution 23

Rs. 14 is the cost of fencing a length = 1m

Rs. 28000 is the cost of fencing the length= 

Perimeter = 4 side = 2000

side = 500 m

Area of a square = 

= 250000 

Cost of mowing the lawn =Question 24

In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 12 cm. Calculate the area of the quadrilateral.

Solution 24

Question 25

Find the area of the quadrilateral ABCD in which AD = 24cm, BAD = 90^o and ΔBCD forms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take 

Solution 25

ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm

By Pythagoras theorem

For area of equilateral DBC, we have

              a = 26 cm

Area of quad. ABCD = Area of ABD + Area of DBC

                            = (120 + 292.37) = 412.37 

Perimeter ABCD = AD + AB + BC + CD

                       = 24 cm + 10 cm + 26 cm + 26 cm

                       = 86 cmQuestion 26

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ACB = 90^o and AC = 15 cm.

Solution 26

Area of quad. ABCD = Area of ABC + Area of ACD

Now, we find area of a ACD

Area of quad. ABCD = Area of ABC + Area of ACD

Perimeter of quad. ABCD = AB + BC + CD + AD

                                    =(17 + 8 + 12 + 9) cm

                                    = 46 cm

Perimeter of quad. ABCD = 46 cmQuestion 27

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29cm, DA = 34 cm and diagonal BD = 20 cm.

Solution 27

Area of quad. ABCD = Area of ABD + Area ofDBC

For area of ABD

Let a = 42 cm, b = 34 cm, and c = 20 cm

For area of DBC

a = 29 cm, b = 21 cm, c = 20 cm

Question 28

Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.Solution 28

Area of the ||gm = (base height) sq. unit

= (25 16.8) Question 29

The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find the distance between the shorter sides.Solution 29

Longer side = 32 cm, shorter side = 24 cm

Distance between longer sides = 17.4 cm

Let the distance between the shorter sides be x cm

Area of ||gm = (longer side distance between longer sides)

= (shorter side distance between the short sides)

distance between the shorter side = 23.2 cmQuestion 30

The area of a parallelogram is 392. If its altitude is twice the corresponding base, determine the base and the altitude.Solution 30

Question 31

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Solution 31

Area ofparallelogram = 2 area of DABC

Opposite sides of parallelogram are equal

AD = BC = 20 cm

And AB = DC = 34 cm

In ABC we have

a = AC = 42 cm

b = AB = 34 cm

c = BC = 20 cm

Then, (s a) = 6 cm, (s b) = 14 cm and (s c) = 28 cm

Question 32

Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also find the perimeter of the rhombus.Solution 32

We know that the diagonals of a rhombus, bisect each other at right angles

OA = OC = 15 cm,

And OB = OD = 8 cm

And AOB = 90

By Pythagoras theorem, we have

Question 33

The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long find (i) the length of the other diagonal and (ii) the area of the rhombus.Solution 33

(i)Perimeter of rhombus = 4 side

4 side = 60 cm

By Pythagoras theorem

OB = 12 cm

OB = OD = 12 cm

BD = OB + OD = 12 cm + 12 cm = 24 cm

                       Length of second diagonal is 24 cm

             (ii) Area of rhombus = 

Question 34

The area of a rhombus is 480 and one of its diagonals measures 48 cm. Find (i) the length of the other diagonals, (ii) the length of each of its sides, and (iii) its perimeter.Solution 34

(i)Area of rhombus = 480 

One of its diagonals = 48 cm

Let the second diagonal =x cm

Hence the length of second diagonal 20 cm

(ii)We know that the diagonals of a rhombus bisect each other at right angles

AC = 48, BD = 20 cm

OA = OC = 24 cm and OB = OD = 10 cm

By Pythagoras theorem , we have

(iii)Perimeter of the rhombus = (4 26) cm = 104 cmQuestion 35

The parallel sides of a trapezium are 12 cm and 9 cm and the distance between them is 8 cm. Find the area of the trapezium.Solution 35

Question 36

The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 , find the depth of the canal.Solution 36

Areaof cross section = 

Question 37

Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the non parallel sides are 15 m and 13 m long.Solution 37

Let ABCD be a given trapezium in which

AB = 25, CD = 11

BC = 15, AD = 13

Draw CE || AD

In ||gm ADCE, AD || CE and AE || CD

AE = CD = 11 cm,

And BE = AB BE

           = 25 11 = 14 cm

In BEC,

Area of BEC = 

Let height of BEC is h

Area of BEC = 

From (1) and (2), we get

7h = 84 h = 12 m

Area of trapezium ABCD

Exercise MCQ

Question 1

The length of a rectangular hall is 5 m more than its

breadth. If the area of the hall is 750 m² then its length

is

1. 15 m
2. 20 m
3. 25 m
4. 30 m

Solution 1

Question 2

The length of a rectangular field is 23 m more than its breadth. If the perimeter of the field is 206 m, then its area is

1. 2420 m²
2. 2520 m²
3. 2480 m²
4. 2620 m²

Solution 2

Question 3

The length of a rectangular field is 12 m and the length of its diagonal is 15 m. Then area of the field is

Solution 3

Question 4

The cost of carpeting a room 15m long with a carpet 75 cm wide, at Rs.70 per metre, is Rs. 8400. Then width of the room is

1. 9 m
2. 8 m
3. 6 m
4. 12 m

Solution 4

Question 5

Solution 5

Question 6

On increasing the length of a rectangle by 20% and decreasing its breadth by 20%, what is the change in its area?

1. 20% increase
2. 20% decrease
3. No change
4. 4% decrease

Solution 6

Question 7

A rectangular ground 80 m x 50 m has a path 1 m wide

outside around it. The area of the path is

1. 264 m²
2. 284 m²
3. 400 m²
4. 464 m²

Solution 7

Question 8

Solution 8

Question 9

The area of a square field is 6050 m². The length of its

diagonal is

1. 135 m
2. 120 m
3. 112 m
4. 110 m

Solution 9

Question 10

The area of a square field is 0.5 hectare. The length of its diagonal is

Solution 10

Question 11

Solution 11

Question 12

Each side of an equilateral triangle is 8 cm. Its area is

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

The base and height of a triangle are in the ratio 3:4 and its area is 216 cm². The height of the triangle is

1. 18 cm
2. 24 cm
3. 21 cm
4. 28 cm

Solution 15

Question 16

The lengths of the sides of a triangular field are 20 m, 21m and 29 m. The cost of cultivating the field at Rs. 9 per m² is

1. Rs. 2610
2. Rs. 3780
3. Rs. 1890
4. Rs. 1800

Solution 16

Question 17

The side of a square is equal to the side of an equilateral triangle. The ratio of their areas is

Solution 17

Question 18

The side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm². The area of the triangle is

Solution 18

Question 19

The area of a rhombus is 480 cm² and the length of one of its diagonals is 20 cm. The length of each side of the rhombus is

1. 24 cm
2. 30 cm
3. 26 cm
4. 28 cm

Solution 19

Question 20

One side of a rhombus is 20 cm long and one of its diagonals measures 24 cm. The area of the rhombus is

1. 192 cm²
2. 480 cm²
3. 240 cm²
4. 384 cm²

Solution 20

Exercise FA

Question 1

In the given figure ABCD is a quadrilateral in which ∠ABC = 90, ∠BDC = 90, AC = 17 cm, BC= 15 cm, BD = 12 cm and CD = 9cm. The area of quad. ABCD is

1. 102 cm²
2. 114 cm²
3. 95 cm²
4. 57 cm²

Solution 1

Question 2

In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CE ^ AB. Area of trap. ABCD is

1. 306 m²
2. 316 m²
3. 296 m²
4. 284 m²

Solution 2

Question 3

The sides of a triangle are in the ratio 12:14:25 and it perimeter is 25.5 cm. The largest side of the triangle is

1. 7 cm
2. 14 cm
3. 12.5 cm
4. 18 cm

Solution 3

Question 4

The parallel sides of trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is

1. 104 cm²
2. 78 cm²
3. 52 cm²
4. 65 cm²

Solution 4

Question 5

Solution 5

Question 6

Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.Solution 6

Question 7

The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall.Solution 7

Question 8

The length of the diagonal of a square is 24 cm. Find its area.Solution 8

Question 9

Find the area of a rhombus whose diagonals are 48 cm and 20 cm long.Solution 9

Question 10

Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm.Solution 10

Question 11

A lawn is in the form of a rectangle whose sides are in the ratio 5:3 and its area is 3375 m². Find the cost of fencing the lawn at Rs.20 per metre.  Solution 11

Question 12

Find the area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm.Solution 12

Question 13

Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.Solution 13

Question 14

The adjacent sides of a ‖ gm AECD measure 34 cm and 20 cm and the diagonal AC is 42 cm long. Find the area of the ‖ gm.

Solution 14

Question 15

The cost of fencing a square lawn at Rs.14 per metre is Rs. 2800. Find the cost of mowing the lawn at Rs.54 per 100 m².Solution 15

Question 16

Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD =29 cm, DA = 39 cm and diag. BD = 20 cm.

Solution 16

Question 17

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of the sides of the ‖gm is 66m long, find its corresponding altitude.Solution 17

Question 18

The diagonal of a rhombus are 48 cm and 20cm long. Find the perimeter of the rhombus.Solution 18

Question 19

The adjacent sides of parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides.Solution 19

Question 20

In a four-sided field, the length of the longer diagonal is 128 m. The length of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field.Solution 20

Exercise Ex. 13A

Question 1

Prove the following identities:

Solution 1

(i)        

LHS = RHS

(ii)      

LHS = RHSQuestion 2

Prove the following identities:

(i)

(ii)

(iii)Solution 2

(i)

LHS = RHS

(ii)

LHS = RHS

(iii)

LHS = RHSQuestion 3

Prove the following identities:

(i)

(ii)Solution 3

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 4

Prove the following identities:

(i) 

_(ii) Solution 4

(i) 

(ii)

LHS = RHSQuestion 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 7

Prove the following identities:

(i)  

(ii)Solution 7

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 8

Prove the following identities:

Solution 8

(i)    LHS = 

(ii)             

Hence, LHS = RHSQuestion 9

Prove the following identity:

Solution 9

LHS = RHSQuestion 10

Prove the following identity:

Solution 10

Question 11

Prove the following identity:

Solution 11

Question 12

Prove the following identity:

Solution 12

RHS = LHSQuestion 13

Prove the following identity:

Solution 13

LHS = 

RHS = LHSQuestion 14

Prove the following identity:

Solution 14

LHS = RHSQuestion 15

Prove the following identity:

Solution 15

RHS = LHSQuestion 16

Prove the following identity:

Solution 16

Question 17

Prove the following identities:

Solution 17

(i)To prove 

     We know, 

      Therefore, LHS = RHS

(ii)

      Therefore, LHS = RHS

(iii)

Question 18

Prove the following identity:

(i) 

(ii) Solution 18

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 19

Prove the following identities:

(i) 

(ii)Solution 19

(i)

LHS = RHS

(ii) LHS =  

Question 20

Prove the following identities:

Solution 20

(i) 

   _{LHS =}

Hence, LHS = RHS

(ii)

LHS = RHSQuestion 21(i)

Solution 21(i)

Question 21(ii)

Solution 21(ii)

Question 21(iii)

Solution 21(iii)

Question 22

Prove the following identity:

Solution 22

LHS = RHSQuestion 23

Prove the following identity:

Solution 23

LHS = RHSQuestion 24

Prove the following identities:

(i)

(ii)Solution 24

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 25

Prove the following identity:

Solution 25

LHS = RHSQuestion 26

Prove the following identities:

Solution 26

(i)

             Further,

                                LHS = RHS

(ii)

             LHS = 

Further,

Question 27

Prove the following identities:

(i) 

(ii)Solution 27

(i)

             On dividing the numerator and denominator of LHS by cos,We get

(ii)

            On dividing the numerator and denominator of LHS by cos,We get

                        LHS = RHS
Question 28

Prove the following identity:

Solution 28

LHS = RHSQuestion 29

Prove the following identity:

Solution 29

Question 30

Prove the following identity:

Solution 30

Question 31

Prove the following identity:

Solution 31

Question 32

Solution 32

Question 33

Prove the following identity:

Solution 33

Question 34

Prove the following identity:

Solution 34

Question 35

Prove the following identity:

Solution 35

Question 36(i)

Show that none of the following is an identity:

cos² θ + cos θ = 1Solution 36(i)

Question 36(ii)

sin² θ + sin θ = 2Solution 36(ii)

Question 36(iii)

tan² θ + sin θ = cos² θSolution 36(iii)

Question 37

Prove that:

(sin θ – 2sin³ θ) = (2cos³ θ – cos θ)tan θSolution 37

Exercise Ex. 13B

Question 1

If a cos + b sin = m and a sin – b cos = n, prove that

.Solution 1

Question 2

If x = a sec  + b tan and y = a tan  + b sec , prove that

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

If prove that Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If (cosec θ – sin θ) = a³ and (sec θ – cos θ) = b³, prove that a²b² (a² + b²) = 1.Solution 9

Question 10

If (2sin θ + 3cos θ) = 2, prove that (3sin θ – 2cos θ) = ± 3.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

If sec θ + tan θ = p, prove that

Solution 13(i)

Question 13(ii)

If sec θ + tan θ = p, prove that

Solution 13(ii)

Question 13(iii)

If sec θ + tan θ = p, prove that

Solution 13(iii)

Question 14

Solution 14

Question 15

Solution 15

Exercise Ex. 13C

Question 1

Write the value of (1 – sin² θ) sec² θ.Solution 1

Question 2

Write the value of (1 – cos²θ) cosec² θ.Solution 2

Question 3

Write the value of (1 + tan² θ) cos² θ.Solution 3

Question 4

Write the value of (1 + cot² θ) sin² θ.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Write the value of sin θ cos (90^ᵒ – θ) + cos θ sin (90^ᵒ – θ).Solution 7

Question 8

Write the value of cosec² (90^ᵒ – θ) – tan² θ.Solution 8

Question 9

Write the value of sec² θ (1 + sin θ)(1 – sin θ).Solution 9

Question 10

Write the value of cosec² θ(1 + cos θ)(1 – cos θ)

Note: Question modifiedSolution 10

Question 11

Write the value of sin² θ cos² θ (1 + tan² θ)(1 + cot² θ).Solution 11

Question 12

Write the value of (1 + tan² θ)(1 + sin θ)(1 – sin θ).Solution 12

Question 13

Write the value of 3cot² θ – 3cosec²θ.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Write the value of tan 10ᵒ tan 20ᵒ tan 70ᵒ tan 80ᵒ.Solution 27

Question 28

Write the value of tan 1ᵒ tan 2ᵒ … tan 89ᵒ.Solution 28

Question 29

Write the value of cos 1ᵒ cos 2ᵒ…cos 180ᵒ.Solution 29

Question 30

Solution 30

Question 31

If sin θ = cos (θ – 45ᵒ),  where θ is a acute, find the value of θ.Solution 31

Question 32

Solution 32

Question 33

Find the value of sin 48ᵒ sec 42ᵒ + cos 48ᵒ cosec 42ᵒ .Solution 33

Question 34

If x = a sin θ and y = b cos θ, write the value of (b²x² + a²y²).Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

If sec θ + tan θ = x , find the value of sec θ.Solution 37

Question 38

Solution 38

Question 39

If sin θ = x, write the value of cot θ.Solution 39

Question 40

If sec θ = x, write the value of tan θ.Solution 40

Exercise MCQ

Question 1

Solution 1

Question 2

(a)0

(b) 1

(c) 2

(d) none of theseSolution 2

Question 3

tan 10° tan 15° tan 75° tan 80° = ?

Solution 3

Question 4

tan 5° tan 25° tan 30° tan 65° tan 85° = ?

Solution 4

Question 5

cos 1° cos 2° cos 3° …… cos 180° = ?

(a) -1

(b) 1

(c) 0

(d) Solution 5

Question 6

Solution 6

Question 7

sin 47° cos 43° + cos 47° sin 43° = ?

(a) sin 4° 

(b) cos 4° 

(c) 1

(d) 0Solution 7

Question 8

sec 70° sin 20° + cos 20° cosec 70° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 8

Question 9

If sin 3A = cos (A – 10^o) and 3A is acute then ∠A = ?

(a) 35° 

(b) 25° 

(c) 20° 

(d) 45° Solution 9

Question 10

If sec 4A = cosec (A – 10°) and 4A is acute then ∠A = ?

(a) 20° 

(b) 30° 

(c) 40° 

(d) 50° Solution 10

Question 11

If A and B are acute angles such that sin A = cos B then (A + B) =?

(a) 45° 

(b) 60° 

(c) 90° 

(d) 180° Solution 11

Question 12

If cos (𝛼 + 𝛽) = 0 then sin (𝛼 – 𝛽) = ?

(a) sin 𝛼 

(b) cos 𝛽 

(c) sin 2𝛼 

(d) cos 2𝛽 Solution 12

Question 13

sin (45° + θ) – cos (45° – θ) = ?

(a) 2 sin θ 

(b) 2 cos θ 

(c) 0

(d) 1Solution 13

Question 14

sec²10° – cot² 80° = ?

(a) 1

(b) 0

Solution 14

Question 15

cosec² 57° – tan² 33° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 15

Question 16

Solution 16

Question 17

(a) 0

(b) 1

(c) 2

(d) 3Solution 17

Question 18

(a) 0

(b) 1

(c) -1

(d) none of theseSolution 18

Question 19

Solution 19

Question 20

(a) 30° 

(b) 45° 

(c) 60° 

(d) 90°Solution 20

Question 21

If 2cos 3θ = 1 then θ = ?

(a) 10° 

(b) 15° 

(c) 20° 

(d) 30° Solution 21

Question 22

(a) 15° 

(b) 30° 

(c) 45° 

(d) 60° Solution 22

Question 23

If tan x = 3cot x then x = ?

(a) 45° 

(b) 60° 

(c) 30° 

(d) 15° Solution 23

Question 24

If x tan 45° cos 60° = sin 60° cot 60° then x = ?

Solution 24

Question 25

If tan² 45° – cos² 30° = x sin 45° cos 45° then x = ?

Solution 25

Question 26

sec² 60° – 1 = ?

(a) 2

(b) 3

(c) 4

(d) 0Solution 26

Correct option: (b)

sec² 60° – 1 = (2)² – 1 = 4 – 1 = 3Question 27

(cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° – cos 45°) =?

Solution 27

Question 28

sin²30° + 4cot² 45° – sec² 60° = ?

Solution 28

Question 29

3cos² 60° + 2cot² 30° – 5sin² 45° = ?

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

If (tan θ + cot θ) = 5 then (tan² θ + cot² θ) = ?

(a) 27

(b) 25

(c) 24

(d) 23Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

If sin A + sin² A = 1 then cos² A + cos⁴ A = ?

(a) 

(b) 1

(c) 2

(d) 3Solution 47

Question 48

If cos A + cos² A = 1 then sin² A + sin⁴ A = ?

(a) 1

(b) 2

(c) 4

(d) 3Solution 48

Question 49

(a) sec A + tan A

(b) sec A – tan A

(c) sec A tan A

(d) none of theseSolution 49

Question 50

(a) cosec A – cot A

(b) cosec A + cot A

(c) cosec A cot A

(d) none of theseSolution 50

Question 51

Solution 51

Question 52

(cosec θ – cot θ)² = ?

Solution 52

Question 53

(sec A + tan A)(1 – sin A) = ?

(a) sin A

(b) cos A

(c) sec A

(d) cosec ASolution 53

Exercise FA

Question 1

Solution 1

Question 2

Solution 2

Question 3

If cos A + cos² A = 1 then (sin² A + sin⁴ A) = ?

(a) 

(b) 2

(c) 1

(d) 4Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.Solution 14

Question 15

If x = a sin θ + b cos θ and y = a cos θ – b sin θ, prove that x² + y² = a² + b².Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

If sec 5A = cosec (A – 36°) and 5A is an acute angle, show that A = 21° Solution 20

Exercise Ex. 14

Question 1

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take ]Solution 1

Let AB be the tower standing on a level ground and O be the position of the observer. Then OA = 20 m and OAB = 90° and AOB = 60°

Let AB = h meters

From the right OAB, we have

Hence the height of the tower is Question 2

A kite is flying at a height of 75m from the level ground, attached to a string inclined at 60° to the horizontal. Find the length of the string assuming that there is no slack in it. Solution 2

Let OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and OAB = 90° and AOB = 60°, let OB = l meters.

From the right OAB, we have

Question 3

An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60°. Find the height of the chimney.Solution 3

Question 4

The angles of elevation of the top of a tower from two points at distances of 5 metres and 20 metres from the base of the tower and in the same straight line with it, are complementary. Find the height of the tower.Solution 4

Question 5

The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. Solution 5

Question 6

From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°. Find (i) the height of the tower, (ii) the depth of the tank.Solution 6

Question 7

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 60°. Find the height of the tower.

Solution 7

Question 8

A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Solution 8

Let SP be the statue and PB be the pedestal. Angles of elevation of S and P are 60° and 45° respectively.

Further suppose AB = x m, PB = h m

In right ABS,

In right PAB,

Thus, height of the pedestal = 2mQuestion 9

The angle of elevation of the top of an unfinished tower at a distance of 75m from its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?Solution 9

Let AB be the unfinished tower and let AC be complete tower.

Let O be the point of observation. Then,

OA = 75 m

AOB = 30° and AOC = 60°

Let AB = h meters

And AC = H meters

Hence, the required height is Question 10

On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it. Solution 10

Let AB be the tower and BC be flagpole, Let O be the point of observation.

Then, OA = 9 m, AOB = 30° and AOC = 60°

From right angled BOA

From right angled OAC

Thus 

Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 mQuestion 11

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the height of each pole and distances of the point P from the poles.Solution 11

Question 12

Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men.

Solution 12

Question 13

From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars.

Solution 13

Question 14

A straight highway leads to the foot of a tower. A man standing on the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower form this point.Solution 14

Question 15

A TV tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.Solution 15

Question 16

The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building.Solution 16

Question 17

The horizontal distance between two towers is 60metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90metres, find the height of the tower.  Solution 17

Let AB and CD be the first and second towers respectively.

Then, CD = 90 m and AC = 60 m.

Let DE be the horizontal line through D.

Draw BF CD,

Then, BF = AC = 60 m

FBD = EDB = 30°

Question 18

The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 metres, find the height of the chimney.

According to pollution control norms, the minimum height of a smoke-emitting chimney should be 100 metres. State if the height of the above-mentioned chimney meets the pollution norms. What value is discussed in this question?Solution 18

Question 19

From the top of a 7-metre-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution 19

Question 20

The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A.Solution 20

Question 21

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.Solution 21

Question 22

Solution 22

Question 23

A man on the deck of a ship, 16m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60 and 30. Calculate the distance of the cliff from the ship and height of the cliff. Solution 23

Let AB be the height of the deck and let CD be the cliff..

Let the man be at B, then, AB= 16 m

Let BE CD and AE CD

Then, EBD = 60 and EBC = 30

CE = AB = 16m

Let CD = h meters

Then, ED = (h 16)m

From right BED, we have

From right CAB, we have

Hence the height of cliff is 64 m and the distance between the cliff and the ship = Question 24

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ.

Solution 24

Question 25

The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane.Solution 25

Question 26

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres. Solution 26

Let AB be the tower and let the angle of elevation of its top at C be 30°. Let D be a point at a distance 150 m from C such that the angle of elevation of the top of tower at D is 60°.

Let h m be the height of the tower and AD = x m

In CAB, we have

Hence the height of tower is 129.9 mQuestion 27

As observed from the top of a lighthouse, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. Solution 27

Let AB be the light house and let C and D be the positions of the ship.

Llet AD =x, CD = y

In BDA,

The distance travelled by the ship during the period of observation = 115.46 mQuestion 28

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river.

Solution 28

Question 29

The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres.Solution 29

Question 30

A ladder of length 6 metres makes an angle of 45° with the floor while leaning against one wall of a room. If the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between two walls of the room.Solution 30

Question 31

From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.Solution 31

Question 32

An electrician has to repair an electric fault on a pole of height 4 metres. He needs to reach a point 1 metre below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of 60° to the horizontal would enable him to reach the required position? Solution 32

Question 33

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30°  and 60° respectively. Find

(i) the horizontal distance between AB and CD,

(ii) the height of the lamp post,

(iii) the difference between the heights of the building and the lamp post.Solution 33

Exercise MCQ

Question 1

If the height of a vertical pole is equal to the length of its shadow on the ground, the angle of elevation of the sun is

(a) 0ᵒ 

(b) 30ᵒ 

(c) 45ᵒ 

(d) 60ᵒ Solution 1

Question 2

(a) 30ᵒ 

(b) 45ᵒ 

(c) 60ᵒ 

(a) 75ᵒ Solution 2

Question 3

(a) 45ᵒ 

(b) 30ᵒ 

(c) 60ᵒ 

(d) 90ᵒ Solution 3

Question 4

(a) 60ᵒ 

(b) 45ᵒ 

(c) 30ᵒ 

(b) 90ᵒ Solution 4

Question 5

The shadow of a 5-m-long stick is 2 m long. At the same time, the length of the shadow of a 12.5-m-high tree is

(a) 3 m

(b) 3.5 m

(c) 4.5 m

(d) 5 mSolution 5

Question 6

A ladder makes an angle of 60ᵒ with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, the length of the ladder is

Solution 6

Question 7

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60ᵒ with the wall then the height of the wall is

Solution 7

Question 8

From a point on the ground, 30 m away from the foot of a tower the angle of elevation of the top of the tower is 30°. The height of the tower is

Solution 8

Question 9

The angle of depression of a car parked on the road from the top of a 150-m-high tower is 30°. The distance of the car from the tower is

Solution 9

Question 10

A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is

(a) 45ᵒ 

(b) 30ᵒ 

(c) 60ᵒ 

(a) 90ᵒ Solution 10

Question 11

From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is

(a) 20 m

(b) 40 m

(c) 60 m

(d) 80 m Solution 11

Question 12

If a 1-5-m-tall girl stands at a distance of 3 m from a lamp post and casts a shadow of length 4.5 m on the ground, then the height of the lamp post is

(a) 1.5 m

(b) 2 m

(c) 2.5 m

(d) 2.8 mSolution 12

Question 13

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun’s elevation is 30ᵒ than when it was 45ᵒ. The height of the tower is

Solution 13

Question 14

(a) 30ᵒ 

(b) 45ᵒ 

(c) 60ᵒ 

(d) 90ᵒ Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

The tops of two towers of heights x and y, standing on a level ground subtend angles of 30ᵒ and 60ᵒ respectively at the centre of the line joining their feet. Then, x : y is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 3

(d) 3 : 1Solution 17

Question 18

The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30ᵒ. The height of the tower is

Solution 18

Question 19

The string of a kite is 100 m long and it makes an angle of 60ᵒ with the horizontal. If there is no slack in the string, the height of the kite from the ground is

Solution 19

Question 20

If the angles of elevation of the top of a tower from two points at distances a and b from the base and in the same straight line with it are complementary then the height of the tower is

Solution 20

Question 21

On the level ground, the angle of elevation of a tower is 30ᵒ. On moving 20 m nearer, the angle of elevation is 60ᵒ. The height of the tower is

Solution 21

Question 22

In a rectangle, the angle between a diagonal and a side is 30ᵒ and the length of this diagonal is 8 cm. The area of the rectangle is ,

Solution 22

Question 23

From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30ᵒ and 45ᵒ. The height of the hill is

Solution 23

Question 24

If the elevation of the sun changes from 30ᵒ to 60ᵒ then the difference between the lengths of shadows of a pole 15 m high, is

Solution 24

Question 25

An observer 1.5 m tall is 28.5 m away from a tower and the angle of elevation of the top of the tower from the eye of the observer is 45ᵒ. The height of the tower is

(a) 27 m

(b) 30 m

(c) 28.5 m  

(d) none of theseSolution 25

Chapter 18 Maxima and Minima Ex 18.1

Question 1

Solution 1

Question 2

Find the maximum and minimum values, if any, without using derivatives of the following function given by f(x) = -(x-1)² + 2 on R.Solution 2

Question 3

Solution 3

Question 4

Find the maximum and minimum values, if any, without using derivatives of the following function given by h(x) = sin(2x) + 5 on R.Solution 4

Question 5

Find the maximum and minimum values, if any,  usingwithout derivatives of the following function given by  on R.Solution 5

Question 6

Solution 6

Question 7

Find the maximum and minimum values, if any, without using derivatives of the following function given by  on R.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 18 Maxima and Minima Ex. 18.2

Question 1

Solution 1

Question 2

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) =x³ – 3xSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) = sinx – cos x, 0 < x < 2πSolution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 18 Maxima and Minima Ex. 18.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Question 4

Show that the function given by f(x)= has maximum value at x = e.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

If f(x) = x³ + ax² + bx + c has a maximum at x = -1 and minimum at x = 3. Determine a, b and c.Solution 7

Question 8

Prove that  has maximum value at Solution 8

Given: 

Differentiating w.r.t x, we get

Take f'(x) = 0

Differentiating f'(x) w.r.t x, we get

At 

Clearly, f”(x) < 0 at 

Thus,  is the maxima.

Hence, f(x) has maximum value at .

Chapter 18 Maxima and Minima Ex. 18.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Find both the absolute maximum and absolute minimum of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3] Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 18 Maxima and Minima Ex. 18.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Divide 15 into two parts such that the square of one multiplied with the cube of the other is maximum.Solution 4

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Also, find this maximum volume.Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

An isosceles triangle of vertical angle 2θ is inscribed in a circle radius a. show that the area of the triangle is maximum when Solution 22

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its semi – circular ends is π : ( π + 2) Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 45

Solution 45

Question 5

Amongst all open (from the top) right circular cylindrical boxes of volume 125π cm³, find the dimensions of the box which has the least surface area.Solution 5

Let r and h be the radius and height of the cylinder.

Volume of cylinder  

 … (i)

Surface area of cylinder  

From (i), we get

Question 23

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is  Solution 23

Let  be an isosceles triangle with AB = AC.

Let  

Here, AO bisects 

Taking O as the centre of the circle, join OE, OF and OD such that

OE = OF = OD = r (radius)

Now, 

In 

Similarly, AF = r cot x

In 

As OB bisect  we have

In 

Similarly, BD = DC = CE = 

We have, perimeter of 

P = AB + BC + CA

 = AE + EC + BD + DC + AF + BF

Differentiating w.r.t x, we get

Taking 

As 

Therefore,  is an equilateral triangle.

Taking second derivative of P, we get

At 

Therefore, the perimeter is minimum when 

Least value of P 

Exercise Ex. 7

Question 1

Without using trigonometric tables, evaluate:

Solution 1

(i)

(ii)              

(iii)

(iv)

(v)

(vi)

Question 2

Without using trigonometric tables, prove that:

(i) 

(ii) 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)Solution 2

(i)                LHS = cos81° – sin9°

= cos(90° -9°)- sin9° = sin9° – sin9°

= 0 = RHS

(ii)              LHS = tan71° – cot19°

=tan(90° – 19°) – cot19° =cot19° – cot19°

=0 = RHS

(iii)               LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10° = 0

= RHS

(iv)               LHS= 

(v)            

(vi)             

(vii)                 

LHS = RHS

(viii) 

(ix)              LHS = (sin65° + cos25°) (sin65° – cos25°)

Question 3(i)

Without using trigonometric tables, prove that:

sin 53^ᵒ cos37^ᵒ + cos53^ᵒ sin37^ᵒ = 1Solution 3(i)

Question 3(ii)

cos 54^ᵒ cos 36^ᵒ – sin 54^ᵒ sin36^ᵒ = 0Solution 3(ii)

Question 3(iii)

sec 70^ᵒ sin 20^ᵒ + cos 20^ᵒ cosec 70^ᵒ = 2Solution 3(iii)

Question 3(iv)

sin 35^ᵒ sin 55^ᵒ – cos 35^ᵒ cos 55^ᵒ = 0Solution 3(iv)

Question 3(v)

(sin 72^ᵒ + cos 18^ᵒ)(sin72^ᵒ – cos18^ᵒ) = 0Solution 3(v)

Question 3(vi)

tan 48^ᵒ tan 23^ᵒ tan 42^ᵒ tan 67^ᵒ = 1Solution 3(vi)

Question 4(i)

Prove that:

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5(i)

Prove that:

sin θ cos (90^ᵒ – θ) + sin (90^ᵒ – θ) cos θ = 1Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

cos 1^ᵒ cos2^ᵒ cos3^ᵒ … cos 180^ᵒ= 0Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 7

Prove that:

(i)

(ii) 

(iii)

(iv)

(v)Solution 7

(i)                  LHS =

= RHS

(ii)                LHS =

= RHS

(iii)                  LHS =

= RHS

(iv)             LHS = cosec(65° + ) – sec(25°-  ) – tan(55° –  ) + cot(35° +  )

= RHS

(v)        LHS =

             = 0 + 1 = 1

             = RHS
Question 8(i)

Express each of the following in terms of T-ratios of angles lying between 0^ᵒ and 45^ᵒ:

sin 67^ᵒ + cos 75^ᵒSolution 8(i)

Question 8(ii)

cot 65^ᵒ + tan 49^ᵒSolution 8(ii)

Question 8(iii)

sec 78^ᵒ + cosec 56^ᵒSolution 8(iii)

Question 8(iv)

cosec 54^ᵒ + sin 72^ᵒSolution 8(iv)

Question 9

If A, B, C are the angles of a triangle ABC, prove that:Solution 9

A + B + C = 180°

So, B + C= 180° – A

Question 10

If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.Solution 10

Question 11

If sec 2A= cosec (A – 42ᵒ), where 2A is an acute angle, find the value of A.Solution 11

Question 12

If sin3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.Solution 12

Question 13

If tan 2A = cot(A – 12^o), where 2A is an acute angle, find the value of A.Solution 13

Question 14

If sec 4A = cosec(A – 15^o), where 4A is an acute angle, find the value of A.Solution 14

Question 15

Prove that:

Solution 15

Exercise Ex. 8A

Question 1

Prove the following identities:

Solution 1

(i)        

LHS = RHS

(ii)      

LHS = RHSQuestion 2

Prove the following identities:

(i)

(ii)

(iii)Solution 2

(i)

LHS = RHS

(ii)

LHS = RHS

(iii)

LHS = RHSQuestion 3

Prove the following identities:

(i)

(ii)Solution 3

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 4

Prove the following identities:

(i) 

_(ii) Solution 4

(i) 

(ii)

LHS = RHSQuestion 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 7

Prove the following identities:

(i)  

(ii)Solution 7

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 8

Prove the following identities:

Solution 8

(i)    LHS = 

(ii)             

Hence, LHS = RHSQuestion 9

Prove the following identity:

Solution 9

LHS = RHSQuestion 10

Prove the following identity:

Solution 10

Question 11

Prove the following identity:

Solution 11

Question 12

Prove the following identity:

Solution 12

RHS = LHSQuestion 13

Prove the following identity:

Solution 13

LHS = 

RHS = LHSQuestion 14

Prove the following identity:

Solution 14

LHS = RHSQuestion 15

Prove the following identity:

Solution 15

RHS = LHSQuestion 16

Prove the following identity:

Solution 16

Question 17

Prove the following identities:

Solution 17

(i)To prove 

     We know, 

      Therefore, LHS = RHS

(ii)

      Therefore, LHS = RHS

(iii)

Question 18

Prove the following identity:

(i) 

(ii) Solution 18

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 19

Prove the following identities:

(i) 

(ii)Solution 19

(i)

LHS = RHS

(ii) LHS =  

Question 20

Prove the following identities:

Solution 20

(i) 

   _{LHS =}

Hence, LHS = RHS

(ii)

LHS = RHSQuestion 21(i)

Solution 21(i)

Question 21(ii)

Solution 21(ii)

Question 21(iii)

Solution 21(iii)

Question 22

Prove the following identity:

Solution 22

LHS = RHSQuestion 23

Prove the following identity:

Solution 23

LHS = RHSQuestion 24

Prove the following identities:

(i)

(ii)Solution 24

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 25

Prove the following identity:

Solution 25

LHS = RHSQuestion 26

Prove the following identities:

Solution 26

(i)

             Further,

                                LHS = RHS

(ii)

             LHS = 

Further,

Question 27

Prove the following identities:

(i) 

(ii)Solution 27

(i)

             On dividing the numerator and denominator of LHS by cos,We get

(ii)

            On dividing the numerator and denominator of LHS by cos,We get

                        LHS = RHS
Question 28

Prove the following identity:

Solution 28

LHS = RHSQuestion 29

Prove the following identity:

Solution 29

Question 30

Prove the following identity:

Solution 30

Question 31

Prove the following identity:

Solution 31

Question 32

Solution 32

Question 33

Prove the following identity:

Solution 33

Question 34

Prove the following identity:

Solution 34

Question 35

Prove the following identity:

Solution 35

Question 36(i)

Show that none of the following is an identity:

cos² θ + cos θ = 1Solution 36(i)

Question 36(ii)

sin² θ + sin θ = 2Solution 36(ii)

Question 36(iii)

tan² θ + sin θ = cos² θSolution 36(iii)

Question 37

Prove that:

(sin θ – 2sin³ θ) = (2cos³ θ – cos θ)tan θSolution 37

Exercise Ex. 8B

Question 1

If a cos + b sin = m and a sin – b cos = n, prove that

.Solution 1

Question 2

If x = a sec  + b tan and y = a tan  + b sec , prove that

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

If prove that Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If (cosec θ – sin θ) = a³ and (sec θ – cos θ) = b³, prove that a²b² (a² + b²) = 1.Solution 9

Question 10

If (2sin θ + 3cos θ) = 2, prove that (3sin θ – 2cos θ) = ± 3.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

If sec θ + tan θ = p, prove that

Solution 13(i)

Question 13(ii)

If sec θ + tan θ = p, prove that

Solution 13(ii)

Question 13(iii)

If sec θ + tan θ = p, prove that

Solution 13(iii)

Question 14

Solution 14

Question 15

Solution 15

Exercise Ex. 8C

Question 1

Write the value of (1 – sin² θ) sec² θ.Solution 1

Question 2

Write the value of (1 – cos²θ) cosec² θ.Solution 2

Question 3

Write the value of (1 + tan² θ) cos² θ.Solution 3

Question 4

Write the value of (1 + cot² θ) sin² θ.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Write the value of sin θ cos (90^ᵒ – θ) + cos θ sin (90^ᵒ – θ).Solution 7

Question 8

Write the value of cosec² (90^ᵒ – θ) – tan² θ.Solution 8

Question 9

Write the value of sec² θ (1 + sin θ)(1 – sin θ).Solution 9

Question 10

Write the value of cosec² θ(1 + cos θ)(1 – cos θ)

Note: Question modifiedSolution 10

Question 11

Write the value of sin² θ cos² θ (1 + tan² θ)(1 + cot² θ).Solution 11

Question 12

Write the value of (1 + tan² θ)(1 + sin θ)(1 – sin θ).Solution 12

Question 13

Write the value of 3cot² θ – 3cosec²θ.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Write the value of tan 10ᵒ tan 20ᵒ tan 70ᵒ tan 80ᵒ.Solution 27

Question 28

Write the value of tan 1ᵒ tan 2ᵒ … tan 89ᵒ.Solution 28

Question 29

Write the value of cos 1ᵒ cos 2ᵒ…cos 180ᵒ.Solution 29

Question 30

Solution 30

Question 31

If sin θ = cos (θ – 45ᵒ),  where θ is a acute, find the value of θ.Solution 31

Question 32

Solution 32

Question 33

Find the value of sin 48ᵒ sec 42ᵒ + cos 48ᵒ cosec 42ᵒ .Solution 33

Question 34

If x = a sin θ and y = b cos θ, write the value of (b²x² + a²y²).Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

If sec θ + tan θ = x , find the value of sec θ.Solution 37

Question 38

Solution 38

Question 39

If sin θ = x, write the value of cot θ.Solution 39

Question 40

If sec θ = x, write the value of tan θ.Solution 40

Exercise MCQ

Question 1

Solution 1

Question 2

(a)0

(b) 1

(c) 2

(d) none of theseSolution 2

Question 3

tan 10° tan 15° tan 75° tan 80° = ?

Solution 3

Question 4

tan 5° tan 25° tan 30° tan 65° tan 85° = ?

Solution 4

Question 5

cos 1° cos 2° cos 3° …… cos 180° = ?

(a) -1

(b) 1

(c) 0

(d) Solution 5

Question 6

Solution 6

Question 7

sin 47° cos 43° + cos 47° sin 43° = ?

(a) sin 4° 

(b) cos 4° 

(c) 1

(d) 0Solution 7

Question 8

sec 70° sin 20° + cos 20° cosec 70° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 8

Question 9

If sin 3A = cos (A – 10^o) and 3A is acute then ∠A = ?

(a) 35° 

(b) 25° 

(c) 20° 

(d) 45° Solution 9

Question 10

If sec 4A = cosec (A – 10°) and 4A is acute then ∠A = ?

(a) 20° 

(b) 30° 

(c) 40° 

(d) 50° Solution 10

Question 11

If A and B are acute angles such that sin A = cos B then (A + B) =?

(a) 45° 

(b) 60° 

(c) 90° 

(d) 180° Solution 11

Question 12

If cos (𝛼 + 𝛽) = 0 then sin (𝛼 – 𝛽) = ?

(a) sin 𝛼 

(b) cos 𝛽 

(c) sin 2𝛼 

(d) cos 2𝛽 Solution 12

Question 13

sin (45° + θ) – cos (45° – θ) = ?

(a) 2 sin θ 

(b) 2 cos θ 

(c) 0

(d) 1Solution 13

Question 14

sec²10° – cot² 80° = ?

(a) 1

(b) 0

Solution 14

Question 15

cosec² 57° – tan² 33° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 15

Question 16

Solution 16

Question 17

(a) 0

(b) 1

(c) 2

(d) 3Solution 17

Question 18

(a) 0

(b) 1

(c) -1

(d) none of theseSolution 18

Question 19

Solution 19

Question 20

(a) 30° 

(b) 45° 

(c) 60° 

(d) 90°Solution 20

Question 21

If 2cos 3θ = 1 then θ = ?

(a) 10° 

(b) 15° 

(c) 20° 

(d) 30° Solution 21

Question 22

(a) 15° 

(b) 30° 

(c) 45° 

(d) 60° Solution 22

Question 23

If tan x = 3cot x then x = ?

(a) 45° 

(b) 60° 

(c) 30° 

(d) 15° Solution 23

Question 24

If x tan 45° cos 60° = sin 60° cot 60° then x = ?

Solution 24

Question 25

If tan² 45° – cos² 30° = x sin 45° cos 45° then x = ?

Solution 25

Question 26

sec² 60° – 1 = ?

(a) 2

(b) 3

(c) 4

(d) 0Solution 26

Correct option: (b)

sec² 60° – 1 = (2)² – 1 = 4 – 1 = 3Question 27

(cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° – cos 45°) =?

Solution 27

Question 28

sin²30° + 4cot² 45° – sec² 60° = ?

Solution 28

Question 29

3cos² 60° + 2cot² 30° – 5sin² 45° = ?

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

If (tan θ + cot θ) = 5 then (tan² θ + cot² θ) = ?

(a) 27

(b) 25

(c) 24

(d) 23Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

If sin A + sin² A = 1 then cos² A + cos⁴ A = ?

(a) 

(b) 1

(c) 2

(d) 3Solution 47

Question 48

If cos A + cos² A = 1 then sin² A + sin⁴ A = ?

(a) 1

(b) 2

(c) 4

(d) 3Solution 48

Question 49

(a) sec A + tan A

(b) sec A – tan A

(c) sec A tan A

(d) none of theseSolution 49

Question 50

(a) cosec A – cot A

(b) cosec A + cot A

(c) cosec A cot A

(d) none of theseSolution 50

Question 51

Solution 51

Question 52

(cosec θ – cot θ)² = ?

Solution 52

Question 53

(sec A + tan A)(1 – sin A) = ?

(a) sin A

(b) cos A

(c) sec A

(d) cosec ASolution 53

Exercise FA

Question 1

Solution 1

Question 2

Solution 2

Question 3

If cos A + cos² A = 1 then (sin² A + sin⁴ A) = ?

(a) 

(b) 2

(c) 1

(d) 4Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.Solution 14

Question 15

If x = a sin θ + b cos θ and y = a cos θ – b sin θ, prove that x² + y² = a² + b².Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

If sec 5A = cosec (A – 36°) and 5A is an acute angle, show that A = 21° Solution 20

Exercise Ex. 7

Question 1

Without using trigonometric tables, evaluate:

Solution 1

(i)

(ii)              

(iii)

(iv)

(v)

(vi)

Question 2

Without using trigonometric tables, prove that:

(i) 

(ii) 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)Solution 2

(i)                LHS = cos81° – sin9°

= cos(90° -9°)- sin9° = sin9° – sin9°

= 0 = RHS

(ii)              LHS = tan71° – cot19°

=tan(90° – 19°) – cot19° =cot19° – cot19°

=0 = RHS

(iii)               LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10° = 0

= RHS

(iv)               LHS= 

(v)            

(vi)             

(vii)                 

LHS = RHS

(viii) 

(ix)              LHS = (sin65° + cos25°) (sin65° – cos25°)

Question 3(i)

Without using trigonometric tables, prove that:

sin 53^ᵒ cos37^ᵒ + cos53^ᵒ sin37^ᵒ = 1Solution 3(i)

Question 3(ii)

cos 54^ᵒ cos 36^ᵒ – sin 54^ᵒ sin36^ᵒ = 0Solution 3(ii)

Question 3(iii)

sec 70^ᵒ sin 20^ᵒ + cos 20^ᵒ cosec 70^ᵒ = 2Solution 3(iii)

Question 3(iv)

sin 35^ᵒ sin 55^ᵒ – cos 35^ᵒ cos 55^ᵒ = 0Solution 3(iv)

Question 3(v)

(sin 72^ᵒ + cos 18^ᵒ)(sin72^ᵒ – cos18^ᵒ) = 0Solution 3(v)

Question 3(vi)

tan 48^ᵒ tan 23^ᵒ tan 42^ᵒ tan 67^ᵒ = 1Solution 3(vi)

Question 4(i)

Prove that:

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5(i)

Prove that:

sin θ cos (90^ᵒ – θ) + sin (90^ᵒ – θ) cos θ = 1Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

cos 1^ᵒ cos2^ᵒ cos3^ᵒ … cos 180^ᵒ= 0Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 7

Prove that:

(i)

(ii) 

(iii)

(iv)

(v)Solution 7

(i)                  LHS =

= RHS

(ii)                LHS =

= RHS

(iii)                  LHS =

= RHS

(iv)             LHS = cosec(65° + ) – sec(25°-  ) – tan(55° –  ) + cot(35° +  )

= RHS

(v)        LHS =

             = 0 + 1 = 1

             = RHS
Question 8(i)

Express each of the following in terms of T-ratios of angles lying between 0^ᵒ and 45^ᵒ:

sin 67^ᵒ + cos 75^ᵒSolution 8(i)

Question 8(ii)

cot 65^ᵒ + tan 49^ᵒSolution 8(ii)

Question 8(iii)

sec 78^ᵒ + cosec 56^ᵒSolution 8(iii)

Question 8(iv)

cosec 54^ᵒ + sin 72^ᵒSolution 8(iv)

Question 9

If A, B, C are the angles of a triangle ABC, prove that:Solution 9

A + B + C = 180°

So, B + C= 180° – A

Question 10

If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.Solution 10

Question 11

If sec 2A= cosec (A – 42ᵒ), where 2A is an acute angle, find the value of A.Solution 11

Question 12

If sin3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.Solution 12

Question 13

If tan 2A = cot(A – 12^o), where 2A is an acute angle, find the value of A.Solution 13

Question 14

If sec 4A = cosec(A – 15^o), where 4A is an acute angle, find the value of A.Solution 14

Question 15

Prove that:

Solution 15

Exercise 9A

Question 1:

Steps of construction:
Step 1 : Draw a line segment AB = 6.5 cm
Step 2: Draw a ray AX making ∠ BAX.
Step 3: Along AX mark (4+7) = 11 points
A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, A₁₀, A₁₁, such that
AA₁ = A₁A₂
Step 4: Join A₁₁ and B.
Step 5: Through A₄ draw a line parallel to A₁₁ B meeting AB at C.
Therefore, C is the point on AB, which divides AB in the ratio 4 : 7
On measuring,
AC = 2.4 cm
CB = 4.1 cm

Question 2:

Steps of Construction:
Step 1 : Draw a line segment PQ = 5.8 cm
Step 2: Draw a ray PX making an acute angle QPX.
Step 3: Along PX mark (5 + 3) = 8 points
A₁, A₂, A₃, A₄, A₅, A₆, A₇ and A₈ such that
PA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A6 = A₆A7 = A₇A₈
Step 4: Join A₈Q.
Step 5: From A₅ draw A₅C || A₈Q meeting PQ at C.
C is the point on PQ, which divides PQ in the ratio 5 : 3
On measurement,
PC = 3.6 cm, CQ = 2.2 cm

Question 3:

Steps of construction:
Step 1: Draw a line segment BC = 6 cm
Step 2: With B as centre and radius equal to 5 cm draw an arc.
Step 3: With C as centre and radius equal to 7 cm draw another arc cutting the previous arc at A.
Step 4: Join AB and AC. Thus, ∆ABC is obtained.
Step 5: Below BC draw another line BX.
Step 6: Mark 7 points B₁B₂B₃B₄B₅B₆B₇ such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 7: Join  B₇C.
Step 8: from B₅, draw B₅D || B₇C.
Step 9: Draw a line DE through D parallel to CA.
Hence ∆ BDE is the required triangle.

Question 4:

Steps of construction:
Step 1: Draw a line segment QR = 6 cm
Step 2: At Q, draw an angle RQA of 60◦.
Step 3: From QA cut off a segment QP = 5 cm.
Join PR. ∆PQR is the given triangle.
Step 4: Below QR draw another line QX.
Step 5: Along QX cut – off equal distances Q₁Q₂Q₃Q₄Q₅
QQ₁ = Q₁Q2 = Q₂Q₃ = Q3Q₄ = Q₄Q₅
Step 6: Join Q₅R.
Step 7: Through Q₃ draw Q₃S || Q₅R.
Step 8: Through S, draw ST || PR.
∆ TQS is the required triangle.

Question 5:

Steps of construction:
Step 1: Draw a line segment BC = 6 cm
Step 2: Draw a right bisector PQ of BC meeting it at M.
Step 3: From QP cut – off a distance MA = 4 cm
Step 4: Join AB, AC.
∆ ABC is the given triangle.
Step 5: Below BC, draw a line BX.
Step 6: Along BX, cut – off 3 equal distances such that
BR₁ = R₁R₂ = R₂R₃ 
Step 7: Join R₂C.
Step 8: Through R₃ draw a line R₃C₁ || R₂C.
Step 9 : Through C₁ draw line C₁A₁ || CA _.
∆ A₁BC₁ is the required triangle.

Question 6:

Steps of Construction:
Step 1: Draw a line segment BC = 5.4 cm
Step 2. At B, draw ∠ CBM = 45°
Step 3: Now ∠ A = 105°, ∠ B = 45°, ∠ C = 180° – (105°+ 45°) = 30°
At C draw ∠ BCA = 30°.
∆ ABC is the given triangle.
Step 4: Draw a line BX below BC.
Step 5: Cut-off equal distances such that  BR₁ = R₁R₂ = R₂R₃ = R₃R₄
Step 6: Join R₃C.
Step 7: Through R₄, draw a line R₄C₁ || R₃C.
Step 8: Through C₁ draw a line C₁A₁ parallel to CA.
∆ A₁BC₁ is the required triangle.

Question 7:

Steps of Construction:
Step 1: Draw a line segment BC = 4 cm
Step 2: Draw a right- angle CBM at B.
Step 3: Cut-off BA = 3cm from BM.
Step 4: Join AC.
ΔABC is the given triangle.
Step 5: Below BC draw a line BX.
Step 6: Along BX, cut-off 7 equal distances such that
BR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅ = R₅R₆ = R₆R₇
Step 7: Join R₅C.
Step 8: Through R₇ draw a line parallel to R₅C cutting BC produced at C₁
Step 9: Through C₁ draw a line parallel to CA cutting BA at A₁
∆ A₁BC₁ is the required triangle.

Question 8:

Steps of Construction:
Step 1: draw a line segment BC = 5 cm
Step 2: With B as centre and radius 7cm an arc is drawn.
Step 3: With C as centre and radius 6 cm another arc is drawn intersecting the previous arc at A.
Step 4: Join AB and AC.
Step 5: ∆ ABC is the given triangle.
Step 6: Draw a line BX below BC.
Step 7: Cut- off equal distances from DX such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 8: join B₅C.
Step 9: Draw a line through B₇ parallel to B₅C cutting BC produced at C’.
Step 10: Through C’ draw a line parallel to CA, cutting BA produced at A’.
Step 11: ∆ A’BC’ is the required triangle.

Question 9:

Steps of construction:
Step 1: Draw a line segment AB = 6.5 cm
Step 2: With B as centre and some radius draw an arc cutting AB at D.
Step 3: With centre D and same radius draw another arc cutting previous arc at E. ∠ ABE = 60°
Step 4: Join BE and produce it to a point X.
Step 5: With centre B and radius 5.5 cm draw an arc intersecting BX at C.
Step 6: Join AC.
∆ ABC is the required triangle.
Step 7: Draw a line AP below AB.
Step 8: Cut- off 3 equal distances such that
AA₁ = A₁A₂ = A₂A₃
Step 9: Join BA₂
Step 10: Draw A₃B’ through A₃ parallel to A₃B.
Step 11: Draw a line parallel to BC through B’ intersecting AY at C’.
∆ AB’C’ is the required triangle.

Question 10:

Steps of construction:
Step 1: Draw a line segment BC = 6.5 cm
Step 2: Draw an angle of 60° at B so that ∠ XBC = 60°.
Step 3: With centre B and radius 4.5cm, draw an arc intersecting XB at A.
Step 4: Join AC.
∆ ABC is the required triangle.
Step 5: Draw a line BY below BC.
Step 6: Cut- off 4 equal distances from BY.
Such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 7: Join CB₄
Step 8: draw B₃C’ parallel to CB₄
Step 9: Draw C’A’ parallel to CA through C’ intersecting BA produced at A’.
∆ A’BC’ is the required similar triangle.

Question 11:

Steps of Construction:
Step 1: Draw a line segment BC = 9 cm
Step 2: with centre B and radius more than 1/2 BC, draw arcs on both sides of BC.
Step 3: With centre C and same radius draw other arcs on both sides of BC intersecting previous arcs at P and Q.
Step 4: join PQ and produce it to a point X. PQ meets BC at M.
Step 5: With centre M and radius 5 cm, draw an arc intersecting MX at A.
Step 6: Join AB and AC.
∆ ABC is the required triangle.
Step 7: Draw a line BY below BC.
Step 8: Cut off 4 equal distances from BY so that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 9: Join CB₄
Step 10: Draw C’B₃ parallel to CB₄
Step 11: Draw C’A’ parallel to CA, through C’ intersecting BA at A’.
∆ A’BC’ is the required similar triangle.

Exercise 9B

https://youtube.com/watch?v=HEdUc-q9_es%3Ffeature%3Doembed

Question 1:

Steps of construction:
Step 1: Draw a circle of radius 5 cm with centre O.
Step 2: A point P at a distance of 8cm from O is taken.
Step 3: A right bisector of OP meeting OP at M is drawn.
Step 4: With centre M radius OM a circle is drawn intersecting the previous circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ and the required tangents. Measuring PT₁ and PT₂
We find, PT₁= PT₂ =6.2 cm

Question 2:

Steps of construction:
Step 1: Two concentric circles with centre O and radii 4 cm and 6 cm are drawn.
Step 2: A point P is taken on outer circle and O, P are joined.
Step 3: A right bisector of OP is drawn bisecting OP at M.
Step 4: With centre M and radius OM a circle is drawn cutting the inner circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ are the required tangents. Further PT₁= PT₂ =4.8 cm

Question 3:

Steps of construction:
Step 1: Draw a circle with centre O and radius 3.5 cm
Step 2: the diameter P₁P₂ is extended to the points A and B such that AO = OB = 7 cm
Step 3: With centre P₁ and radius 3.5 cm draw a circle cutting the first circle at T₁ and T₂
Step 4: join AT₁ and AT₂
Step 5: With centre P₂ and radius 3.5 cm draw another circle cutting the first circle at T₃ and T₄
Step 6: Join BT₃ and BT₄ . Thus AT₁, AT₂ and BT₃, BT₄ are the required tangents to the given circle from A and B.

Question 4:

Steps of construction:
(i) A circle of radius 4.2 cm at centre O is drawn.
(ii) A diameter AB is drawn.
(iii) With OB as base, an angle BOC of 45° is drawn.
(iv) At A, a line perpendicular to OA is drawn.
(v) At C, a line perpendicular to OC is drawn.
(vi) These lines intersect each other at P.
PA and PC are the required tangents.

Question 5:

Steps of construction:
(i) A line segment AB = 8,5 cm is drawn.
(ii) Draw a right bisector of AB which meets AB at M.
(iii) With M as centre AM as radius a circle is drawn intersecting the given circles at T₁, T₂, T₃ and T₄
(iv) Join AT₃, AT₄ and BT₁, BT₂.
Thus AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 6:

Steps of construction:
(i) Draw a line segment AB = 7cm
(ii) Taking A as centre and radius 3 cm, a circle is drawn.
(iii) With centre B and radius 2.5 cm, another circle is drawn.
(iv) With centre A and radius more than 1/2 AB, arcs are drawn on both sides of AB.
(v) With centre B and the same radius, [as in step (iv)] arcs are drawn on both sides of AB intersecting previous arcs at P and Q.
(vi) Join PQ which meets AB at M.
(vii) With centre M and radius AM, a circle is drawn which intersects circle with centre A at T₁ and T₂ and the circle with centre B at T₃ and T₄
(viii) Join AT₃, AT₄, BT₁ and BT₂
Thus, AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 7:

Steps of construction:
(i) A circle of radius 3 cm with centre O is drawn.
(ii) A radius OC is drawn making an angle of 60° with the diameter AB.
(iii) At C, ∠OCP = 90° is drawn.
CP is required tangent.

Question 8:

Steps of construction :
(i) Draw a circle of radius 4 cm with centre O.
(ii) With diameter AB, a line OC is drawn making an angle of 30° i.e., ∠BOC = 30°
(iii) At C a perpendicular to OC is drawn meeting OB at P.
PC is the required tangent.

Practice Test – MCQs test series for Term 2 ExamsENROLL NOW×Contact UsOrRequest a callAsk a DoubtLog InSign Up

Exercise Ex. 8A

Question 1

Find the length of tangent drawn to a circle with radius 8 cm from a point 17 cm away from the centre of the circle.Solution 1

PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.

In PAO, A = 90

By Pythagoras theorem:

Hence, the length of the tangent = 15 cm.Question 2

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Findthe radius of the circle.Solution 2

PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm

In PAO, A = 90,

By Pythagoras theorem:

Hence, the radius of the circle is 7 cm.Question 3

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution 3

Question 4

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.

Solution 4

Question 5

In the given figure, PA and PB are the tangent segments to a circle with centre O. Show that the points A, O, B and P are concyclic.

Solution 5

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.

OAP = 90

And OBP = 90

So, OAP = OBP = 90

OBP + OAP = (90 + 90) = 180

Thus, the sum of opposite angles of quad. AOBP is 180

AOBP is a cyclic quadrilateralQuestion 6

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Solution 6

Question 7

From an external point P, tangents PA and PB are drawn to a circle with centre O. if CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of PCD.

Solution 7

Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.

Since the tangents from an external point are equal, we have

PA = PB,

Also, CA = CE and DB = DE

Perimeter of PCD = PC + CD + PD

=(PA – CA) + (CE + DE) +(PB – DB)

= (PA – CE) + (CE + DE) + (PB – DE)

= (PA + PB) = 2PA = (2 14) cm

= 28 cm

Hence, Perimeter of PCD = 28 cmQuestion 8

A circle is inscribed in a ABC, touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7cm and CR = 5 cm, find the length of BC.

Solution 8

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.

Also, AB = 10 cm, AR = 7cm, CR = 5cm

AR, AP are the tangents to the circle

AP = AR = 7cm

AB = 10 cm

BP = AB – AP = (10 – 7)= 3 cm

Also, BP and BQ are tangents to the circle

BP = BQ = 3 cm

Further, CQ and CR are tangents to the circle

CQ = CR = 5cm

BC = BQ + CQ = (3 + 5) cm = 8 cm

Hence, BC = 8 cm
Question 9

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm and CD = 4 cm. Find AD.

Solution 9

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively

We know that the length of tangents drawn from an exterior point to a circle are equal

AP = AS —-(1)    {tangents from A}

BP = BQ —(2)     {tangents from B}

CR = CQ —(3)    {tangents from C}

DR = DS—-(4)    {tangents from D}

Adding (1), (2) and (3) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AD = (AB + CD) – BC = {(6 + 4) – 7} cm = 3 cm

Hence, AD = 3 cm
Question 10

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Solution 10

Question 11

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB up to one place of decimal

Solution 11

Given O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10cm. Join OA, OB and OP.

Then, OB = 4 cm, OA= 6 cm and PA = 10 cm

In triangle OAP,

Hence, BP = 10.9 cm
Question 12

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm² then find the lengths of sides AB and AC.

Solution 12

Question 13

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Solution 13

Question 14

Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.Solution 14

Question 15

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.

Solution 15

Question 16

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA: AT = 2:1.

Solution 16

Exercise Ex. 8B

Question 1

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD.

Solution 1

Question 2

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50° then what is the measure of ∠OAB.

Solution 2

Question 3

In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.

Solution 3

Question 4

In the given figure, common tangents AB and CD to the two circles with centres O₁ and O₂ intersect at E. Prove that AB = CD.

Solution 4

Question 5

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.

Solution 5

Question 6

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ∆ABC = 21 cm² then find the lengths of sides AB and AC.

Solution 6

Question 7

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.Solution 7

Question 8

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.Solution 8

Question 9

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ =120°, then prove that OR = PR + RQ.

Solution 9

Question 10

In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.

Solution 10

Question 11

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.

Solution 11

Question 12

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.Solution 12

Question 13

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.

Solution 13

Question 14

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then find the measure of ∠OAB.

Solution 14

Exercise MCQ

Question 1

The number of tangents that can be drawn from an external circle is

(a) 1

(b) 2

(c) 3

(d) 4Solution 1

Correct option : (b)

We can draw only 2 tangents from an external point to a circle.Question 2

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to

(a) 2.5 cm

(b) 3 cm

(c) 5 cm

(d) 8 cmSolution 2

Question 3

In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If 0 is the centre of the circle, then length OP = ?

(a) 30 cm

(b) 28 cm

(c) 25 cm

(d) 18 cmSolution 3

Question 4

Which of the following pairs of lines in a circle cannot be parallel?

(a) two chords

(b) a chord and a tangent

(c) two tangents

(d) two diametersSolution 4

Correct option: (d)

The diameter of the circle always passes through the centre. This means all the diameters of a given circle will intersect at the centre, and hence they cannot be parallel.Question 5

The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is

Solution 5

Question 6

In the given figure, PT is a tangent to the circle with centre O. If OT = 6 cm and OP =10 cm, then the length of tangent PT is

(a) 8 cm

(b) 10 cm

(c) 12 cm

(d) 16 cmSolution 6

Question 7

In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then, the radius of the circle is

(a) 10 cm

(b) 12 cm

(c) 13 cm

(d) 15 cmSolution 7

Question 8

PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°Solution 8

Question 9

In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°.

Then, ∠BOC is equal to

(a) 80°

(b) 100°

(c) 120°

(d) 140°

Solution 9

Question 10

If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is

(a) 30° 

(b) 60° 

(c) 90° 

(d) 120° Solution 10

Question 11

In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of chord AB is

(a) 8 cm

(b) 14 cm

(c) 16 cm

(d) 136 cm

Solution 11

Question 12

In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA =17 cm, then the length of AC (in cm) is

(a) 9

(b) 15

(c)  

(d) 25

Solution 12

Question 13

In the given figure, O is the centre of a circle, AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A then ∠BAT=?

(a) 40°

(b) 50°

(c) 60°

(d) 65° 

Solution 13

Question 14

In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then ∠TPQ is equal to

(a) 35°

(b) 45°

(c) 55°

(d) 70°

Solution 14

Question 15

In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then, AT= ?

(a) 4 cm

(b) 2 cm

(c)  

(d) 

Solution 15

Question 16

If PA and PB are two tangents to a circle with centre O such that ∠AOB =110° then ∠APB is equal to

(a) 55°

(b) 60°

(c) 70°

(d) 90°

Solution 16

Question 17

In the given figure, the length of BC is

(a) 7 cm

(b) 10 cm

(c) 14 cm

(d) 15 cm

Solution 17

Question 18

In the given figure, if ∠AOD = 135° then ∠BOC is equal to

(a) 25°

(b) 45°

(c) 52.5°

(d) 62.5°

Solution 18

Question 19

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord such that ∠QPT = 50° then ∠POQ = ?

(a) 100°

(b) 90°

(c) 80°

(d) 75° 

Solution 19

Question 20

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then ∠OAB is

(a) 15°

(b) 30°

(c) 60°

(d) 90° 

Solution 20

Question 21

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm then the length of each tangent is

Solution 21

Question 22

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27° then ∠QAR equals

(a) 63°

(b) 117°

(c) 126°

(d) 153° 

Solution 22

Question 23

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Solution 23

Question 24

If PA and PB are two tangents to a circle with centre O such that ∠APB = 80°. Then, ∠AOP=?

(a) 40°

(b) 50°

(c) 60°

(d) 70° 

Solution 24

Question 25

In the given figure, Q is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58° then the measure of ∠PQB is

(a) 32°

(b) 58°

(c) 122°

(d) 132° 

Solution 25

Question 26

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30° then ∠CPB + ∠ACP is equal to

(a) 60°

(b) 90°

(c) 120°

(d) 150° 

Solution 26

Question 27

In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If ∠PAB = 67°, then the measure of ∠AQB is

(a) 73°

(b) 64°

(c) 53°

(d) 44° 

Solution 27

Question 28

In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is

(a) 45°

(b) 60°

(c) 90°

(d) 120° 

Solution 28

Question 29

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quad. PQOR is

(a) 60 cm²

(b) 32.5 cm²

(c) 65 cm²

(d) 30 cm²

Solution 29

Question 30

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR such that ∠BQR = 70°. Then, ∠AQB = ?

(a) 20°

(b) 35°

(c) 40°

(d) 45°

Solution 30

Question 31

The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

Solution 31

Question 32

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30° then ∠PTA =?

(a) 60°

(b) 30°

(c) 15°

(d) 45°

Solution 32

Question 33

In the given figure, a circle touches the side DF of ∆EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of ∆EDF is

(a) 9 cm

(b) 12 cm

(c) 13.5 cm

(d) 18 cm

Solution 33

Question 34

To draw a pair of tangents to a circle, which are inclined to each other at an angle of 45°, we have to draw tangents at the end points of those two radii, the angle between which is

(a) 105°

(b) 135°

(c) 140°

(d) 145°Solution 34

Question 35

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, ∠QSR =?

(a) 40°

(b) 50°

(c) 60°

(d) 70°

Solution 35

Question 36

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of ∆PQR = 189 cm² then the length of side PQ is

(a) 17.5 cm

(b) 20 cm

(c) 22.5 cm

(d) 25 cm

Solution 36

Question 37

In the given figure, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is (a) 1.9 cm

(b) 3.8 cm

(c) 5.7 cm

(d) 7.6 cm

Solution 37

Question 38

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC =7 cm and CS = 3 cm. Then, the length AB =?

(a) 9 cm

(b) 10 cm

(c) 12 cm

(d) 8 cm

Solution 38

Question 39

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm then perimeter of quad. ABCD is

(a) 18 cm

(b) 27 cm

(c) 36 cm

(d) 32 cm

Solution 39

Question 40

In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB =100° then ∠BAT is equal to

(a) 40°

(b) 50°

(c) 90°

(d) 100°

Solution 40

Question 41

In a right triangle ABC, right-angled at B, BC = 12 cm and AB =5 cm. The radius of the circle inscribed in the triangle is

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cmSolution 41

Question 42

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is

(a) 11 cm

(b) 15 cm

(c) 20 cm

(d) 21 cm

Solution 42

Question 43

In the given figure, ∆ABC is right-angled at B such that BC= 6 cm and AB = 8 cm. A circle with centre O has been inscribed inside the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC. If OP = OQ =OR= x cm then x = ?

(a) 2 cm

(b) 2.5 cm

(c) 3 cm

(d) 3.5 cm

Solution 43

Question 44

Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC =7 cm, and CD = 4 cm then the length of AD is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 7 cmSolution 44

Question 45

In the given figure, PA and PB are tangents to the given circle such that PA = 5 cm and ∠APB = 60°. The length of chord AB is

Solution 45

Question 46

In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF then the radius of the circle is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Solution 46

Question 47

In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm then the radius of the circle with centre A is

(a) 1.5 cm

(b) 2 cm

(c) 2.5 cm

(d) 3 cm

Solution 47

Question 48

In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is

(a) 15 cm

(b) 10 cm

(c) 9 cm

(d) 7.5 cm

Solution 48

Question 49

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA =12 cm then PB is equal to

Solution 49

Question 50

Which of the following statements is not true?

a. If a point P lies inside a circle, no tangent can be drawn to the circle, passing through P.

b. If a point P lies on the circle, then one and only one tangent can be drawn to the circle at P.

c. If a point P lies outside the circle, then only two tangents can be drawn to the circle from P.

d. A circle can have more than two parallel tangents, parallel to a given line.Solution 50

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since we can draw only parallel tangents on either side of the diameter, which would be parallel to a given line.Question 51

Which of the following statements is not true?

1. A tangent to a circle intersects the circle exactly at one point.
2. The point common to the circle and its tangent is called the point of contact.
3. The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
4. A straight line can meet a circle at one point only.

Solution 51

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since a straight line can meet a circle at two points even as shown below.

Question 52

Which of the following statements is not true?

1. A line which intersects a circle in two points, is called a secant of the circle.
2. A line intersecting a circle at one point only, is called a tangent to the circle.
3. The point at which a line touches the circle, is called the point of contact.
4. A tangent to the circle can be drawn from a point inside the circle.

Solution 52

Correct option: (d)

Options (a), (b) and (c) are true.

However, option (d) is false since it is not possible to draw a tangent from a point inside a circle.Question 53

Assertion-and-Reason

Type Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
At a point P of a circle with centre O and radius 12 cm, a tangent  PQ of length 16 cm is drawn. Then, OQ = 20 cm.	The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The correct answer is (a)/(b)/(c)/(d).Solution 53

Question 54

Assertion (A)	Reason (R)
If two tangents are drawn to a circle from an external point then they subtend equal angles at the centre.	A parallelogram circumscribing a circle is a rhombus.

The correct answer is (a)/(b)/(c)/(d).Solution 54

Question 55

Assertion (A)	Reason (R)
In the given figure, a quad. ABCD is drawn to circumscribe a given circle, as shown.Then, AB + BC = AD + DC	In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

The correct answer is (a) / (b) / (c) / (d).Solution 55

Exercise FA

Question 1

In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?

(a) 130°

(b) 100°

(c) 90°

(d) 75°

Solution 1

Question 2

If the angle between two radii of a circle is 130° then the angle between the tangents at the ends of the radii is

(a) 65°

(b) 40°

(c) 50°

(d) 90°Solution 2

Question 3

If tangents PA and PB from a point P to a circle with centre O are drawn so that ∠APS = 80° then ∠POA = ?

(a) 40°

(b) 50°

(c) 80°

(d) 60°

Solution 3

Question 4

In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5 cm then perimeter of  ∆ABC is

(a) 15 cm

(b) 10 cm

(c) 22.5 cm

(d) 20 cm

Solution 4

Question 5

In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, find x.

Solution 5

Question 6

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Solution 6

Question 7

In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If ∠PBA = 65°, find ∠OAB and ∠APB.

Solution 7

Question 8

Two tangent segments BC and BD are drawn to a circle with centre O such that ∠CBD = 120°. Prove that OB = 2BC.

Solution 8

Question 9

Fill in the blanks.

i. A line intersecting a circle in two distinct points is called a ______ .

ii. A circle can have ________  parallel tangents at the most.

iii. The common point of a tangent to a circle and the circle is called the ________ .

iv. A circle can have _________ tangents.Solution 9

1. A line intersecting a circle in two distinct points is called a secant.
2. A circle can have two parallel tangents at the most.
3. This is since we can draw only parallel tangents on either side of a diameter.
4. The common point of a tangent to a circle and the circle is called the point of contact.
5. A circle can have infinitely many tangents.

Question 10

Prove that the lengths of two tangents drawn from an external point to a circle are equal.Solution 10

Question 11

Prove that the tangents drawn at the ends of the diameter of a circle are parallel.Solution 11

Question 12

In the given figure, if AB = AC, prove that BE = CE.

Solution 12

Question 13

If two tangents are drawn to a circle from an external point, show that they subtend equal angles at the centre.Solution 13

Question 14

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.Solution 14

Question 15

Prove that the parallelogram circumscribing a circle, is a rhombus.Solution 15

Question 16

Two concentric circles are of radii 5 cm and 3 cm respectively. Find the length of the chord of the larger circle which touches the smaller circle.Solution 16

Question 17

A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal.Solution 17

Question 18

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.Solution 18

Question 19

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.Solution 19

Question 20

PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.Solution 20