Author Archives: Swati

Chapter 14 – Co-ordinate Geometry Exercise Ex. 14.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Chapter 14 – Co-ordinate Geometry Exercise Ex. 14.2

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 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Show that the points A(1,-2), B(3,6), C(5,10) and D(3,2) are the vertices of a parallelogram.Solution 7

Question 8

Prove that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of triangle ABC right angled at B. Find the values of a and hence the area of ⧍ABC.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

Find a point which is equidistant from the points

A (-5, 4) and B (-1, 6). How many such points are there?Solution 21

Question 22

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter units.Solution 22

Question 23

Ayush starts walking from his house to office, Instead of going to the office directly, he goes to bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8) school at (13, 14) and office at (13, 26) and coordinates are in kilometer.Solution 23

Question 24

Solution 24

Question 25

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior (ii) exterior of the triangle.Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29(i)

Show that the points A(5,6), B(1,5), C(2,1) and D(6,2) are the vertices of a square.Solution 29(i)

Question 29(ii)

Prove that the points A(2, 3), B(-2, 2), C(-1, -2), and D (3, -1) are the vertices of a square ABCD.Solution 29(ii)

Question 29(iii)

Solution 29(iii)

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

If the point P(x,y) is equidistant from the points A(5,1) and B(1,5), prove that x = y.Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

If the point P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.Solution 36

Question 37

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.Solution 37

Question 38

Solution 38

Question 39

Find the equation of the perpendicular bisector of the line segment joining points (7,1) and (3,5).Solution 39

Question 40

Prove that the points (3,0), (4,5), (-1,4) and (-2,-1), taken in order, form a rhombus. Also, find its area.Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

If a point A(0,2) is equidistant from the points B(3,p) and C(p,5), then find the value of p.Solution 44

Question 45

Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.Solution 45

Question 46

If the point P(x, 3) is equidistant from the points A(7, -1) and B(6, 8), find the value of x and the find the distance AP.Solution 46

Question 47

If A(3, y) is equidistant from the points P(8, – 3) and Q(7, 6), find the value of y and find the distance AQ.Solution 47

Question 48

If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.Solution 48

For an equilateral triangle, the perpendicular bisector of any side passes through the opposite vertex.

Both the points, (0, -3) and (0, 3), lie on the y-axis equidistant from the origin. Hence, the perpendicular bisector joining these two points is the x-axis.

Any point on the x-axis has the coordinates (a, 0).

The distance between (0, -3) and (0, 3) is 6.

Hence, the distance between (a, 0) and (0, 3) should also be 6.

6² = (a – 0)² + (0 – 3)²

36 = a² + 9

a² = 27

Question 49

If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also, find the length of AP.Solution 49

Question 50

Show that ∆ABC, where A (-2, 0), B (2, 0) C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.Solution 50

Using the distance formula,

AB=

AC=

BC=

PQ=

PR=

QR=

Now,

ΔABC ~ ΔPQR

by the SSS test.Question 51

Solution 51

Question 52

Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6).Solution 52

Question 53

Solution 53

Question 54

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Chapter 14 – Co-ordinate Geometry Exercise Ex. 14.3

Question 1

Solution 1

Question 2

Find the points of trisection of the line segment joining the points:

(i) (5, -6) and (-7, 5), (ii) (3, -2) and (-3,-4), (iii) (2,-2) and (-7,4).Solution 2

(i)



(ii)



(iii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

If P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.Solution 5

Question 6

If (a, b) is the mid-point of the line segment joining the points A (10, -6), B(k, 4) and a – 2b = 18, find the value of k and the distance AB.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If the points P,Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B (7, 10) in 5 equal parts, find x, y and p.Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

  *Note: (i) Answer given in the book is incorrect.Question 15

Solution 15

Question 16

Prove that (4,3), (6,4), (5,6) and (3,5) are the angular points of a square.Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Find the ratio in which the line segment joining the points A (3, -3) and B (-2, 7) is divided by x- axis. Also, find the coordinates of the point of division.Solution 19

Let the point on the x-axis be (a, 0).

Let this point divide the line segment AB in the ratio of r : 1.

Using the section formula for the y-coordinate, we get

Question 20

Find the ratio in which the point P(x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.Solution 20

Question 21

Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also find the value of y.Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence, find m.Solution 29

Let P divides the line segment AB is the ratio k: 1.

So, the ratio is 1:1.

Also, 

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Show that the points A(1,0) B(5,3), C(2,7) and D(-2,4) are the vertices of a parallelogram.Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q, and R.Solution 38

The difference between the x-coordinates of A and B is 6 – 1 = 5

Similarly, the difference between the y-coordinates of A and B is 7 – 2 = 5

Hence, if the line segment joining A(1, 2) and B(6, 7) is divided into 5 equal parts by the points P, Q, R and S, then the coordinates of P, Q, R and S can be found out by increasing the x and the y coordinates of A by 1 successively.

Hence, the coordinates of P are (1 + 1, 2 + 1) = (2, 3)

The coordinates of Q are (2 + 1, 3 + 1) = (3, 4)

The coordinates of R are (3 + 1, 4 + 1) = (4, 5)Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

If two vertices of a parallelogram are (3,2), (-1,0) and the diagonals cut at (2,-5), find the other vertices of the parallelogram.Solution 46

Question 47

If the coordinates of the mid-points of the sides of a triangle are (3,4), (4,6), and (5,7), find its vertices.Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50 (i)

Solution 50 (i)

Question 50 (ii)

Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.Solution 50 (ii)

Given: ABCD is a parallelogram

We know that, diagonals of a parallelogram bisect each other.

Therefore, midpoints of diagonals coincide

The midpoints of AC and BD coincide.

Question 51

Solution 51

Question 52 (i)

Points P and Q trisect the line segment joining the points A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of P and Q.Solution 52 (i)

As P and Q trisect AB and P is near to A.

Therefore, P divides AB in the ratio 1:2.

Also, Q divides AB in the ration 2:1.

Question 52 (ii)

Solution 52 (ii)

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

A point P divides the line segment joining the points A(3,-5) and B(-4,8), such that  . If P lies on the line x + y = 0, then find the value of k.Solution 56

Given points are A(3,-5) and B(-4,8).

P divides AB in the ratio k : 1.

Using the section formula, we have:

Coordinate of point P are {(-4k+3/k+1)(8k-5/k+1)}

Now it is given, that P lies on the line x+y = 0

Therefore,

-4k+3/k+1 + 8k-5/k+1 =0

=> -4k+3+8k-5 =0

=> 4k -2 =0

=> k=2/4

=> k=1/2

Thus, the value of k is 1/2.Question 57

The mid – point P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the points C (-9, -4) and D (-4, y). Find the ratio in which P divides CD. Also, find the value of y.Solution 57

Question 58

If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio 3 : 4, find the value of x² + y².Solution 58

Question 59

ABCD is a parallelogram with vertices A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃). Find the coordinates of the fourth vertex D in terms of x₁, x₂, x₃, y₁, y₂ and y₃.Solution 59

Question 60

The points A (x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are the vertices of ⧍ABC.

 i. The median from A meets BC at D. Find the coordinates of the point D.

 ii. Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

 iii. Find the points of coordinates Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

 iv. What are the coordinates of the centroid of the triangle ABCSolution 60

Chapter 14 – Co-ordinate Geometry Exercise Ex. 14.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 14 – Co-ordinate Geometry Exercise Ex. 14.5

Question 1

Solution 1

Question 1 (iv)

Find the area of a triangle whose vertices are

(1, -1), (-4, 6) and (-3, -5)Solution 1 (iv)

Area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 24 sq. units.Question 1 (v)

Find the area of a triangle whose vertices are

(-5, 7), (-4, -5) and (4, 5)Solution 1 (v)

Area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 53 sq. units.Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Find the area of the quadrilaterals, the coordinates of whose vertices are

(-4,-2), (-3,-5), (3,-2), (2,3)Solution 2(iii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4).Solution 6

Question 7

In ⧍ABC, the coordinates of vertex A are (0, -1) and D(1, 0) and E(0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of ⧍DEF.Solution 7

Question 8

Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).Solution 8

Question 9

If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.Solution 9

Question 10

If A (-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13 (i)

If the vertices of a triangle are (1,-3), (4,p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.Solution 13 (i)

Question 13 (ii)

Find the value of k so that the area of triangle ABC with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.Solution 13 (ii)

Area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

Hence, the value of k is 3.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

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(-2, 6) and C (3, 1) is 10 sequare units.Solution 20

Question 21

If a ≠ b ≠ 0, prove that the points (a, a²), (b, b²), (0, 0) are never collinear.Solution 21

Let the points (a, a²), (b, b²), (0, 0) represent a triangle. If we can prove that the area of the triangle so formed is not equal to zero, then we can prove that the points (a, a²), (b, b²), (0, 0) are never collinear.

Area of a triangle is given by

Now b≠a≠0.

So,

b – a≠0,

Δ≠0

Thus, points (a, a²), (b, b²), (0, 0) are never collinear.Question 22

The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (7/2, y), find y.Solution 22

Area of a triangle is given by Question 23

Solution 23

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

Find the value (s) of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k – 2) are collinear.Solution 30

Let the points be A, B and C respectively.

If A, B and C are collinear, then the area of ∆ABC is zero.

Question 31

If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c.Solution 31

Question 32

If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a – b = 1, find the values of a and b.Solution 32

Question 33

If the points A(1, -2), B(2, 3), C(a, 2) and D(-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.Solution 33

Question 34

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ⧍ADE.Solution 34

Question 35

If D(-1/5, 5/2), E(7, 3) and F(7/2, 7/2) are the mid-points of sides of ⧍ABC, find the area of ⧍ABCSolution 35

Chapter 14 – Co-ordinate Geometry Exercise 14.63

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

A line Segement is of length 10 units. If the coordinates of its one end are (2,3) and the abscissa of the other end is 10, then its ordinate is

(a) 9,6

(b) 3,-9

(c) -3,9

(d) 9,-6Solution 6

Question 7

Solution 7

So, the correct option is (d).Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 14 – Co-ordinate Geometry Exercise 14.64

Question 11

Solution 11

Question 12

If (-1,2),(2,-1) and (3,1) are any three vertices of a parallelogram, then

(a) a = 2, b = 0

(b) a = -2, b = 0

(c) a = -2, b = 6

(d) a = 6, b = 2Solution 12

Note: The answer does not match the options.Question 13

If A (5,3), B(11,-5) and P(12,y) are the vertices of a right triangle right angled at P, then y = 

(a) – 2,4

(b) -2 ,4

(c) 2, -4

(d) 2,4Solution 13

Question 14

The area of the triangle formed by (a,b+c), (b,c+a) and (c,a+b) is 

(a)  a+b+c

(b) abc

(c) (a+b+c)²

(d) 0Solution 14

Question 15

If (x,2),(-3,-4) and (7,-5) are collinear, then x = 

(a) 60

(b) 63

(c) -63

(d) -60Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

The line segment joining points (-3,4), and (1,-2) is divided by y – axis in the ratio

(a) 1 : 3

(b) 2 : 3

(c) 3 : 1

(d) 2: 3Solution 20

Question 21

The ratio in which (4,5) divides the join of (2,3) and (7,8) is 

(a) -2 : 3

(b) -3 : 2 

(c) 3 : 2

(d) 2 : 3Solution 21

Question 22

The ratio in which the x-axis divides the segment joining (3,6) and (12,-3) is

(a) 2:1

(b) 1 :2

(c) -2 : 1

(d) 1 : -2 Solution 22

Question 23

If the centrroid  of the triangle formed by the points (a,b),(b,c) and (c,a) is at the origin, then a³ + b³ + c³ = 

(a) abc

(b) 0

(c) a+b+c

(d) 3abcSolution 23

Question 25

If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =

(a) (4, 5)

(b) (5, 4)

(c) (-5, -2)

(d) (5, 2)Solution 25

Question 24

If points A(3, 1), B(5, p) and C(7, -5) are collinear, then p =

(a) -2

(b) 2

(c) -1

(d) 1Solution 24

As the points A, B and C are collinear, then the area formed by these three points is 0.

Area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

Hence, the value of p is -2.

Chapter 14 – Co-ordinate Geometry Exercise 14.65

Question 26

The distance of the point (4, 7) from the x-axis is

(a) 4

(b) 7

(c) 11

(d) Solution 26

We know, distance of point (x, y) from x-axis is y.

Hence, distance of point (4, 7) from x-axis is 7.

Hence, correct option is (b).Question 27

The distance of the point (4, 7) from the y-axis is

(a) 4

(b) 7

(c) 11

(d) Solution 27

We know distance of point (x, y) from y-axis is x.

Hence distance of point (4, 7) from y-axis is 4.

Hence, correct option is (a).Question 28

If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are

(a) (0, 3)

(b) (3, 0)

(c) (0, 0)

(d) (0, -3)Solution 28

Given P is a point on x-axis

Hence P = (x, 0)

Distance from the origin is 3

Hence P = (3, 0)

Given Q is a point on y-axis

So Q is (0, y)

Given that OP = OQ

implies OQ = 3

Distance of Q from the origin is 3

Hence y = 3

implies Q = (0, 3)

Hence, correct option is (a).Question 29

If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =

(a) ±5

(b) ±3

(c) 0

(d) ±4Solution 29

Question 30

If the point P (x, y) is equidistant from A(5, 1) and B (-1, 5), then

(a) 5x = y

(b) x = 5y

(c) 3x = 2y

(d) 2x = 3ySolution 30

Question 31

If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =

(a) 7

(b) 3

(c) 6

(d) 8Solution 31

Question 32

Solution 32

Question 33

The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are

(a) (0, 2)

(b) (3, 0)

(c) (0, 3)

(d) (2, 0)Solution 33

Question 34

If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, – 1), then k =

(a) 3

(b) 1

(c) 2

(d) 4Solution 34

Question 35

If (-2, 1) is the Centroid of the triangle having its vertices at (x, 2), (10, -2), (-8, y), then x, y satisfy the relation

a. 3x + 8y = 0

b. 3x – 8y = 0

c. 8x + 3y = 0

d. 8x = 3ySolution 35

The Centroid of the triangle is given by

x=-8 and y=3 satisfy(a) 3x + 8y = 0.Question 36

The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are

(a) (3, 0)

(b) (0, 2)

(c) (2, 3)

(d) (3, 2)Solution 36

Question 37

The length of a line segment joining A(2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be

(a) 3 or -9

(b) -3 or 9

(c) 6 or 27

(d) -6 or -27Solution 37

Question 38

The ratio in which the line segment joining P(x₁, y₁) and Q(x₂, y₂) is divided by x-axis is

(a) y₁ : y₂

(b) -y₁ : y₂

(c) x₁ : x₂

(d) -x₁ : x₂Solution 38

Question 39

The ratio in which the line segment joining points A(a₁, b₁) and B(a₂, b₂) is divided by y-axis is

(a) -a₁ : a₂

(b) a₁ : a₂

(c) b₁ : b₂

(d) -b₁ : b₂Solution 39

Chapter 6 – Co-ordinate Geometry Exercise 6.66

Question 40

Solution 40

Question 41

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are

(a) (-6, 7)

(b) (6, -7)

(c) (6, 7)

(d) -6, -7)Solution 41

Question 42

The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are

(a) (2, 4)

(b) (3, 5)

(c) (4, 2)

(d) (5, 3)Solution 42

Question 43

In the figure, the area of ΔABC (in square units) is

(a) 15

(b) 10

(c) 7.5

(d) 2.5Solution 43

Question 44

The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is

(a) (0, 2)

(b) (2, 0)

(c) (3, 0)

(d) (0, 3)Solution 44

Question 45

Solution 45

Question 46

If P(2, 4), Q(0,3), R (3, 6) and S (5, y) are the vertices of a parallelogram PQRS, then the value of y is

(a) 7

(b) 5

(c) -7

(d) -8Solution 46

Question 47

If A(x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is

(a) -63

(b) 63

(c) 60

(d) -60Solution 47

Question 48

Solution 48

Chapter 14 – Co-ordinate Geometry Exercise 14.67

Question 49

If the point P(2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then

Solution 49

Question 50

A line intersects the y-axis and x-axis at P and Q, respectively, If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are respectively

a. (0, -5) and (2, 0)

b. (0, 10) and (-4, 0)

c. (0, 4) and (-10, 0)

d. (0, -10) and (4, 0)Solution 50

Question 51

If the point (k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2, then the value of k is

(a) 1

(b) 2

(c) -2

(d) -1Solution 51

As the point (k, 0) divides the line segment AB in the 1:2

Hence, option (d) is correct.

Chapter 13 – Probability Exercise Ex. 13.1

Question 1

Solution 1

Question 2

A die is thrown. Find the probability of getting:

(i) a prime number

(ii) 2 or 4

(iii) a multiple of 2 or 3

(iv) an even prime number

(v) a number greater than 5

(vi) a number lying between 2 and 6

(vii) A die is thrown. Find the probability of getting: a composite number.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 6(xix)

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is neither a king nor a queen.Solution 6(xix)

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

Solution 14

Question 15

If the probability of winning a game is 0.3, what is the probability of loosing it?Solution 15

Question 16

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) red (ii) black or white (iii) not blackSolution 16

Question 17

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) white (ii) red (iii) not black (iv) red or whiteSolution 17

Question 18

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 18

Question 19(i)

Five cards-ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.

(i) What is the probability that the card is a queen?


Solution 19(i)

Question 19(ii)

Five cards – ten, jack, queen, king, and an ace of diamonds are shuffled face downwards.  One card is picked at random.

If a king is drawn first and put aside, what is the probability that the second card picked up is the

(i)ace? (ii)king?Solution 19(ii)

Question 20

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) black?Solution 20

Question 21

Solution 21

Question 22

In A Class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixed  thoroughly. A child is asked to one card from the basket. What is the probability that the name written on the card is

(i) The name of a girl  

(ii) The name of a boySolution 22

Question 23

Solution 23

A coin has only two options-head and tail and both are equally likely events i.e. the probability of occurrence of both is same. Hence, a coin is a fair option to decide which team will choose ends in the game.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

Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that gets at least one head?Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Fill in the blanks:

(i) Probability of a sure event is ________.

(ii) Probability of an impossible event is _______.

(iii) The probability of an event (other than sure and impossible event) lies between ______.

(iv) Every elementary event associated to a random experiment has _______ probability.

(v) Probability of an event A + Probability of an event ‘not A’ = ___________.

(vi) Sum of the probabilities of each outcome in an experiment is ____________.Solution 41

(i) 1

(ii) 0

(iii) 0 and 1

(iv) equal

(v) 1

(vi) 1Question 42

Solution 42

Question 43

A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card.Solution 43

Question 44

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 drawn card is a prime number.Solution 44

Question 45

A group consists of 12 persons, of which 3 are extremely patient, 6 are extremely honest and the 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 would you prefer more?Solution 45

Question 46

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from this bag. Find the probability that the number on the drawn card is,

(i) Not divisible by 3

(ii) A prime number greater than 7

(iii) Not a perfect square number.Solution 46

Question 47

A piggy bank contains hundred 50 paise coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, find the probability that the coin which falls out

(i) is a 50 paise win

(ii) is of value more than Rs. 1

(iii) is of value less than Rs. 5

(iv) is a Rs. 1 or Rs. 2 coinSolution 47

Question 48

A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the card thoroughly. Find the probability that the number on the drawn card is

(i) an odd number

(ii) a multiple of 5

(iii) a perfect square

(iv) an even prime numberSolution 48

Question 49

A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is

1. divisible 2 or 3
2. a prime number

Solution 49

Question 50

Solution 50

  Question 50(xvi)

In a simultaneous throw of a pair of dice, find the probability that:

2 will come up at least onceSolution 50(xvi)

Question 50(xvii)

In a simultaneous throw of a pair of dice, find the probability that:

2 will not come either timeSolution 50(xvii)

Question 51

What is the probability that an ordinary year has 53 Sundays?Solution 51

Question 52

Solution 52

Question 53

(viii) that the product of numbers appearing on the top of the dice is less than 9.

(ix) that the difference of the numbers appearing on the top of the two dice is 2.

(x) that the numbers obtained have a product less than 16.Solution 53

(x)

When a black die and a white die are thrown at the same time, the sample space is given by

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1)(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

n(S) = 36

Let A be the event that the numbers obtained have a product less than 16.

A = {{(1,1),(1,2),(1,3),(1,4),(1,5),

(1,6),(2,1)(2,2),(2,3),(2,4),(2,5),

(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),

(4,1),(4,2),(4,3), (5,1),(5,2),

(5,3),(6,1),(6,2)}}

n(A) = 25

P(A) =  Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

(iv) card  (v) diamondSolution 61

Question 62

Solution 62

Question 63

A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.Solution 63

Question 64

A dice is rolled twice. Find the probability that

(i) 5 will not come up either time.

(ii) 5 will come up exactly once.Solution 64

Question 65

All the black face cards are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a

(i) Face card

(ii) Red card

(iii) Black card

(iv) KingSolution 65

Question 66

Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn cards is

(i) an odd number

(ii) a perfect square number

(iii) divisible by 5

(iv) a prime number less than 20Solution 66

Question 67

All kings and queens are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is randomly drawn from it. Find the probability that this card is

(i) a red face card

(ii) a black cardSolution 67

Question 68

All jacks, queens and kings are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is randomly drawn from it. Find the probability that this card is

(i) a black face card

(ii) a red cardSolution 68

Question 69

Red queens and black jacks are removed from a pack of 52 playing  is drawn at random from the remaining cards, after reshuffling them. Find the probability that the card drawn is

(i) a king

(ii) of red colour

(iii) a face card

(iv) a queenSolution 69

Question 70

In a bag there are 44 identical cards with figure of circle or square on them. There are 24 circles, of which 9 are blue and rest are green and 20 squares of which 11 are blue and rest are green. One card is drawn from the bag at random. Find the probability that it has the figure

1. square
2. green colour,
3. blue circle and
4. green square.

Solution 70

Question 71

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

1. a red card
2. a face card and
3. a card of clubs.

Solution 71

Question 72

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) different days? (iii) consecutive days?Solution 72

Question 3(i)

Three coins are tossed together. Find the probability of getting :Exactly two headsSolution 3(i)

Sample space when three coins are tossed together is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let A be the event ofgetting exactly two heads.

A = {HHT, THH, HHT}

n (A) = 3

P(A) =  Question 3(ii)

Three coins are tossed together. Find the probability of getting :At most two headsSolution 3(ii)

Sample space when three coins are tossed together is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let A be the event of getting at most two heads.

A = {TTT, HTT, THT, TTH, HHT, THH, HHT}

n (A) = 7

P(A) =  Question 3(iii)

Three coins are tossed together. Find the probability of getting :At least one head and one tail.Solution 3(iii)

Sample space when three coins are tossed together is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let A be the event of getting at least one head and one tail.

A = {HHT, HTH, THH, HTT, THT, TTH }

n (A) = 6

P(A) =  Question 3(iv)

Three coins are tossed together. Find the probability of getting :No tailsSolution 3(iv)

Sample space when three coins are tossed together is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let A be the event ofgetting no tails.

A = {HHH}

n (A) = 1

P(A) =  

Chapter 13 – Probability Exercise Ex. 13.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

A target shown in Fig. 16.11 consists of three concentric circles of radii 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region?

Solution 3

Assume first circle to be the circle with the smallest radius, that is 3. Similarly, second circle to be the circle with radius 7 and third circle to be the circle with radius 9.

Question 4

In Fig. 16.12, points A, B, C and D are the centres of four circles that each have a radius of length one unit. If a point is selected at random from the interior of square ABCD. What is the probability that the point will be chosen from the shaded region?

Solution 4

Question 5

In the fig., JKLM is a square with sides of length 6 units. Points A and B are the mid-points of sides KL and LM respectively. If a point is selected at random from the interior of the square. What is the probability that the point will be chosen from the interior of JAB?
Solution 5

Question 6

In the fig., a square dart board is shown. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square. What is the probability that it will land in the interior of the smaller square?

Solution 6

Chapter 13 – Probability Exercise 13.35

Question 1

If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, is

Solution 1

n(E) = total numbers

       = 9

n(0) = odd numbers {1, 3, 5, 7, 9}

       = 5

So, the correct option is (b).Question 2

In Q. No. 1, the probability that the digit is even, is

Solution 2

n(E) = 9

n(4) = no. is even {2, 4, 6, 8}

       = 4

So, the correct option is (a).

Chapter 13 – Probability Exercise 13.36

Question 3

In Q. No. 1, the probability that the digit is a multiple of 3 is

Solution 3

n(E) = 9

n(A) = no. is multiple of 3 {3, 6, 9}

       = 3

So, the correct option is (a).Question 4

If three coins are tossed simultaneously, then the probability of getting at least two heads, is

Solution 4

3 coins are tossed simultaneously.

Hence sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Event (E) = at least two Heads

              = {HHH, HHT, HTH, THH}

n(s) = 8

n(E) = 4

So, the correct option is (c).Question 5

In a single throw of a die, the probability of getting a multiple of 3 is

Solution 5

sample space (s) = {1, 2, 3, 4, 5, 6}

n(s) = 6

Event (E) = getting a multiple of 3

              = {3, 6}

n(E) = 2

So, the correct option is (b).Question 6

Solution 6

Question 7

A bag contains three green marbles, four blue marbles, and two orange marbles. If a marble is picked at random, then the probability that it is not an orange marble is

Solution 7

Question 8

A number is selected at random from the numbers 3, 5, 5, 7, 7, 7, 9, 9, 9, 9 The probability that the selected number is their average is

Solution 8

Question 9

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

Solution 9

Sample space (S) = {1, 2, 3, 4, 5, 6}

n(S) = 6

Event (E) = getting number greater than 2

              = {3, 4, 5, 6}

n(E) = 4

So, the correct option is (c).Question 10

A card is accidently dropped from a pack of 52 playing cards. The probability that it is an ace is

Solution 10

n(S) = 52

no. of ace in a pack of 52 cards = 4

n(E) = 4

So, the correct option is (b).Question 11

A number is selected from numbers 1 to 25. The probability that it is prime is

Solution 11

n(S) = 25

Event (E) = prime numbers between 1 to 25

              = {2, 3, 5, 7, 11, 13, 17, 19, 23}

n(E) = 9

Note: The answer does not match the options in the question.Question 12

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

Solution 12

We know probability P(E) of an event lies between 0 < P(E) < 1     ……(1)

(a), (c), (d) satisfies the (1) but (b) is a negative number. It can’t be the probability of an event.

So, the correct option is (b).Question 13

If P(E) = 0.05, then P(not E) =

(a) – 0.05

(b) 0.5

(c) 0.9

(d) 0.95Solution 13

We know

P(E) + P(not E) = 1

given P(E) = 0.05

so P(not E) = 1 – 0.05

                 = 0.95

So, the correct option is (d).Question 14

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

(a) 0.2

(b) 0.4

(c) 0.8

(d) 1.6Solution 14

We know 0 < P(E) < 1

(a), (b), (c) fullfill the condition. But (d) doesn’t 

Hence (d) is correct option.

So, the correct option is (d).Question 15

The probability of a certain event is

(a) 0

(b) 1

(c) 1/2

(d) no existentSolution 15

An event that is certain to occur is called Certain event.

Probability of certain event is 1.

Ex: If it is Monday, the probability that tomorrow is Tuesday is certain and therefore probability is 1.

So, the correct option is (b).

Chapter 13 – Probability Exercise 13.37

Question 16

The probability of an impossible event is

(a) 0

(b) 1

(c) 1/2

(d) non-existentSolution 16

Events that are not possible are impossible event.

Probability of impossible event is 0.

So, the correct option is (a).Question 17

Aarushi sold 100 lottery tickets in which 5 tickets carry prizes. If Priya purchased a ticket, what is the probability of Priya winning a prize?

Solution 17

Question 18

A number is selected from first 50 natural numbers. What is the probability that it is a multiple of 3 or 5 ?

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

Solution 27

Chapter 13 – Probability Exercise 13.38

Question 28

Solution 28

Question 29

A number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3 the probability that |x| < 2 is

Solution 29

Sample space (s) = {-3, -2, -1, 0, 1, 2, 3}

n(s) = 7

Event (E) = |x| < 2

             = {-1, 0, 1}

n(E) = 3

So, the correct option is (c).Question 30

If a number x is chosen from the numbers 1, 2, 3, and a number y is selected from the number 1, 4, 9. Then, P(xy < 9)

Solution 30

Question 31

The probability that a non-leap year has 53 Sundays, is

Solution 31

There are 365 days in a non-leap year.

52 complete weeks and 1 spare day.

so This day can be any out of 7 day of week.

Hence n(s) = 7

Now, year already have 52 Sundays. so for a total of 53 Sundays in a calendar year, this spare day must be a Sunday.

Hence n(E) = 1

So, the correct option is (d).Question 32

In a single throw of a pair of dice, the probability of getting the sum a perfect sqaure is

Solution 32

We know on throwing a pair of die there are a total of 36 possible outcomes.

n(S) = sum is a perfect square

        = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}

n(E) = 7

So, the correct option is (b).Question 33

What is the probability that a non-leap year has 53 Sundays ?

Solution 33

There are 365 days in a non-leap year.

52 complete weeks and 1 spare day.

So this day can be any out of 7 day of a week.

Hence n(s) = 7

Now, a non-leap year already has 52 Sundays. So for a total of 53 Sundays in a calendar year, this spare day must be a Sunday.

Hence n(E) = 1

So, the correct option is (b).Question 34

Solution 34

Question 35

Two dice are rolled simultaneously. The probability that they show different faces is

Solution 35

Question 36

What is the probability that a leap year has 52 Mondays?

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Chapter 13 – Probability Exercise 13.39

Question 39

Solution 39

Question 40

Solution 40

Chapter 11 – Constructions Exercise Ex. 11.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Draw a line segment of length 8 cm and divide it internally in the radio 4:5.Solution 4

1. Draw segment AB of length 8 cm.

2. Draw any ray AX making acute angle ∠BAX with AB.

3. Draw BY parallel to AX by making ∠ABY = ∠BAX.

4. Mark four points A₁, A₂,…..,A₄ on AX and five points B₁, B₂,…..,B₅ on BY such that AA₁=A₁A₂=…..= A₃A₄=BB₁=B₁B₂=….=B₄B₅

5. Join A₄B₅; it intersects AB at P. P divides AB in the ratio 4:5. 

Chapter 11 – Constructions Exercise Ex. 11.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Construct a triangle with sides 5 cm, 6cm and 7 cm and then another triangle whose sides are  of the corresponding sides of the first triangle. Solution 5

Steps of construction

(1) Draw a line segment BC with 6 cm.

(2) Taking centres B and C and radii 5 cm and 7 cm respectively, draw two arcs (one from each centre) which intersect each other at A.

(3) Join AB and AC.

(4) At B, draw an angle CBX of any acute measure.

(5) Starting from B, cut 3 equal parts on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅ = X₅X₆ = X₆X₇.

(6) Join X₇C.

(7) Through X₅, draw X₅Q ‖ X₇C.

(8) Through Q, draw QP ‖ CA.

∴ ∆PBQ ∼ ∆ABC.Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Construct a traingle similar to a given XYZ with its sides equal to (3/4)^th of the corresponding sides of XYZ. Write the steps of construction.Solution 11

Question 12

Solution 12

Question 13

Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle.Solution 13

Steps of construction:

1. Draw a line segment AB = 6.5 cm
2. With A as the centre and radius AC = 5.5 cm, draw an arc.
3. With B as the centre and radius BC = 5 cm, draw an arc intersecting the arc drawn in step 2 at C
4. Join AC and BC to obtain ∆ABC.
5. Below AB make an acute ∠BAX.
6. Along AX, mark off five points as the sides of triangle to be 3/5^th of original triangle, the sides to be divided into five equal parts. Mark A₁, A_2, A_3, A_4, A₅ along AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
7. Join A₅B.
8. Consider three parts out of five equal parts on AX. From point A₃, draw A₃B’ ∥ A₅B meeting AB at B’ such that ∠AA₅B = ∠AA₃B’.
9. From B’, draw B’C’ ∥ BC meeting AC at C’ such that  ∠ABC = ∠AB’C’
10. ∆AB’C’ is the required triangle, each of whose sides is 3/5^th of the corresponding sides of ∆ABC

Question 14

Construct a ∆PQR with side QR = 7 cm, PQ = 6 cm and m∠PQR = 60°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ∆PQR.Solution 14

Steps of construction:

1. Draw a line segment PQ = 6 cm
2. To construct ∠ PQR = 60°, taking Q as the centre and with an arbitrary radius draw an arc cutting PQ on S.
3. Taking S as the centre, and the same radius, draw an arc cutting the previous arc at T. Join Q and T, extend QT further.
4. With Q as the centre and a radius of 7 cm, draw an arc on the extended line segment QT at R.
5. Join PR and QR to obtain ∆PQR.
6. Below PQ make an acute angle QPX.
7. Along PX, mark off five points as the sides of triangle to be 3/5^th of original triangle, the sides to be divided into five equal parts. Mark P₁, P_2, P_3, P_4, P₅ along PX, such that PP₁ = P₁P₂ = P₂P₃ =P₃P₄ = P₄P₅.
8. Join P₅Q.
9. Consider three parts out of the five equal parts on PX. From point P₃, draw P₃Q’ ∥ P₅Q meeting PQ at Q’ by making ∠QP₅P = ∠Q’P₃P
10. From Q’, draw Q’R’ ∥ QR by making ∠PQR = ∠ PQ’R’
11. ∆PQ’R’ is the required triangle, each of whose sides is 3/5^th of the corresponding sides of ∆PQR

Question 15

Draw a ∆ ABC in which base BC = 6cm, AB = 5 cm and ∠ABC = 60^o. Then construct another triangle whose sides are of the corresponding sides of ∆ABC.Solution 15

1. Draw ∆ABC with base BC = 6cm, AB = 5 cm and∠ABC = 60^˚.

2. Draw any ray BX making acute angle ∠ABX with BC.

3. Mark four points B₁, B₂,B₃,B₄ on BY such that BB₁=B₁B₂=B₂B₃=B₃B₄.

4. Join CB₄ and draw a line parallel to CB₄, through B₃, to intersect AC at C’.

5. Join AC, and draw a line parallel to AC, through C’, to intersect AB at A’.

6. ∆A’BC’ is the required triangle.

Question 16

Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now, construct another triangle whose sides aretimes the corresponding sides of the given triangle.Solution 16

1. Draw AB = 4cm.

2. Draw RA perpendicular to AB, and mark C on it such that AC = 3 cm.

3. Join BC. DABC is the required triangle.

4. Draw any ray AX making acute angle ∠BAX with AB.

5. Mark five points A₁, A₂,…..,A₅ on AX.

6. Join BA₅, and draw a line parallel to it passing from A₃, intersecting AB at B’.

7. Draw line parallel to CB through B’ to intersect AC at C’.

8. ∆AB’C’ is the required triangle.

Question 17

Construct an equilateral triangle with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆ABC.Solution 17

Steps of construction

(1) Draw a line segment BC with 5 cm.

(2) Taking centres B and C with equal radii 5 cm, draw two arcs (one from each centre) which intersect each other at A.

(3) Join AB and AC.

(4) At B, draw an angle CBX of any acute measure.

(5) Starting from B, cut 3 equal parts on BX such that BX₁ = X₁X₂ = X₂X₃.

(6) Join X₃C.

(7) Through X₂, draw X₂Q ‖ X₃C.

(8) Through Q, draw QP ‖ CA.

∴ ∆PBQ ∼ ∆ABC.

Chapter 11 – Constructions Exercise Ex. 11.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2cm from its centre.Solution 4

Steps of construction:

1. Construct a line segment OP of length = 6.2 cm
2. Taking O as the centre, construct a circle of radius3.5 cm.
3. Taking O and P as the centres, draw arcs of circles above and below OP intersecting each other at points R and S respectively.
4. Draw the perpendicular bisector of OP by joining A and  B. Mark the midpoint of OP as Q.
5. Taking Q as the centre draw a circle of radius OQ or PQ and mark the points of intersection of the two circles as A and B.
6. Join PA and extend it on both the sides.

Join PB and extend it on both the sides

PA and PB are the required tangents.

Question 5

Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.Solution 5

1. Draw a circle having a centre O and a radius of 4.5 cm.
2. Take point P on the circle and join OP.
3. Angle between the tangents = 45°Hence, the angle at the centre= 180° – 45° = 135° (supplement of the angle between the tangents)∴Construct m∠POQ = 135°
4. Keeping a radius of 4.5 cm, draw arcs of circle taking the points P, and Q as the centres.
5. Name the points of intersection of arcs and circle as A and C respectively.
6. Taking A as the centre, and with the same radius mark B such that OA = AB.
7. Similarly, taking C as the centre and with the same radius mark D such that OC = CD.
8. Taking A and B as the centres and the same radius draw two arcs intersecting each other at U.
9. Join P, S and U and extend it on both the sides to draw a tangent at point P.
10. Taking C and D as the centres and the same radius draw two arcs intersecting each other at V.
11. Join Q, T and V and extend it on both the sides to draw a tangent at point Q.
12. Extended tangents at P and Q intersect at R.
13. Hence, the required tangents are UR and VR such that the angle between them is 45°.

Question 6

Draw a right ∆ABC in which AB = 6 cm, BC = 8 cm and m∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.Solution 6

Steps of construction:

1. Draw a line segment BC = 8 cm.
2. Measure m∠B = 90˚ at point B.
3. Taking B as the centre and a radius of 6 cm, draw an arc and hence, construct AB = 6 cm
4. Join AC. Hence, ∆ABC right angled at B is constructed.
5. Taking B as the centre and a radius more than BD, draw two arcs cutting AC at points E and F.
6. Taking E and F as centres draw arcs to intersect each-other at P and Q respectively. Join and extend PQ on both the sides. This is the perpendicular BD from B on AC.
7. BD⊥CD, hence, ∆BCD is a right angled triangle. Therefore the circle passing through points B, C and D has the hypotenuse BC as the diameter.
8. Construct a circle by taking O, the midpoint of BC as the centre (BC is the diameter of the circle)
9. OB ⊥ AB, extend AB on both the sides. OB is one of the tangents passing through point A and tangent at point B.
10. Join OA. Take the centre of OA by drawing intersecting arcs from point O and A and which intersect at points R and S respectively.
11. Join RS. The point at which RS intersects OA is G.
12. Taking G as the centre and radius OG, construct a circle.
13. The point of intersection of the two circles is H.
14. Join H and A. Now extend AH on both the sides. AH is the second tangent drawn from point A.

Question 7

Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.Solution 7

Chapter 10 – Circles Exercise Ex. 10.1

Question 1

Fill in the blanks:

(i) The common point of a tangent and the circle is called ……. .

(ii) A circle may have ……. parallel tangents.

(iii) A tangent to a circle intersects it in ……. point(s).

(iv) A line intesecting a circle in two points is called a ……. .

(v) The angle between tangent at a point on a circle and the radius through the point is ……. .Solution 1

Fill in the blanks:

(i) The common point of a tangent and the circle is called point of contact .

(ii) A circle may have two parallel tangents.

(iii) A tangent to a circle intersects it in one point(s).

(iv) A line intesecting a circle in two points is called a secant .

(v) The angle between tangent at a point on a circle and the radius through the point is 90^o .Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 10 – Circles Exercise Ex. 10.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.Solution 4

Question 5

Solution 5

Question 6

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.Solution 6

Question 7

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.Solution 7

Question 8

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.Solution 8

Question 9

If AB, AC, PQ are tangents in Fig. 8.56 and AB = 5 cm, find the perimeter of ΔAPQ.

Solution 9

Question 10

Solution 10

Question 11

In Fig., PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m∠PRS.

Solution 11

Question 12

Solution 12

Question 13

In a right triangle ABC in which ∠B =90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. Solution 13

Question 14

From an external point P, tangents PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn, which intersects PA and PB at C and D respectively. If PA = 14 cm, find the perimeter of Δ PCD.Solution 14

Question 15

In fig., ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incirde.

Solution 15

Question 16

Prove that the tangent drawn at the mid-point an arc of a circle is parallel to the chord joining the end points of the arc.Solution 16

Question 17

Solution 17

Question 18

Two tangent segments PA and PB are drawn to a circle with centre O such that APB = 120^o. Prove that OP = 2 AP.Solution 18

Question 19

Solution 19

Question 20

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC =BDSolution 20

Question 21

In fig., a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.Solution 24

AP = AQ , BP = PR and CR = CQ (tangents from an external point)

Perimeter of ∆ABC = AB + BR + RC + CA

= AB + BP + CQ + CA

= AP + AQ

= 2AP

∆APO is a right-angled triangle. AO² = AP² + PO²

13² = AP² + 5²

AP² = 144

AP = 12

∴ Perimeter of ∆ABC = 24 cmQuestion 25

In Fig., a circle is inscribed in a quadrilateral ABCD in which ∠B= 90°. If AD=23 cm, AB =29 cm and DS =5 cm, find the radius r of the circle.

Solution 25

Question 26

In Fig., there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP.

Solution 26

Question 27

In Fig., AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA.

Solution 27

Question 28

In Fig., PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN.

Solution 28

Question 29

In Fig., BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the Calculate (i) AF (ii) radius of the circle.

Solution 29

Question 30

Solution 30

Question 31

In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.

Solution 31

Since RS is drawn parallel to the tangent PQ,

∠SRQ = ∠PQR

Also, PQ = PR

⇒ ∠PQR = ∠PRQ

In ∆PQR,

∠PQR + ∠PRQ + ∠QPR = 180°

⇒∠PQR + ∠PQR + 30° = 180°

⇒2∠PQR = 150°

⇒∠PQR = 75°

⇒∠SRQ =∠PQR = 75° (alternate angles)

Also, ∠RSQ =∠RQP = 75° (the angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.)

In ∆RSQ,

∠RSQ + ∠SRQ + ∠RQS = 180°

⇒75° + 75° + ∠RQS = 180°

⇒ ∠RQS =30° Question 32

From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear.Solution 35

Question 36

In Fig, common tangents PQ and RS to two circles intersects at A. Prove that PQ = RS.

Solution 36

Question 37

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle.Solution 37

Since AC is the tangent to the circle with radius 9 cm, we have OB ⊥ AC.

Hence, by applying the Pythagoras Theorem, we have,

OA² = OB² + AB²

⇒ 15² = 9² + AB²

⇒ AB² = 15² – 9²

⇒ AB² = 225 – 81 = 144

∴ AB = 12 cm

We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

Here, OB is the perpendicular and AC is the length of the chord of the circle with radius 15 cm.

So,

AC = 2 × AB = 2 × 12 = 24 cm

Length of the chord of the larger circle which touches the smaller circle = 24 cm.Question 38

AB and CD are common tangents to two circles of equal radii. Prove that AB =CD.Solution 38

Question 39

A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆ PQR is 336 cm², find the sides PQ and PR.Solution 39

Let PA = PB = x

Tangents drawn from an external point are equal in length. QB = QT = 14 cm , RA = RT = 16 cm

PR = (x + 16) cm, PQ = (x + 14)cm,

QR = 30 cm

= x + 30

Area of ∆PQR

Area of ∆PQR = 336 cm²

Side PR = (12 + 16) = 28 cm

Side PQ = (12 + 14) = 26 cmQuestion 40

In Fig., the tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA =110°, find ∠CBA.

Solution 40

Question 41

AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.

Solution 41

Question 42

In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ ABC is 84 cm².

Solution 42

Let M and N be the points where AB and AC touch the circle respectively.

Tangents drawn from an external point to a circle are equal

⇒ AM=AN

BD=BM=8 cm and DC=NC=6 cm

Question 43

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.

Solution 43

∠AOQ=58° (given)

In right ∆BAT,

∠ABT + ∠BAT + ∠ATB=180°

29° + 90° + ∠ATB=180° 

∠ATB = 61° 

that is, ∠ATQ = 61° Question 44

In Fig., OQ : PQ = 3 : 4 and perimeter of ΔPOQ = 60 cm. Determine PQ, QR and OP.

Solution 44

Question 45

Solution 45

Question 46

In Fig., BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABCSolution 46

In ∆AOP,

OA = OP (radii) ∆AOP is an isosceles triangle. OE is a median.

In an isosceles triangle,the median drawn∴∠OEA = 90^o

In ∆AOE and ∆ABC,

∠ABC = ∠OEA = 90^o

∠A is common.

∆AEO ~ ∆ABC…(AA test)Question 47

In fig., PO ⊥ QO . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Solution 47

Question 48

In fig., O is the centre of the circle and BCD is tangent to it at C. Prove that BAC + ACD = 90^o.

Solution 48

Question 49

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.Solution 49

Question 50

In Fig., there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS.

Solution 50

PR = PQ…(tangents fromexternal points)

PQ = 5 cm

Also,

OQ is perpendicular to PS …(tangent is perpendicular to the radius)

Now, in a circle,a perpendicular drawn from the centre of a circle bisects the chord.

So, OQ bisects PS.

PQ = QS

QS = 5 cm

PS = 10 cmQuestion 51

In Fig., PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If ∠POR = 130˚ and S is a point on the circle, find ∠1 + ∠2.

Solution 51

In DPQR,

∠POR is an external angle.

So,

∠POR = ∠PQO + ∠OPQ

Now, PQ is tangent to the circle with radius OQ.

∠PQO = 90^o

130˚ = 90˚ + ∠OPQ

∠OPQ = 40^o

∠1 = 40^o

Now,

Minor arc RT subtends a 130˚ angle at the centre.

So, it will subtend a 65˚angle at any other point on the circle.∠RST = 65˚

∠2 = 65˚

∠1 + ∠2 =105^˚Question 52

In Fig., PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Solution 52

AP = PB = 12 cm, AC = CQ = 3 cm and QD = DB = 3 cm …(tangent from external point)

PA = 12 cm, PC + CA = 12

PC + 3 = 12

PC= 9 …(i)

Now,

PB = 12

PD + DB = 12

PD + 3 = 12

PD = 9 …(ii)

PC + PD = 18 cm…from (i) and (ii)

Chapter 10 – Circles Exercise 10.48

Question 1

A tangent PQ at a point P of a circle of radius 5 cm meets line through the centre O at a point Q such that OQ = 12 cm. Length PQ is

(a) 12 cm

(b) 13 cm

(c) 8.5 cm

(d) Solution 1

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

So, the correct option is (d).Question 2

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cmSolution 2

PQ is a tangent to the circle

So, OP² + PQ² = OQ²

OP² = OQ² – PQ²

      = (25)² – (24)²

      = 49

OP = 7

So, the correct option is (a).Question 3

The length of tangent from point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is

(a) 

(b) 7 cm

(c) 5 cm

(d) 25 cmSolution 3

Given OP = 3 cm

         PA = 4 cm

Hence, OA² = OP² + PA²

OA² = 3² + 4²

      = 25

OA = 5 cm

So, the correct option is (c).Question 4

Solution 4

Chapter 10 – Circles Exercise 10.49

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Tangents from same point to circle have equal length.

Hence Bb = Ba

          bC = Cc

          Ac = Aa

Let Ba = x    then Bb = x

bc = 6 – x     and Aa = 8 – x

and Cc = 6 – x and Ac = 8 – x

So AC = AC + cC

         = 6 – x + 8 – x

AC = 14 – 2x       ……(1)

Also AC² = AB² + BC²

             = 8² + 6² 

             = 100

AC = 10    …..(2)

from (1) & (2)

14 – 2x = 10

4 = 2x

x = 2           also aB = Ob = radius = 2 cm

So, the correct option is (b).Question 9

Solution 9

Question 10

If four sides of a quadrilateral ABCD are tangential to a circle, then

(a) AC + AD = BD + CD

(b) AB + CD = BC + AD

(c) AB + CD = AC + BC

(d) AC + AD = BC + DBSolution 10

Tangents from same point are of equal length.

AP = AS, PB = BQ

QC = CR, RD = DS

AB = AP + PB      …..(1)

BC = BQ + QC   ……(2)

CD = CR + RD   …..(3)

AD = AS + DS …..(4)

Adding (1) & (3)

AB + CD = AP + BP + CR + RD

            = AS + BQ + CQ + DS

            = (AS + DS) + (BQ + CQ)

from (2) & (4)

AB + CD = AD + BC

So, the correct option is (b).Question 11

The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is

(a) 

(b) 

(c) 10 cm

(d) 5 cmSolution 11

Given OQ = 8 cm

         OP = 6 cm

OP² + PQ² = OQ²

6² + PQ² = 8²

       PQ² = 64 – 36

             = 28

PQ = 

So, the correct option is (b).Question 12

AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cmSolution 12

DA and DC are tangents to circle from same point

so, DA = DC ……(1)

similarly DB = DC   ……(2)

(1) + (2)

2DC = DA + DB

2DC = AB

AB = 2 × 4

     = 8 cm

So, the correct option is (c).Question 13

In Figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA

(b) 2AD = AB + BC + CA

(c) 3AD = AB + BC + CA

(d) 4AD = AB + BC + CASolution 13

AD = AE        …….(1)

CD = CF    ……(2)

BF = BE   …..(3)

from (1)

2AD = 2AE

       = AE + AD

       = (AB + BE) + (AC + CD)

       = AB + BF + AC + CF

       = AB + AC + BC

So, the correct option is (b).Question 14

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

(a) 8 cm

(b) 3 cm

(c) 2.5 cm

(d) 5 cmSolution 14

Chapter 10 – Circles Exercise 10.50

Question 15

Solution 15

Question 16

Solution 16

Question 17

AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =

(a) 12 cm

(b) 18 cm

(c) 24 cm

(d) 36 cmSolution 17

AP = PQ    ….(1)

and OA² = OP²  + PA²

      (15)² = (9)² + AP²

       AP² = 225 – 81

             = 144

       AP = 12

AP + AQ = 2AP

             = 24 cm

So, the correct option is (c).Question 18

At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cmSolution 18

Question 19

Solution 19

Chapter 10 – Circles Exercise 10.51

Question 20

In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =

(a) 5 cm

(b) 4 cm

(c) 6 cm

(d) 7 cmSolution 20

AB = 12 cm

BC = 8 cm

AC = 10 cm

Let AD = x  

      AF = x

     BD = 12 – x

and BE = BD = 12 – x

CE = BC – BE

     = 8 – (12 – x)

     = x – 4

and CE = CF = x – 4

AC = AF + FC

     = x + x – 4

AC = 2x – 4

Given, AC = 10 cm

so 2x – 4 = 10

2x = 14

x = 7 cm

AD = 7 cm

So, the correct option is (d).Question 21

In figure, if AP = PB, then

(a) AC = AB

(b) AC = BC

(c) AQ = QC

(d) AB = BCSolution 21

AP = BP     given

and AP = AQ

also BP = BR

from this, we conclude that

AQ = BR     …..(1)

We know CR = CQ    …..(2)

from (1) & (2)

AQ + CR = BR + CR

AQ + CQ = BR + CR

AC = BC

So, the correct option is (b).Question 22

In figure, if AP = 10 cm, then BP = 

Solution 22

AP = 10 cm

AO = 6 cm

OB = 3 cm

AP² + OA² = OP²

OP² = 10² + 6²

OP² = 136

Also OB² + BP² = OP² 

        3² + BP² = 136

        BP² = 136 – 9

So, the correct option is (b).Question 23

Solution 23

Chapter 10 – Circles Exercise 10.52

Question 24

In Figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =

(a) PQ

(b) QR

(c) PR

(d) PSSolution 24

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).Question 25

In figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR = 

(a) 9 cm

(b) 18 cm

(c) 15 cm

(d) 13.5 cmSolution 25

PQ = PT     …..(1)

and PT = PR     …..(2)

so from (1) & (2)

PQ = PR

PQ = PR = 4.5 cm

QR = PQ + PR

     = 4.5 + 4.5 = 9 cm

So, the correct option is (a).Question 26

Solution 26

Question 27

Solution 27

Chapter 10 – Circles Exercise 10.53

Question 28

In figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cmSolution 28

radius of circle 1 = 3 cm

radius of circle 2 = 5  cm

OP² = OQ² + QP²         and    O’S² + SR² = O’R² 

OP² = 4² + 3²                         O’R² = 5² + 12²     

      = 16 + 9                          O’R² = 169                       

      = 25                                O’R’ = 13 cm

OP = 5 cm

OO’ = OK + KO’

       = 3 + 5

       = 8 cm

PR = PO + OK + KO’ + O’R

     = 5 + 3 + 5 + 13

     = 26 cm

So, the correct option is (b).Question 29

Solution 29

Question 30

Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 10 cmSolution 30

OB = OC = OA = 5 cm

OQ = OP = 3 cm

OB² = OQ² + BQ²

BQ² = OB² – OQ²

      = 5² – 3²

      = 16

BQ = 4 cm

also BQ = BP

BP = 4 cm

In ΔOPC,

OP² + PC² = OC²

PC²  = OC² – OP²

       = 5² – 3²

       = 16

PC = 4 cm

BC = BP + PC = 4 + 4 = 8 cm

So, the correct option is (c).Question 31

In figure, there are two concentric circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to

(a) 10 cm

(b) 12 cm

(c) 15 cm

(d) 18 cmSolution 31

Given PR = 7.5 cm

so PR = PQ

PQ = 7.5 cm

PS is the chord to the larger circle. We know that, perpendicular drawn from centre bisect the chords.

Hence PQ = QS

PS = PQ + QS

     = 2PQ

     = 2 × 7.5

     = 15 cm

So, the correct option is (c).

Chapter 10 – Circles Exercise 10.54

Question 32

In figure, if AB = 8 cm and PE = 3 cm, then AE = 

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cmSolution 32

AC = AB      …..(1)

BD = DP     ……(2)

PE = EC    ……(3)

AB = 8 so AC = 8 cm

PE = 3 so EC = 3 cm

AE = AC – EC = 8 – 3 = 5 cm

So, the correct option is (c).Question 33

Solution 33

Question 34

Solution 34

Chapter 10 – Circles Exercise 10.55

Question 35

In figure, the sides AB, BC and CA of a triangle ABC, touch a circle at P,Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is

(a) 11 cm

(b) 10 cm

(c) 14 cm

(d) 15 cmSolution 35

PA = AR

AR = 4 cm

BP = BQ and QC = RC

BQ = 3 cm

Given AC = 11

AR + RC = 11

4 + RC = 11

RC = 7

so QC = 7 cm

BC = BQ + QC

     = 3 + 7

     = 10 cm

So, the correct option is (b).Question 36

Solution 36

EK = 9 cm

and EK = EM

Hence EM = 9 cm         …..(1)

Also EK = ED + DK

and DK = DH

EK = ED + HD     …….(2)

EM = EF + FM

and FM = FH

EM = EF + FH       ……(3)

(2) + (3)

EK + EM = ED + EF + DH + HF

18 = ED + DF + EF

perimeter = 18 cm

So, the correct option is (a).Question 37

Solution 37

Chapter 10 – Circles Exercise 10.56

Question 38

Solution 38

AB = 29 cm

AD = 23

DS = 5 cm

DS = DR

so DR = 5 cm

AR = AD – DR

     = 23 – 5

     = 18 cm

AR = AQ 

AQ = 18 cm

BQ = AB – AQ

     = 29 – 18

BQ = 11 cm

As OP || BQ and OQ || PB 

Hence, OP = BQ

OP = 11 cm

So, the correct option is (a).Question 39

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 (in cm) is

(a) 4

(b) 3

(c) 2

(d) 1Solution 39

AB = 5 cm

BC = 12 cm

AB²  + BC² = AC²

AC² = 5² + 12²

      = 169

AC = 13 cm

Let BQ = x

AQ = AR = 5 – x

CR = AC – AR

      = 13 – (5 – x)

      = x + 8

And CP = CR = x + 8

so BP = BC – PC

         = 12 – (x + 8)

         = 4 – x

But BP = BQ = x

4 – x = x

x = 2

and BQ || OP and OQ || PB

so BQ = PO

PO = 2 cm

So, the correct option is (c).Question 40

Solution 40

Question 41

Solution 41

Question 42

In 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 (in cm) is

(a) 3.8

(b) 7.6

(c) 5.7

(d) 1.9Solution 42

PT = 3.8 cm

We know

PQ = PT and PT = PR

Hence PQ = 3.8 cm and PR = 3.8 cm

Now, QR = QP + PR

             = 3.8 + 3.8

QR = 7.6 cm

So, the correct option is (b).

Chapter 10 – Circles Exercise 10.57

Question 43

In 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, then x =

(a) 10

(b) 9

(c) 8

(d) 7Solution 43

AB = x cm

BC = 7 cm

CR = 3 cm

AS = 5 cm

CR = CQ

CQ = 3 cm

given BC = 7 cm

BQ = BC – QC

       = 7 – 3

       = 4 cm

And BQ = BP

so BP = 4 cm

Also AS = AP

Hence AP = 5 cm

AB = AP + BP

     = 5 + 4

     = 9 cm

x = 9 cm

So, the correct option is (b).Question 44

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

a. 90°

b. 50°

c. 70°

d. 40°Solution 44

Question 45

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

Solution 45

Question 46

If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is

a. 3 cm

b. 6 cm

c. 9 cm

d. 1 cmSolution 46

Question 47

At one end A of a diameter AB of a circle of radius 5 cm, Tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

a. 4 cm

b. 5 cm

c. 6 cm

d. 8 cmSolution 47

Question 48

From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

a. 60 cm²

b. 65 cm²

c. 30 cm²

d. 32.5 cm²Solution 48

Question 49

If PA, PB are tangents to the circle with centre O such that ∠APB  = 50°, then ∠OAB is equal to

a. 25°

b. 30°

c. 40°

d. 50°Solution 49

Question 50

The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is

a. 10 cm

b. 7.5 cm

c. 5 cm

d. 2.5 cmSolution 50

Chapter 10 – Circles Exercise 10.58

Question 51

In figure, if ∠AOB = 125° , then ∠COD is equal to

a. 45°

b. 35°

c. 55°

d. 62°

Solution 51

Question 52

In figure, if PQR is tangent to a circle at Q whole centre is O , AB is a chord parallel to PR and ∠BQR = 70°, then∠AQB is equal to

a. 20°

b. 40°

c. 35°

d. 45°

Solution 52

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.1

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)

Write the first five terms for the sequence whose nth terms is:

a_n = n² – n + 1Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Find the indicated terms in the sequence whose nth terms are :

a_n = (n – 1) (2 – n) (3 + n); a₁, a₂, a₃Solution 2 (iv)

Question 2 (v)

Find the indicated terms in the sequence whose nth terms are:

a_n = (-1)ⁿ n; a₃, a₅, a₈Solution 2 (v)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5

Solution 5

Question 6(i)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 2n – 1Solution 6(i)

Question 6(ii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 3n² + 5Solution 6(ii)

Question 6(iii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 1 + n + n²Solution 6(iii)

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.3

Question 1

Solution 1

Question 2

Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = -3

(ii) a = -1, d = 1/2

(iii) a = -1.5, d = -0.5Solution 2

Question 3 (i)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.


Solution 3 (i)

Question 3 (ii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.Solution 3 (ii)

Question 3(iii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

Divya deposited Rs.1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …., and so on.Solution 3(iii)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5 (i)

Solution 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)

Find out whether of the given sequence is an arithemtic progressions. If it is an arithmetic progressions, find out the common difference.

p, p + 90, p + 180, p + 270, … where p = (999)⁹⁹⁹Solution 5 (vi)

Question 5 (vii)

Solution 5 (vii)

Question 5 (viii)

Solution 5 (viii)

Question 5 (ix)

Solution 5 (ix)

Question 5 (x)

Solution 5 (x)

Question 5 (xi)

Solution 5 (xi)

Question 5 (xii)

Solution 5 (xii)

Question 6

Solution 6

Question 7

Solution 7

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.4

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 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Which term of the A.P. 4, 9, 14, … is 254?Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Solution 2 (v)

Question 2 (vi)

Which term of the A.P. -7, -12, -17, -22,… will be -82? Is -100 any term of the A.P.?Solution 2 (vi)

The given A.P. is -7, -12, -17, -22,…

First term (a) = -7

Common difference = -12 – (-7) = -5

Suppose nth term of the A.P. is -82.

So, -82 is the 16^th term of the A.P.

To check whether -100 is any term of the A.P., take a_n as -100.

So, n is not a natural number.

Hence, -100 is not the term of this A.P.Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9 (i)

Solution 9 (i)

Question 9 (ii)

Find the 12^th term from the end of the following arithmetic progressions:

3, 8, 13, …, 253Solution 9 (ii)

A.P. is 3, 8, 13, …, 253

We have:

Last term (l) = 253

Common difference (d) = 8 – 3 = 5

Therefore,

12^th term from end

       = l – (n – 1)d

       = 253 – (12 – 1) (5)

       = 253 – 55

       = 198Question 9 (iii)

Solution 9 (iii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The 26^th, 11^th and last term of an A.P. are 0, 3 and , respectively. Find the common difference and the number of terms.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 34. Find the first term and the common difference of the A.P.Solution 18

Question 19

Solution 19

Question 20 (i)

Solution 20 (i)

Question 20 (ii)

Solution 20 (ii)

Question 21

Solution 21

Question 22 (i)

Solution 22 (i)

Question 22 (ii)

Solution 22 (ii)

Question 22 (iii)

Solution 22 (iii)

Question 22 (iv)

Solution 22 (iv)

Question 23

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15^th term.Solution 23

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

The 17^th term of an A.P. is 5 more than twice its 8^th term. If the 11^th term of the A.P. is 43, find the n^th term.Solution 37

Thus, n^th term is given by

a_n = a + (n – 1)d
a_n = 3 + (n – 1)4
a_n = 3 + 4n – 4
a_n = 4n – 1Question 38

Find the number of all three digit natural numbers which are divisible by 9.Solution 38

The smallest three digit number divisible by 9 = 108

The largest three digit number divisible by 9 = 999

Here let us write the series in this form,

108, 117, 126, …………….., 999

a = 108, d = 9

t_n = a + (n – 1)d

999= 108 + (n – 1)9

⇒ 999 – 108 = (n – 1)9

⇒ 891 = (n – 1)9

⇒ (n – 1) = 99

⇒ n = 99 + 1

∴ n = 100

Number of terms divisible by 9

Number of all three digit natural numbers divisible by 9 is 100.Question 39

The 19^th term of an A.P. is equal to three times its sixth term. If its 9^th term is 19, find the A.P.Solution 39

Question 40

The 9^th term of an A.P. is equal to 6 times its second term. If its 5^th term is 22, find the A.P.Solution 40

Question 41

The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is 4 times its 15^th term.Solution 41

Let the first term be ‘a’ and the common difference be ‘d’

t₂₄ = a + (24 – 1)d = a + 23d

t_10­ = a + (10 – 1)d = a + 9d

t₇₂ = a + (72 – 1)d = a + 71d

t₁₅ = a + (15 – 1)d = a + 14d

t₂₄ = 2t₁₀

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 23d – 18d = 2a – a

∴ 5d = a

t₇₂ = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t₁₅

∴t₇₂ = 4t₁₅Question 42

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.Solution 42

L.C.M. of 2 and 5 = 10

3- digit number after 100 divisible by 10 = 110

3- digit number before 999 divisible by 10 = 990

Let the number of natural numbers be ‘n’

990 = 110 + (n – 1)d

⇒ 990 – 110 = (n – 1) × 10

⇒ 880 = 10 × (n – 1)

⇒ n – 1 = 88

∴ n = 89

The number of natural numbers between 110 and 999 which are divisible by 2 and 5 is 89.Question 43

If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)^rd term.Solution 43

Let the first term be ‘a’ and the common difference be ‘d’.

Question 44

The sum of 5^th and 9^th terms of an A.P. is 30. If its 25^th term is three times its 8^th term, find the A.P.Solution 44

Question 45

Find where 0(zero) is a term of the A.P. 40, 37, 34, 31, …Solution 45

Let the first term be ‘a’ and the common difference be ‘d’.

a = 40

d = 37 – 40 = – 3

Let the n^th term of the series be 0.

t_n = a + (n – 1)d

⇒ 0 = 40 + (n – 1)( – 3)

⇒ 0 = 40 – 3(n – 1)

⇒ 3(n – 1) = 40

∴ No term of the series is 0.Question 46

Find the middle term of the A.P. 213, 205, 197, …, 37.Solution 46

Given A.P. is 213, 205, 197, …, 37.

Here, first term = a = 213

And, common difference = d = 205 – 213 = -8

a_n = 37

n^th term of an A.P. is given by

a_­n = a + (n – 1)d

⇒ 37 = 213 + (n – 1)(-8)

⇒ 37 = 213 – 8n + 8

⇒ 37 = 221 – 8n

⇒ 8n = 221 – 37

⇒ 8n = 184

⇒ n = 23

So, there are 23 terms in the given A.P.

⇒ The middle term is 12^th term.

⇒ a₁₂ = 213 + (12 – 1)(-8)

= 213 + (11)(-8)

= 213 – 88

= 125

Hence, the middle term is 125.Question 47

If the 5^th term of an A.P. is 31 and 25^th term is 140 more than the 5^th term, find the A.P.Solution 47

Let a be the first term and d be the common difference of the A.P.

Then, we have

a₅ = 31 and a₂₅ = a₅ + 140

⇒ a + 4d = 31 and a + 24d = a + 4d + 140

⇒ a + 4d = 31 and 20d = 140

⇒ a + 4d = 31 and d = 7

⇒ a + 4(7) = 31 and d = 7

⇒ a + 28 = 31 and d = 7

⇒ a = 3 and d = 7

Hence, the A.P. is a, a + d, a + 2d, a + 3d, ……

i.e. 3, 3 + 7, 3 + 2(7), 3 + 3(7), ……

i.e. 3, 10, 17, 24, …..Question 48

Find the sum of two middle terms of the A.P.

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?Solution 51

Question 52

Find the 12^th term from the end of the A.P. -2, -4, -6, …, -100.Solution 52

Question 53

For the A.P.: -3, -7, -11, …, can we find a₃₀ – a₂₀ without actually finding a₃₀ and a₂₀? Give reason for your answer.Solution 53

Question 54

Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10^th terms is the same as the difference between their 21^st terms, which is the same as the difference between any two corresponding terms. Why?Solution 54

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.Solution 8

Question 9

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.Solution 9

Let the first three terms of an A.P. be a – d, a, a + d

As per the question,

(a – d) + a + (a + d) = 18

∴ 3a = 18

∴ a = 6

Also, (a – d)(a + d) = 5d

∴ (6 – d)(6 + d) = 5d

∴ 36 – d² = 5d

∴ d² + 5d – 36 = 0

∴ d² + 9d – 4d – 36 = 0

∴ (d + 9)(d – 4) = 0

∴ d = -9 or d = 4

Thus, the terms will be 15, 6, -3 or 2, 6, 10.Question 10

Spilt 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.Solution 10

Question 11

The angles of a triangle are in A.P. the greatest angle is twice the least. Find all the angles.Solution 11

Question 12

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.Solution 12

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.6

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 2

Solution 2

Question 3

Solution 3

Question 4

Find the sum of last ten terms of the A.P: 8, 10, 12, 14,.., 126.Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Find the sum of the first 15 terms of each of the following sequences having n^th term as

y_n = 9 – 5nSolution 5 (iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10 (i)

Solution 10 (i)

Question 10 (ii)

How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?Solution 10 (ii)

Question 10 (iii)

How many terms of the A.P. 9, 17, 25, … must be taken so that their sum is 636?

Remark* – Question modified.Solution 10 (iii)

Question 10 (iv)

Solution 10 (iv)

Question 10(v)

How many terms of the A.P. 27, 24, 21… should be taken so that their sum is zero?Solution 10(v)

Question 10 (vi)

How many terms of the A.P. 45, 39, 33 … must be taken so that their sum is 180? Explain the double answer.Solution 10 (vi)

Let the required number of terms be n.

As the given A.P. is 45, 39, 33 …

Here, a = 45 and d = 39 – 45 = -6 

The sum is given as 180

∴ S_n = 180

When n = 10,

When n = 6,

Hence, number of terms can be 6 or 10.Question 11 (i)

Solution 11 (i)

Question 11 (ii)

Solution 11 (ii)

Question 11 (iii)

Solution 11 (iii)

Question 12(i)

Find the sum of

The first 15 multiples of 8.Solution 12(i)

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

Question 12(ii)

Find the sum of

The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.Solution 12(ii)

a) divisible by 3 3,6,9,… Now, n=40, a=3, d=3

b) divisible by 5 5,10,15,… Now, n=40, a=5, d=5

c)divisible by 6 6,12,18,… Now, n=40, a=3, d=6Question 12(iii)

Find the sum of

All 3 – digit natural numbers which are divisible by 13.Solution 12(iii)

Three-digit numbers divisible by 13 are 104,117, 130,…988. Now, a=104, l=988Question 12(iv)

Find the sum of

All 3 – digit natural numbers, which are multiples of 11. Solution 12(iv)

Three-digit numbers which are multiples of 11 are 110,121, 132,…990. Now, a=110, l=990Question 12(v)

Find the sum of

All 2 – digit natural numbers divisible by 4. Solution 12(v)

Two-digit numbers divisible by 4 are 12,16,…96. Now, a=12, l=96Question 12 (vi)

Find the sum of first 8 multiples of 3.Solution 12 (vi)

The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, …

These are in A.P. with,

first term (a) = 3 and common difference (d) = 3

To find S₈ when a = 3, d = 3

Hence, the sum of first 8 multiples of 3 is 108.Question 13 (i)

Solution 13 (i)

Question 13 (ii)

Solution 13 (ii)

Question 13 (iii)

Solution 13 (iii)

Question 13 (iv)

Solution 13 (iv)

Question 13 (v)

Solution 13 (v)

Question 13 (vi)

Solution 13 (vi)

Question 13 (vii)

Solution 13 (vii)

Question 13 (viii)

Find the sum:

Solution 13 (viii)

Let ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

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 (i)

Solution 22 (i)

Question 22 (ii)

If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.Solution 22 (ii)

Sum of first n terms of an AP is given by

As per the question, S₄ = 40 and S₁₄ = 280

Also,  

Subtracting (i) from (ii), we get, 10d = 20

Therefore, d = 2

Substituting d in (i), we get, a = 7

Sum of first n terms becomes

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.Solution 26

Let the number of terms be ‘n’, ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

⇒ 49 = 7 + (n – 1)d

⇒ 42 = (n – 1)d…..(i)

∴ 840 = n[14 + (n – 1)d]……(ii)

Substituting (ii) in (i),

840 = n[14 + 42]

⇒ 840 = 56n

∴ n = 15

Substituting n in (i)

42 = (15 – 1)d

Common difference, d =3Question 27

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.Solution 27

Let the number of terms be ‘n’, ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

⇒ 45 = 5 + (n – 1)d

⇒ 40 = (n – 1)d…..(i)

∴ 800 = n[10 + (n – 1)d]……(ii)

Substituting (ii) in (i),

800 = n[10 + 40]

⇒ 800 = 50n

∴ n = 16

Substituting n in (i)

40 = (16 – 1)d

Question 28

The sum of first q terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1 : 2. Find the first and 15^th term of the A.P.Solution 28

Question 29

If the 10^th term of an A.P. is 21 and the sum of its first ten terms is 120, find its n^th term.Solution 29

Let ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

t₁₀ = a + (10 – 1)d

⇒ 21 = a + 9d……(i)

120 = 5[2a + 9d]

24 = 2a + 9d………(ii)

(ii) – (i) ⇒ 

a = 3

Substituting a in (i), we get

a + 9d = 21

⇒ 3 + 9d = 21

⇒ 9d = 18

∴d = 2

t_n = a + (n – 1)d

= 3 + (n – 1)2

= 3 + 2n – 2

∴ t_n= 2n + 1Question 30

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^th term of this A.P.Solution 30

Let the first term be ‘a’ and the common difference be ‘d’  

63 = 7[a + 3d]

9 = a + 3d……….(i)

Sum of the next 7 terms = 161

Sum of the first 14 terms = 63 + 161 = 224

224 = 7[2a + 13d]

32 = 2a + 13d………..(ii)

Solving (i) and (ii), we get

d = 2, a = 3

28^th term of the A.P., t₂₈ = a + (28 – 1)d

= 3 + 27 × 2

= 3 + 54

= 57

∴ The 28^th of the A.P. is 57.Question 31

The sum of first seven terms of an A.P. is 182. If its 4^th and the 17^th terms are in the ratio 1: 5, find the A.P.Solution 31

Let the first term be ‘a’ and the common difference be ‘d’.

26 × 2 = [2a + (7 – 1)d]

52 = 2a + 6d

26 = a + 3d……..(i)

From (i) and (ii),

⇒ 13d = 104

∴d = 8

From (i), a = 2

The A.P. is 2, 10, 18, 26,…….Question 32

The n^th term of an A.P. is given by (- 4n + 15). Find the sum of the first 20 terms of this A.P.Solution 32

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.

t_n = – 4n + 15

t₁ = – 4 × 1 + 15 = 11

t₂ = – 4 × 2 + 15 = 7

t₃ = – 4 × 3 + 15 = 3

Common Difference, d = 7 – 11 = -4

= 10 × (-54)

= – 540

*Note: Answer given in the book is incorrect.Question 33

Solution 33

Question 34

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^th term.Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Find the number of terms of the A.P. – 12, – 9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.Solution 37

Let the number of the terms be ‘n’.

Common Difference, d = – 9 + 12 = 3

t_n = a + (n – 1)d

⇒ 21 = – 12 + 3(n – 1)

⇒ 21 + 12 = 3(n – 1)

⇒ 3(n – 1) = 33

⇒ n – 1 = 11

∴n = 12

Number of terms of the series = 12

If 1 is added to each term of the above A.P.,

– 11, – 8, – 5,…….,22

Number of terms in the series, n₁ = 12

Sum of all the terms,

The sum of the terms = 66Question 38

The sum of the first n terms of an A.P. is 3n² + 6n. Find the nth term of this A.P.Solution 38

Sum of n terms of the A.P., S_n = 3n² + 6n

S₁ = 3 × 1² + 6 × 1 = 9 = t₁ ……(i)

S₂ = 3 × 2² + 6 × 2 = 24 = t₁ + t₂ …….(ii)

S₃ = 3 × 3² + 6 × 3 = 45 = t₁ + t₂ + t₃ ……..(iii)

From (i), (ii) and (iii),

t₁ = 9, t₂ = 15, t₃ = 21

Common difference, d = 15 – 9 = 6

n^th of the AP, t_n = a + (n – 1)d

= 9 + (n – 1) 6

= 9 + 6n – 6

= 6n + 3

Thus, the n^th term of the given A.P. = 6n + 3Question 39

The sum of the first n terms of an A.P. is 5n – n². Find the n^th term of this A.P.Solution 39

S_n = 5n – n²

S₁ = 5 × 1 – 1² = 4 = t₁………..(i)

S₂ = 5 × 2 – 2² = 6 = t₁ +t₂………..(ii)

S₃ = 5 × 3 – 3² = 6 = t₁ + t₂ + t_3­……….(iii)

From (i), (ii) and (iii),

t₁ = 4, t₂ = 2, t₃ = 0

Here a = 4, d = 2 – 4 = – 2

t_n = a + (n – 1)d

= 4 + (n – 1)( -2)

= 4 – 2n + 2

= 6 – 2nQuestion 40

The sum of the first n terms of an A.P. is 4n² + 2n. find the n^th term of this A.P.Solution 40

S_n = 4n² + 2n

S₁ = 4 × 1² + 2 × 1 = 6 = t₁………….(i)

S₂ = 4 × 2² + 2 × 2 = 20 = t₁ + t₂……….(ii)

S₃ = 4 × 3² + 2 × 3 = 42 = t₁ + t₂ + t₃………..(iii)

From (i), (ii) and (iii),

t₁ = 6, t₂ = 14, t₃ = 22

Here a = 6, d = 14 – 6 = 8

t_n = a + (n – 1)d

t_n = 6 + (n – 1)8

= 6 + 8n – 8

= 8n – 2

*Note: Answer given in the book is incorrect.Question 41

The sum of first n terms of an A.P. is 3n² + 4n. find the 25^th term of this A.P.Solution 41

Sum of n terms of the A.P., S_n = 3n² + 4n

S₁ = 3 × 1² + 4 × 1 = 7 = t₁………(i)

S₂ = 3 × 2² + 4 × 2 = 20 = t₁ + t_2­…….(ii)

S₃ = 3 × 3² + 4 × 3 = 39 = t₁ + t₂ + t₃ …….(iii)

From (i), (ii), (iii)

t₁ = 7, t₂ = 13, t­₃ = 19

Common difference, d = 13 – 7 = 6

25^th of the term of this A.P., t₂₅ = 7 + (25 – 1)6

= 7 + 144 = 151

∴The 25^th term of the A.P. is 151.Question 42

The sum of first n terms of an A.P. is 5n² + 3n. If its m^th term is 168, find the value of m. Also, find the 20^th term of this A.P.Solution 42

Sum of the terms, S_n = 5n² + 3n

S₁ = 5 × 1² + 3 × 1 = 8 = t₁………..(i)

S₂ = 5 × 2² + 3 × 2 = 26 = t₁ + t₂…………..(ii)

S₃ = 5 × 3² + 3 × 3 = 54 = t₁ + t₂ + t₃…………(iii)

From(i), (ii) and (iii),

t₁ = 8, t₂ = 18, t₃ = 28

Common difference, d = 18 – 8 = 10

t_m = 168

⇒ a + (m – 1)d = 168

⇒ 8 + (m – 1)×10 = 168

⇒ (m – 1) × 10 = 160

⇒ m – 1 = 16

∴m = 17

t₂₀ = a + (20 – 1)d

= 8 + 19 × 10

= 8 + 190

= 198Question 43

The sum of first q terms of an A.P. is 63q – 3q². If its pth term -60, find the value of p. Also, find the 11^th term of this A.P.

Remark* – Question modified.Solution 43

Let the first term be ‘a’ and the common difference be ‘d’.

Sum of the first ‘q’ terms, S_q = 63q – 3q²

S₁ = 63 × 1 – 3 × 1² = 60 = t₁……..(i)

S₂ = 63 × 2 – 3 × 2² = 114 = t₁ + t₂…..(ii)

S₃ = 63 × 3 – 3 × 3² = 162 = t₁ + t₂ + t₃ …..(iii)

From (i), (ii) and (iii),

t₁ = 60

t₂ = 54

t₃ = 48

Common difference, d = 54 – 60 = – 6

t_p = a + (p – 1)d

⇒ -60 = 60 + (p – 1)( – 6)

⇒ – 120 = – 6(p – 1)

⇒ p – 1 = 20

∴p = 21

t₁₁ = 60 + (11 – 1)( – 6)

=60 + 10( – 6)

= 60 – 60

= 0

The 11^th term of the A.P. is 0.Question 44

The sum of first m terms of an A.P. is 4m² – m. If its nth term is 107, find the value of n. Also, find the 21^st term of this A.P.Solution 44

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’

Sum of m terms of the A.P., S_m = 4m² – m

S₁ = 4 × 1² – 1 = 3 = t₁ …….(i)

S₂ = 4 × 2² – 2 = 14 = t₁ + t₂……..(ii)

S₃ = 4 × 3² – 3 = 33 = t₁ + t₂ + t_3­ …….(iii)

From (i), (ii) and (iii)

t₁ = 3, t₂ = 11, t₃ = 19

Common difference, d = 11 – 3 = 8

t_n = 107

⇒ a + (n – 1)d = 107

⇒ 3 + (n – 1)8 = 107

⇒ 8(n – 1) = 104

⇒ n – 1 = 13

∴n = 14

t₂₁ = 3 + (21 – 1)8 = 3 + 160 = 163Question 45

Solution 45

Question 46

If the sum of first n terms of an A.P. is then find its n^th term. Hence write its 20^th term.Solution 46

Question 47 (i)

Solution 47 (i)

Question 47 (ii)

If the sum of first n terms of an A.P. is n², then find its 10^th term.Solution 47 (ii)

Sum of first n terms of an AP is given by

As per the question, S_n = n²

Hence, the 10^th term of this A.P. is 19.Question 48

Solution 48

Question 49

Solution 49

Question 50 (i)

Solution 50 (i)

Question 50 (ii)

Solution 50 (ii)

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55 (i)

Solution 55 (i)

Question 55(ii)

Find the sum of all integers between 100 and 550 which are not divisible by 9.Solution 55(ii)

Question 55(iii)

Find the sum of all integers between 1 and 500 which are multiplies 2 as well as of 5.Solution 55(iii)

Question 55(iv)

Find the sum of all integers from 1 to 500 which are multiplies 2 as well as of 5.Solution 55(iv)

Question 55(v)

Find the sum of all integers from 1 to 500 which are multiples of 2 or 5.Solution 55(v)

Question 56 (i)

Solution 56 (i)

Question 56 (ii)

Solution 56 (ii)

Question 56 (iii)

Solution 56 (iii)

Question 56 (iv)

Solution 56 (iv)

Question 56 (v)

Solution 56 (v)

Question 56 (vi)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find

n and a_n, if a = 2, d = 8 and S_n = 90.Solution 56 (vi)

Question 56(vii)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find k, if S_n = 3n² + 5n and a_k = 164.Solution 56(vii)

Question 56 (viii)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find S₂₂, if d = 22 and a₂₂ = 149.Solution 56 (viii)

The nth term of an A.P. is given by a_n = a + (n – 1)d

Sum of first n terms of an AP is given by

Question 57

If S_n denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ – S₄).Solution 57

Question 58

A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?Solution 58

Question 59

The sums of first n terms of three A.P.s are S₁, S₂ and S₃. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S₁ + S₃ = 2S₂.Solution 59

Question 60

Resham wanted to save at least Rs. 6500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her saying by Rs.20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?Solution 60

Question 61

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.Solution 61

Trees planted by the student in class 1 = 2 + 2 = 4

Trees planted by the student in class 2 = 4 + 4 = 8

Trees planted by the students in class 3 = 6 + 6 = 12

…….

Trees planted by the students in class 12 = 24 + 24 = 48

∴ the series will be 4, 8, 12,………., 48

a = 4, Common Difference, d = 8 – 4 = 4

Let ‘n’ be the number of terms in the series.

48 = 4 + (n – 1)4

⇒ 44 = 4(n – 1)

⇒ n – 1 = 11

∴n = 12

Sum of the A.P. series,

Number of trees planted by the students = 312Question 62

Ramkali would need Rs. 1800 for admission fee and books etc., for her daughter to start going to school from the next year. She saved Rs. 50 in the first month of this year and increased her monthly saving by Rs. 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school?Solution 62

Since, the difference between the savings of two consecutive months is Rs. 20, therefore the series is an A.P.

Here, the savings of the first month is Rs. 50

First term, a = 50, Common difference, d = 20

No. of terms = no. of months

No. of terms, n = 12

= 6[100 + 220]

= 6×320

= 1920

After a year, Ramakali will save Rs. 1920.

Yes, Ramakali will be able to fulfill her dream of sending her daughter to school.Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

If S_n denotes the sum of the first n terms of an A.P., prove that S₃₀ = 3(S₂₀ – S₁₀).Solution 70

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.

R.H.S.

= 3(S₂₀ – S₁₀)

= 3(10[2a + 19d] – 5[2a + 9d])

= 3(20a + 190d – 10a – 45d)

= 3(10a + 145d)

= 3 × 5(2a + 29d)

= 15[2a + (30 – 1)d]

= S₃₀

= L.H.S.Question 71

Solve the equation

(-4) + (-1) + 2 + 5 + …. + x = 437.Solution 71

Question 72

Which term of the A.P. -2, -7, -12,…, will be -77? Find the sum of this A.P. upto the term -77.Solution 72

Question 73

The sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.Solution 73

Question 74

The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?Solution 74

Chapter 8 Quadratic Equations Exercise Ex. 8.1

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 1 (xiii)

Solution 1 (xiii)

Question 1 (xiv)

Solution 1 (xiv)

Question 1 (xv)

Is X(x+1)+8=(x+2)(x-2) a quadratic equation?Solution 1 (xv)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Are x = 2 and x = 3, solutions of the equation 2x² – x + 9 = x² + 4x + 3 , Solution 2 (v)

= 2x² – x + 9 – x² + 4x + 3

= x² – 5x + 6 = 0

Here, LHS = x² – 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x² – 5x + 6

= (2)² – 5(2) + 6

=10-10

=0

= RHS

= x² – 5x + 6

= (3)2 – 5(3) + 6

= 9 – 15 + 6

=15 – 15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.Question 2 (vi)

Solution 2 (vi)

Question 2 (vii)

Solution 2 (vii)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Question 4

Solution 4

Question 5

Solution 5

Chapter 8 Quadratic Equations Exercise Ex. 8.2

Question 1

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.Solution 1

Question 2

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.Solution 2

Question 3

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.Solution 3

Question 4

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.Solution 4

Question 5

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.Solution 5

Question 6

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.Solution 6

Chapter 8 Quadratic Equations Exercise Ex. 8.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 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

Solve the following quadratic equation by factorisation:

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following quadratic equation by factorisation:

Solution 15

Question 16

Solution 16

Question 17

9x² – 6b²x – (a⁴ – b⁴) = 0Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solve the following quadratic equation by factorisation:

2x² + ax – a² = 0Solution 20

Question 21

Solve the following quadratic equation by factorisation:

Solution 21

Question 22

Solve the following quadratic equations by factorization:

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solve the following quadratic equation by factorisation:

Solution 29

Question 30

Solve the following quadratic equation by factorisation:

Solution 30

Question 31

Solve the following quadratic equation by factorisation:

Solution 31

Question 32

Solve the following quadratic equation by factorisation:

Solution 32

Question 33

Solve the following quadratic equation by factorisation:

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solve the following quadratic equation by factorisation:

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solve the following quadratic equations by factorization:

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solve the following quadratic equations by factorization:

Solution 47

Question 48

Solve the following quadratic equations by factorization:

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solve the following quadratic equation by factorisation:

Solution 58

Question 59

Solve the following quadratic equation by factorisation:

Solution 59

Question 60

Solve the following quadratic equation by factorisation:

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Chapter 8 Quadratic Equations Exercise Ex. 8.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find the roots of the following quadratic equations (if they exist) by the method of completing the square

x² – 8x + 18 = 0Solution 6

Given equation is x² – 8x + 18 = 0

x² – 2 × x × 4 + + 4² – 4² + 18 = 0

(x – 4)² – 16 + 18 = 0

(x – 4)² = 16 – 18

(x – 4)² = -2

Taking square root on both the sides, we get 

Therefore, real roots does not exist.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 8 Quadratic Equations Exercise Ex. 8.5

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)

Write the discriminant of the following quadratic equations:

(x + 5)² = 2(5x – 3)Solution 1 (vii)

x² + 2 × x × 5 + 5² = 10x – 6

x² + 10x + 25 = 10x – 6

x² + 31 = 0

Here, a = 1, b = 0 and c = 31

Therefore, the discriminant is

D = b² – 4ac

 = 0 – 4 × 1 × 31

 = -124Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Solution 2 (v)

Question 2 (vi)

Solution 2 (vi)

Question 2 (vii)

Solution 2 (vii)

Question 2 (viii)

Solution 2 (viii)

Question 2 (ix)

Solution 2 (ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

3x² – 5x + 2 = 0Solution 2(xii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solve for x:

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solve for x:

Solution 3(iv)

Question 3(v)

Solve for x:

Solution 3(v)

Chapter 8 Quadratic Equations Exercise Ex. 8.6

Question 1(i)

Determine the nature of the roots of the following quadratic equations:

2x² – 3x + 5 = 0Solution 1(i)

Question 1(ii)

Determine the nature of the roots of the following quadratic equations:

2x² – 6x + 3 = 0Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1 (vi)

Determine the nature of the roots of the following quadratic equations:

Solution 1 (vi)

Given quadratic equation is 

Here, 

Therefore, we have

As D = 0, roots of the given equation are real and equal.Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

Question 2(xiii)

Solution 2(xiii)

Question 2(xiv)

Solution 2(xiv)

Question 2(xv)

Solution 2(xv)

Question 2(xvi)

Solution 2(xvi)

Question 2(xvii)

Solution 2(xvii)

Question 2(xviii)

Find the values of k for which the roots are real and equal in each of the following equations:

4x² – 2 (k + 1)x + (k + 1) = 0 Solution 2(xviii)

4x² – 2 (k + 1)x + (k + 1) = 0 Comparing with ax² + bx + c = 0, we get a = 4, b = -2(k + 1), c = k + 1 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 3(i)

Solution 3(i)

Question 3(ii)

In the following, determine the set of values of k for which the given quadratic equation has real roots:

2x² + x + k = 0Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Find the values of k for which the following equations have real and equal roots:

x² + k(2x + k – 1) + 2 = 0 Solution 4(iv)

x² + k(2x + k – 1) + 2 = 0 x² + 2kx + k(k – 1) + 2 = 0 Comparing with ax² + bx + c = 0, we get a = 1, b = 2k, c = k(k – 1) + 2 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 5(i)

Find the values of k for which the roots are real and equal in each of the following equations:

2x² + kx + 3 = 0Solution 5(i)

Question 5(ii)

Find the values of k for which the roots are real and equal in each of the following equations:

kx (x – 2) + 6 = 0Solution 5(ii)

Question 5(iii)

Find the values of k for which the roots are real and equal in each of the following equations:

x² – 4kx + k = 0Solution 5(iii)

Question 5(iv)

Find the value of k for which the roots are real and equal in the following equation:

Solution 5(iv)

Question 5(v)

Find the value of p for which the roots are real and equal in the following equation:

px(x – 3) + 9 = 0Solution 5(v)

Question 5(vi)

Find the values of k for which the following equations have real roots.

4x² + kx + 3 = 0 Solution 5(vi)

4x² + kx + 3 = 0 Comparing with ax² + bx + c = 0, we get a = 4, b = k, c = 3 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.Solution 9(i)

Question 9 (ii)

Write all the values of k for which the quadratic equation x² + kx + 16 = 0 has equal roots. Find the roots of the equation so obtained.Solution 9 (ii)

Given quadratic equation is x² + kx + 16 = 0

As it has equal roots, the discriminant will be 0.

Here, a = 1, b = k, c = 16

Therefore, D = k² – 4(1)(16) = 0

i.e. k² – 64 = 0

i.e. k = ± 8

When k = 8, the equation becomes x² + 8x + 16 = 0

 or x² – 8x + 16 = 0

As D = 0, roots of the given equation are real and equal.Question 10

Find the values of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also, find these roots.Solution 10

Question 11

 If -5 is a root of the quadratic equation, 2x² + px – 15 = 0, and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.Solution 11

Question 12

 If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k.Solution 12

Question 13

If 1 is root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b = 0 has equal roots, find the value of b.Solution 13

Question 14

Find the value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots. Hence, find the roots of equation.Solution 14

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

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

Chapter 8 Quadratic Equations Exercise Ex. 8.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 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

The sum of a number and its square is 63/4. Find the numbers.Solution 11

Question 12

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

Solution 22

Question 23

Solution 23

Question 24

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.Solution 24

Let x be the natural number.

As per the question, we have

Therefore, x = 8 as x is a natural number.

Hence, the required natural number is 8.Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Find two consecutives odd positive integers, sum of whose squares is 970.Solution 34

Question 35

The difference of two natural numbers is 3 and the difference of their reciprocal is . Find the numbers.Solution 35

Question 36

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.Solution 36

Question 37

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.Solution 37

Question 38

The sum of the squares of two consecutive even numbers is 340. Find the numbers.Solution 38

Question 39

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is. Find the original fraction.Solution 39

Question 40

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.Solution 40

Chapter 8 Quadratic Equations Exercise Ex. 8.8

Question 1

Solution 1

Question 2

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?Solution 8

Question 9

Solution 9

Question 10

Solution 10

Concept Insight: Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.Question 11

Solution 11

Question 12

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.Solution 12

Question 13

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.Solution 13

Question 14

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream than to return downstream to the same spot. Find the speed of the stream.Solution 14

Question 15

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.Solution 15

Let the speed of a car be x km/hr. According to the question, time is hr. Distance = Speed × Time 2592 =  x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.Question 16

A motor boat whose speed instill water is 9 km/hr, goes 15 km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.Solution 16

Let x km/hr be the speed of the stream.

Therefore, we have

Downstream speed = (9 + x) km/hr

Upstream speed = (9 – x) km/hr

Distance covered downstream = distance covered upstream

Total time taken = 3 hours 45 minutes = hours

Therefore, x = 3 as the speed can’t be negative.

Hence, the speed of the motor boat is 3 km/hr.

Chapter 8 Quadratic Equations Exercise Ex. 8.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?Solution 8

Question 9

At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.Solution 9

Chapter 8 Quadratic Equations Exercise Ex. 8.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?Solution 4

Chapter 8 Quadratic Equations Exercise Ex. 8.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm². Find the sides of the squares.Solution 3

Question 4

Solution 4

Question 5

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.Solution 5

Question 6

Solution 6

Question 7

Sum of the areas of two squares is 640 m². If the difference of their perimeter is 64 m, find the sides of the two squares.Solution 7

Question 8

Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of two squares.Solution 8

Question 9

The area of a rectangular plot is 528 m². The length of the plot (in meters) is one meter more than twice its breadth. Find the length and the breadth of the plot.Solution 9

Question 10

In the centre of a rectangular lawn of dimension 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m². Find the length and breadth of the pond.Solution 10

Chapter 8 Quadratic Equations Exercise Ex. 8.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?Solution 5

Let us assume that the larger pipe takes ‘x’ hours to fill the pool.

So, as per the question, the smaller pipe takes ‘x + 10’ hours to fill the same pool.

Question 6

Two water taps together can fill a tank in  hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. Solution 6

Let the tap with smaller diameter takes x hours to completely fill the tank.

So, the other tap takes (x – 2) hours to fill the tank completely.

Total time taken to fill the tank  hours

As per the question, we have

When x = 5, then (x – 2) = 3

When  which can’t be possible as the time becomes negative.

Hence, the smaller diameter tap fills in 5 hours and the larger diameter tap fills in 3 hours.

Chapter 8 Quadratic Equations Exercise Ex. 8.13

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180. Find his marks in the two subjects.Solution 9

Question 10

Solution 10

Question 11

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.Solution 11

Question 12

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than  minutes. Find t.Solution 12

Chapter 8 Quadratic Equations Exercise 8.82

Question 1

If the equation x² + 4x + k = 0 has real and distinct root, then

(a) k < 4

(b) k > 4

(c) k ≥ 4

(d) k ≤ 4Solution 1

We know for the quadratic equation

ax² + bx + c = 0

condition for roots to be real and distinct is

D = b² – 4ac > 0       ……….(1)

for the given question

x² + 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 – 4k > 0

k < 4

So, the correct option is (a).

Chapter 8 Quadratic Equations Exercise 8.83

Question 2

If the equation x² – ax + 1 = 0 has two distinct roots, then

(a) |a| = 2

(b) |a| < 2

(c) |a| > 2

(d) None of theseSolution 2

For the equation x² – ax + 1 = 0 has two distinct roots, condition is

(-a)² – 4 (1) (1) > 0

a² – 4 > 0

a² > 4

|a| > 2

So, the correct option is (c).Question 3

Solution 3

Question 4

Solution 4

Question 5

If the equation ax² + 2x + a = 0 has two distinct roots, if

(a) a = ± 1

(b) a = 0

(c) a = 0, 1

(d) a = -1, 0Solution 5

For any quadratic equation

ax²  + bx + c = 0

having two distinct roots, condition is

b² – 4ac > 0

For the equation ax² + 2x + a = 0 to have two distinct roots,

(2)² – 4 (a) (a) > 0

4 – 4a² > 0

4(1 – a²) > 0

1 – a² > 0 since 4 > 0

that is, a² – 1 < 0

Hence -1 < a < 1, only integral solution possible is a = 0

So, the correct option is (b).Question 6

The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is

(a) 4

(b) 8

(c) 12

(d) 16Solution 6

For any quadratic equation

ax² + bx + c = 0

having real roots, condition is

b² – 4ac ≥ 0     …….(1)

According to question 

x² + kx + 64 = 0 have real root if

k² – 4 × 64 ≥ 0

k² ≥ 256

|k|≥ 16             ……..(2)

Also, x² – 8x + k = 0 has real roots if

64 – 4k ≥ 0

k ≤ 16    ………(3)

from (2), (3) the only positive solution for k is

k = 16

So, the correct option is (d).Question 7

Solution 7

Question 8

If 2 is a root of the equation x² + bx + 12 = 0 and the equation x² + bx + q = 0 has equal roots, then q =

(a) 8

(b) -8

(c) 16

(d) -16Solution 8

It is given that 2 is a root of equation x²  + bx + 12 = 0

Hence

(2)² + b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8        ………(1)

It is also given that x² + bx + q = 0 has equal root 

so, b² – 4(q) = 0         ……..(2)

from (1) & (2)

(-8)² – 4q =0

q = 16

So, the correct option is (c).Question 9

Solution 9

For any quadratic equation

ax² + bx + c = 0

Having equal roots, the condition is

b²  – 4ac = 0

For the equation

(a² + b²) x² – 2 (ac + bd)x + c² + d² = 0

to have equal roots, we have

(-2(ac + bd))² – 4 (c² + d²) (a² + b²) = 0

4 (ac + bd)² – 4 (a²c² + b²c² + d²a² + b²d²) = 0

(a²c² + b²d² + 2abcd) – (a²c² + b²c² + a²d²+ b²d²) = 0

2abcd – b²c² – a²d² = 0

b²c² – a²d² – 2abcd = 0

(bc – ad)² = 0

bc = ad

So, the correct option is (b).Question 10

Solution 10

For any quadratic equation

ax² + bx + c = 0

having equal roots, condition is

b² – 4ac = 0

According to question, quadratic equation is

(a² + b²)x² – 2b(a + c)x + b² + c² = 0

having equal roots, so

(2b(a + c))² – 4(a² + b²) (b² + c²) = 0

4b² (a + c)² – 4(a²b² + a²c² + b⁴ + b²c²) = 0

b²(a² + c² + 2ac) – (a²b² + a²c² + b²c² + b⁴) = 0

a²b² + b²c² + 2acb² – a²b² – a²c² – b²c² – b⁴ = 0

2acb² – a²c² – b⁴ = 0

a²c² + b⁴ – 2acb² = 0

(ac – b²)² = 0

ac = b²

So, the correct option is (b).Question 11

If the equation x² – bx + 1 = 0 does not possess real roots, then

(a) -3 < b < 3

(b) -2 < b < 2

(c) b < 2

(d) b < -2Solution 11

For any quadratic equation ax² + bx + c = 0 having no real roots, condition is

b² – 4ac < 0

For the equation, x² – bx + 1 = 0 having no real roots

b² – 4 < 0

b² < 4

-2 < b < 2

So, the correct option is (b).Question 12

If x = 1 is a common root of the equations ax² + ax + 3 = 0 and x² + x + b = 0, then ab =

(a) 3

(b) 3.5

(c) 6

(d) -3Solution 12

Question 13

If p and q are the roots of the equation x² -px + q = 0, then

(a) p = 1, q = -2

(b) b = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1Solution 13

If p, q are the roots of equation x² – px + q = 0, then p and q satisfies the equation

Hence

(p)² – p(p) + q = 0

p² – p² + q = 0

q = 0

and (q)² – p(q) + q = 0

q² – p(q) + q = 0

0 = 0

p can take any value

p = -2 and q = 0

So, the correct option is (c).Question 14

If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax² + bx + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12Solution 14

For the ax² + bx + 1 = 0 having real roots condition is

b² – 4(a) (1) ≥ 0

b² ≥ 4a

For a = 1

b² ≥ 4

b ≥ 2

b can take value 2, 3, 4

Here, 3 possible solutions are possible            …….(1)

For a = 2

b² ≥ 8

Here, b can take value 3, 4            ……(2)

Here, 2 solutions are possible

For a = 3

b² ≥ 12

possible value of b is 4

Hence, only 1 possible solution      ……(3)

For a = 4

b² ≥ 16

possible value of b is 4

Hence, only 1 possible solution    …….(4)

from (1), (2), (3), (4)

Total possible solutions are 7

So, the correct option is (b).Question 15

The number of quadratic equations having real roots and which do not change by squaring their roots is

(a) 4

(b) 3

(c) 2

(d) 1Solution 15

Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.

Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.

So, the correct option is (c).

Chapter 8 Quadratic Equations Exercise 8.84

Question 16

If (a² + b²) x² + 2(ab + bd) x + c² + d² = 0 has no real roots, then

(a) ad = bc

(b) ab = cd

(c) ac = bd

(d) ad ≠ bcSolution 16

If any quadratic equation ax² + bx + c has no real roots then b² – 4ac < 0    ……(1)

According to the question, the equation is

(a² + b²) x² + 2(ac + bd) x + c² + d² = 0

from (1)

4(ac + bd)² – 4(a² + b²) (c² + d²) < 0

a²c² + b²d² + 2abcd – (a²c² + a²d² + b²c² + b²d²) < 0

a²c² + b²d² + 2abcd – a²c² – a²d² – b²c² – b²d² < 0

2abcd – a²d² – b²c² < 0

-(ad – bc)² < 0

(ad – bc)² > 0

For this condition to be true ad ≠ bc

So, the correct option is (d).Question 17

Solution 17

Question 18

If x = 1 is a common root of ax² + ax + 2 = 0 and x² + x + b = 0 then, ab =

(a) 1

(b) 2

(c) 4

(d) 3Solution 18

It is given that x = 1 is root of equation ax² + ax + 2 = 0

Hence, a(1)² + a(1) + 2 = 0

           2a + 2 = 0

           a = -1         ……(1)

It is given that x = 1 is also root of x² + x + b = 0

Hence, (1)² + (1) + b = 0

           b = -2     ……(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

If sin α and cos α are the roots of the equation ax² + bx + c = 0, then b² = 

(a) a² – 2ac

(b) a² + 2ac

(c) a² – ac

(d) a² + acSolution 22

Question 23

If 2 is a root of the equation x² + ax + 12 = 0 and the quadratic equation x²  + ax + q = 0 has equal roots, then q =

(a) 12

(b) 8

(c) 20

(d) 16Solution 23

Given, 2 is a root of equation x² + ax + 12 = 0

so (2)² + a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8     …..(1)

Given x² + ax + q = 0 has equal roots so

a² – 4q = 0   ……(2)

from (1) & (2)

(-8)² – 4q = 0

4q = 64

q = 16

So, the correct option is (d).Question 24

If the sum of the roots of the equation x² – (k + 6)x + 2(2k – 1) = 0 is equal to half of their product, then k =

(a) 6

(b) 7

(c) 1

(d) 5Solution 24

Question 25

If a and b are roots of the equation x² + ax + b = 0, then a + b =

(a) 1

(b) 2

(c) -2

(d) -1Solution 25

If a and b are roots of the equation x² + ax + b = 0

Then, sum of roots = -a

a + b = -a

2a + b = 0          …….(1)

product of roots = b

ab = b

ab – b = 0

b(a – 1) = 0         ……..(2)

from (1) and (2)

-2a(a – 1) = 0…(From (1), we have b = -2a)

a(a – 1) = 0

a = 0 or a = 1

if a = 0  b = 0

if a = 1  b = -2

Now a and b can’t be zero at same time, so correct solution is

a = 1 and b = -2

a + b = -1

So, the correct option is (d).Question 26

A quadratic equation whose one root is 2 and the sum of whose roots is zero, is

(a) x² + 4 = 0

(b) x² – 4 = 0

(c) 4x² – 1 = 0

(d) x² – 2 = 0Solution 26

Given sum of roots is zero and one root is 2.

So the other root must be -2

so any quadratic equation having root 2 and -2 is

(x – 2) (x – (-2)) = 0

(x – 2) (x + 2) = 0

x² – 4 = 0

So, the correct option is (b).Question 27

If one root of the equation ax² + bx + c = 0 is three times the other, then b² : ac =

(a) 3 : 1

(b) 3 : 16

(c) 16 : 3

(d) 16 : 1Solution 27

Question 28

If one root of the equation 2x² + kx + 4 = 0 is 2, then the other root is

(a) 6

(b) -6

(c) -1

(d) 1Solution 28

Question 29

If one root of the equation x² + ax + 3 = 0 is 1, then its other root is

(a) 3

(b) -3

(c) 2

(d) -2Solution 29

x² + ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

Chapter 8 Quadratic Equations Exercise 8.85

Question 30

If one root of the equation 4x² – 2x + (λ – 4) = 0 is a reciprocal of the other, then λ =

(a) 8

(b) -8

(c) 4

(d) -4Solution 30

Question 31

Solution 31

Question 32

Solution 32

Any quadratic equation, ax² + bx + c = 0 has real and equal roots if b² – 4ac = 0

For the question, equation is 16x² + 4kx + 9 = 0,

(4k)² – 4 × 16 × 9 = 0

16k² – 36 × 16 = 0

k² – 36 = 0

k² = 36

k = ± 6

So, the correct option is (c).