Category: Mathematics

Circles Exercise 10.5 – Class 10

Circles Exercise 10.5 – Questions:

  1. Draw a circle of radius 6 cm and construct tangents to it from an external point 10 cm away from the centre. Measure and verify the length of the tangents.
  2. Construct a pair of tangents to a circle of radius 3.5 cm from a point 3.5 cm away from the circle.
  3. Construct a tangent to a circle of radius 5.5 cm from a point 3.5 cm away from it.
  4. Draw a pair of perpendicular tangents of length 5 cm to a circle.
  5. Construct tangents to two concentric circles of radii 2 cm and 4 cm from a point 8 cm away from the centre.

Circles Exercise 10.5 – Solutions:

  1. Draw a circle of radius 6 cm and construct tangents to it from an external point 10 cm away from the centre. Measure and verify the length of the tangents.

Solution:

Circles Exercise 10.5

Lent of the tangents, t = √(d2 – r2)

Length of the tangents, t = √(102 – 62) = √(100 – 36) =  √64  = 8 cm


  1. Construct a pair of tangents to a circle of radius 3.5 cm from a point 3.5 cm away from the circle.

Solution:

Circles Exercise 10.5


  1. Construct a tangent to a circle of radius 5.5 cm from a point 3.5 cm away from it.

Solution:

3


  1. Draw a pair of perpendicular tangents of length 5 cm to a circle.

Solution:

Circles Exercise 10.5


  1. Construct tangents to two concentric circles of radii 2 cm and 4 cm from a point 8 cm away from the centre.

Solution:

Circles Exercise 10.5

Circles Exercise 10.4 – Chapter Circles – Class 10

Circles Exercise 10.41.In the figure PQ, PR and BC are the tangents to the circle BC touches the circles at X. If PQ = 7cm find the perimeter of ∆PBC

Solution:

The perimeter of ∆PBC = PC + PB + BC

= PC + PB + BX + CX

But Cx = CR and BX = BQ [Since, tangents drawn from an external point]

Therefore, PC + PB + BQ + CR = (PC + CR) + (PB + BQ)

= PR + PQ

= 7 + 7 [Since PR = PQ tangents drawn from an external point]

= 14 cm


  1. Two concentric circles of radii 13 cm and 5 cm are drawn. find the length of the chord of the outer circle which touches the outer circle.

Solution:

Circles Exercise 10.4

Given, two concentric circles of radii 13 cm and 5 cm. We have to find the length of the chord of the outer circle which touches the outer circle.

Let O be the centre of the circles. AB is the tangent drawn to an inner circle through the point P.

We have AP = PB [Since the perpendicular drawn from the point of contact devices the chord equally]

In ∆OAP, ∠OPA = 90˚

AP2 + OP2 = OA2 [ Pythagoras Theorem]

AP2 + 52 = 132

AP2 + 25 = 169

AP2 = 169 – 25 = 144

AP = 12 cm

Therefore, AB = AP + PB = 12 + 12 = 24 cm


  1. Circles Exercise 10.4In the given ∆ABC AB = 12cm, BC = 8 cm and AC = 10 cm Find AF . BD and CE

Solution:

In ∆ABC,

AB = 12cm ; BC = 8cm ; AC = 10 cm

AB + BC + CA = 12 + 8 + 10 = 30

AD +  BD + BE + CE + AF + CF = 30

But AF = AD = x [ Tangents drawn from an external point]

BE = BD = y [ Tangents drawn from an external point]

CE = CF = z[ Tangents drawn from an external point]

x + y + y + z + z + x = 30

2x + 2y + 2z = 30

x + y + z = 15 …………..(1)

AB = x + y = 12 cm ; BC = x + y = 8cm ; AC = x + z = 10cm

From (1), 12 + z = 15

⇒ z = 15 – 12 = 3 cm

From (1), x + 8 = 15

⇒ x = 15 – 8 = 7 cm

From (1), y + 10 = 15

⇒ y = 15 – 10 = 5 cm

AF = x = 10 cm

BD = y = 5 cm

CE = z = 3cm


  1. Circles Exercise 10.4In the given quadrilateral ABCD, BC = 38 cm, QB = 27 cm , DC = 25 cm and AD⊥DC find the radius of the circle.

Solution:

In the figure OPDS,

DS = DP [Tangents drawn from an external point]

OP = OS [ Radius of circle]

∠D = 90˚ [AD⊥DC]

Therefore, OPDS is a square.

In the figure,

BQ = BR = 27 cm [Tangents drawn from an external point]

CR = 38 – 27 = 11 cm = CS [Tangents drawn from an external point]

DS = 25 – 11 = 14 cm = DP [Tangents drawn from an external point]

Radius of the circle = OP = OS = 14 cm [OPDS is a square]

 

  1. Circles Exercise 10.4In the given figure AB = BC, ∠ABC = 68˚. DA and DB are the tangents to the circle with centre are the tangents to the circle with centre O. Calculate the measure of

(i) ∠ACB

(ii) ∠AOB

(iii) ∠ADB

Solution:

In the figure, AB = BC, ∠ABC = 68˚

(i) ∠ABC + ∠ACB + ∠BAC = 180˚

∠ACB + ∠BAC = 180˚ – ∠ABC

∠ACB + ∠ACB = 180˚ – ∠ABC [AB = BC]

2∠ACB = 180˚ – 68˚

∠ACB = 180˚ – 68˚/2 = 112˚/2 = 56˚

(ii) ∠AOB = 2∠ACB = 2 x 56˚ = 112˚

(iii) ∠ADB = 180˚ – ∠AOB

= 180˚ – 112˚

= 68˚

 

2. Riders based on tangent properties:

1. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD +BC.

Solution:

Circles Exercise 10.4AP = AS = l

BS = BR = m

CR = CQ = n

DP = DQ = o

AB + CD = AS + SB +  CQ + QD = l + m + n + o …………….(1)

AD + BC = AP + PD + BR + RC

l + m + n +  o = l + o + m +  n ……………(2)

From (1) and (2),

AB + CD = AD + BC

 

  1. Circles Exercise 10.4Tangents AP and AQ are drawn to circle with centre O, from an external point A . Prove that ∠PAQ = 2. ∠OPQ

Solution:

∠POQ + ∠PAQ = 180˚ [Angle between tangents + angle in the centre = 180˚] …………(1)

In ∆POQ , ∠POQ + ∠OPQ + ∠OQP = 180˚ [Sum of the angles of a triangle = 180˚]

Therefore,

∠POQ + 2∠OPQ = 180˚ [ Since ∠OPQ = ∠OQP] ………(2)

From (1) and (@)

∠POQ + ∠PAQ = ∠POQ + 2∠OPQ

∠PAQ = 2∠OPQ

 

  1. Circles Exercise 10.4In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that,

(a) tangent at P bisects AB at Q

(b) ∠APB = 90˚

Solution:

(i)

In the figure

QA = QP [tangents drawn from an external point] ………..(1)

QB = QP [tangents drawn from an external point] ………..(2)

From (1) and (2),

QA = QB

Therefore, P bisects AB at Q.

(ii)

In ∆APB,

∠QAP = ∠APQ = x [QA = QP]

∠QPB = ∠BPQ = y [QB = QP]

Therefore, In ∆APB x + x + y + y = 180˚ [Sum of the angles of a triangle]

2x + 2y = 180˚

x + y = 90˚

∠APB = 90˚

 

  1. Circles Exercise 10.4A pair of perpendicular tangents are drawn to a circle from an external point. Prove that length of each tangent is equal to the radius of the circle.

Solution:

PA = PB [tangents drawn from an external point]

OA = OB {Radius of the circle]

∠APB = 90˚ [Given]

∠OAP = ∠OBP = 90˚ [The sum of a quadrilateral = 360˚]

Therefore, OABP is a square

Hence, tangents PA and PB = Radius of the circle OA and OB

 

  1. Circles Exercise 10.4If the sides of a parallelogram touch a circle. Prove that the parallelogram is a rhombus.

Solution:

Let ABCD be a parallelogram

Therefore, AB||CD, AB = CD [Given]

AD||BC , AD = BC[Given]

AP = AS, PB = BQ [tangents drawn from an external point]

DS = DR, QC = RC

AB + CD = AP + PB + CR + DR

AB + CD = AS + BQ + QC + DS

 

  1. Circles Exercise 10.4In the figure if AB = AC prove that BQ = QC

Solution:

AP = AR …………….(1) [tangents drawn from an external point]

and AB = AC ……….(2)[Given]

From (1) and (2),

AB – AP = AC – AR

BP = CR

But BQ = BP and CQ = CR [tangents drawn from an external point]

Therefore, BQ = CQ.

Circles Exercise 10.2 Solutions – Chapter Circles – Class 10

Property related to the chords of a circle – Circle

  • Equal chords of a circle are equidistant from the centre
  • If the chords of a circle are at equal distance from the centre, then they are equal in length
  • If the length of the chord increases, its perpendicular distance from the centre decreases.
  • If the length of the chord decreases, its perpendicular distance from the centre increases
  • The perpendicular distance between the biggest chord and the centre is zero.

Segment of circle – Circle Exercise 10.2

Angles in a segment:

  • In a circle, an angle in the minor segment is an obtuse angle.
  • In a circle, an angle in the major segment is an acute angle.
  • An angle in a semi circle is a right angle.
  • In a circle, angles in the same segment are equal.

Circles Exercise 10.2 – Questions:

  1. Draw three consecutive circles of radii 1.5 cm, 2.5 cm and 3.5 cm with O as centre.
  2. With O1 and O2 as centres draw two circles of same radii 3 cm and with the distance between the two centres equal to 5 cm
  3. Draw a line segment AB = 8 cm and mark its midpoint as C. With 2 cm as radius draw three circles having A, B and C as centres. Draw another circle with C as centre and 4 cm radius draw another circle. Identify the name and the concentric circles and congruent circles
  4. Draw a circle of radius 4 cm and construct a chord of 6 cm length in it. draw two angles in major segment and two angles in minor segment. Verify that the angles by measuring them.
  5. Draw a circle with centre O and radius 3.5 cm and draw a diameter AOB in it. Draw angles in semi- circles on either side of the diameter. Measure the angles and verify that they are right angles.

Circles Exercise 10.2 – Solutions:

  1. Draw three concentric circles of radii 1.5 cm, 2.5 cm and 3.5 cm with O as centre.

Circles Exercise 10.2

2. With O1 and O2 as centres draw two circles of same radii 3 cm and with the distance between the two centres equal to 5 cm

Solution:

Circles Exercise 10.2

3. Draw a line segment AB = 8 cm and mark its midpoint as C. With 2 cm as radius draw three circles having A, B and C as centres. Draw another circle with C as centre and 4 cm radius draw another circle. Identify the name and the concentric circles and congruent circles

Solution:

Circles Exercise 10.2

4. Draw a circle of radius 4 cm and construct a chord of 6 cm length in it. draw two angles in major segment and two angles in minor segment. Verify that the angles by measuring them.

Solution:

Circles Exercise 10.2

5. Draw a circle with centre O and radius 3.5 cm and draw a diameter AOB in it. Draw angles in semi- circles on either side of the diameter. Measure the angles and verify that they are right angles.

Solution:

Circles Exercise 10.2

Circles Exercise 10.1 – Class 10

Construction of a chord of given length – Circles Exercise 10.1

Circles Exercise 10.1 – Questions:

  1. Draw a circle of radius 3.5 cm and construct a chord of length 6 cm in it. Measure the distance between the centre and the chord.
  2. Construct two chords of length 6 cm and 8 cm on the same side of the centre of a radius 4.5 cm. Measure the distance between the centre and the chords.
  3. Construct two chords of length 6.5 cm on either side of the centre of a circle of radius 4.5 cm. Measure the distance between the centre and the chords.
  4. Construct two chords of length 9 cm and 7 cm on either side of the centre of a circle of radius 5 cm. Measure the distance between the centre and the chords.

Circles Exercise 10.1 – Solutions:

  1. Draw a circle of radius 3.5 cm and construct a chord of length 6 cm in it. Measure the distance between the centre and the chord.

Circles Exercise 10.1

2. Construct two chords of length 6 cm and 8 cm on the same side of the centre of a radius 4.5 cm. Measure the distance between the centre and the chords.

Solution:

Circles Exercise 10.1

3. Construct two chords of length 6.5 cm on either side of the centre of a circle of radius 4.5 cm. Measure the distance between the centre and the chords.

Circles Exercise 10.1

4. Construct two chords of length 9 cm and 7 cm on either side of the centre of a circle of radius 5 cm. Measure the distance between the centre and the chords.

Circles Exercise 10.1

Quadratic Equations Exercise 9.11 – Class X

Quadratic Equations Exercise 9.11 – Questions

  1. Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.
  2. Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
  3. The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15
  4. A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number
  5. Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.
  6. The ages of Kavya and Karthik are 11 years and 14 years. In how many years times will the product of their ages be 304?
  7. The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.
  8. The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.
  9. The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2 . Find its base and height.
  10. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.
  11. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, fibd the lengths of all three sides of the triangle.
  12. A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
  13. A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
  14. Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?

Quadratic Equations Exercise 9.11 – Solutions:

  1. Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.

Solution:

Let one of the odd positive number be x.

The other odd positive number be (x + 2).

The sum of the numbers = x2 + (x + 2)2 = 130

x2 + x2 + 2x + 4 = 130

2x2 + 2x + 4 = 130

2x2 + 2x – 126 = 0

x2 + x – 63 = 0

x2 + 9x – 7x – 63 = 0

x(x + 9) -7(x + 9) = 0

(x – 7)(x + 9) = 0

x = 7 or x = -9

Therefore, one of the odd positive number, x = 7 and the other odd positive number is x + 2 = 7 + 2 = 9

Therefore, two consecutive positive odd numbers 7 and 9.

  1. Find the whole number such that four times the number subtracted from three times the square of the number makes 15.

Solution:

Let the whole number be x.

Four times the number be 4x and three times the number be 3x2

Therefore,

4x – 3x2 + 15 = 0

-3x2 + 4x + 15 = 0

-3x2 + 9x – 5x + 15 = 0

3x(x – 3)-5(x – 3) = 0

(x – 3)(3x – 5) = 0

x = 3 or 3x = 5

x = 3 or x = 5/3

Therefore the whole number is 3.

  1. The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15

Solution:

The sum of two natural numbers is 8. i.e., x + y = 8

x = 8 – y

The sum of their reciprocals is 8/15 i.e., 1x + 1/y = 8/15

1x + 1/y = 8/15

y + x/xy = 8/15

(x + y)15 = 8xy

15 x 8 = 8 xy

xy = 15

(8 – y)y = 15

8y – y2 = 15

-y2 + 8y – 15 = 0

-y2 + 3y + 5y – 15 = 0

-y(y – 3)+5(y – 3) = 0

(y – 3)(-y + 5) = 0

y = 3 or –y + 5 = 0

y = 3 or y = 5

Therefore the numbers be 3, 5 .

  1. A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number

Solution:

Let ten’s digit of the number = x and its unit digit = y. Then two digit number is 10x + y

xy = 12 and 10x + y + 36 = 10y + x

y = 12/x and 9x + 36 = 9y i.e., x + 4 = y

Substituting y = 12/x in x + 4 = 4; we get,

x + 4 = 12/x

x2 +  4x – 12 = 0

x2 +6x – 2x – 12 = 0

x(x + 6)-2(x + 6) = 0

(x + 6)(x – 2) = 0

x = -6 and x = 2

Since x is a digit, therefore, x = 2 and y = 12/x = 12/2 = 6

Therefore the required number is 10x + y = 10×2 + 6 = 26

  1. Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.

Solution:

Let three consecutive numbers be x, x + 1, x + 2

x2 +(x+1)(x+2) = 154

x2 + (x2 + 2x + x + 2) = 154

2x2 + 3x + 2 = 154

2x2 + 3x – 152 = 0

2x2 + 19x – 16x – 152 = 0

x(2x + 19) – 8(2x +19) = 0

(2x + 19)(x – 8) = 0

2x + 19 = 0 or x – 8 = 0

2x = -19 or x = 8

x = –19/2 or x = 8

Therefore, three consecutive numbers be 8, 9 and 10.

  1. The ages of Kavya and Karthik are 11 years and 14 years. In how many years time will the product of their ages be 304?

Solution:

The age of Kavya be 11 years and the age of the Karthik be 14 years.

In x years the product of their ages will be 304. i.e., (x + 11)(x + 14) = 304

x2 + 14x + 11x + 154 = 304

x2 + 25x + 154 – 304 = 0

x2 + 25x + 150 = 0

x2 + 30x – 5x + 150 = 0

x(x + 30) – 5(x + 30) = 0

(x + 30)(x – 5) = 0

x + 30 = 0 or x – 5 = 0

x = -30 or x = 5

Therefore, in 5 years the product of their ages will be 304.

  1. The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.

Solution:

Let the age of the mother be x.

Hence the present age of son = 2x2

Given the age of the mother is twice the square of the age of her son i.e., x = 2x2

Age of son after 8 years = (x + 8)years

Age of mother after 8 years = (2x2 + 8) years

Given that, (2x2 + 8) = 3(x + 8) + 4

2x2 + 8 = 3x + 24 + 4

2x2 – 3x + 8 – 28 = 0

2x2 – 3x – 20 =0

2x2 – 8x + 5x – 20 = 0

2x(x – 4) +5(x – 4) = 0

(2x + 5)(x – 4) = 0

x – 4 = 0 or 2x + 5 = 0

x = 4 or 2x = -5

x = 4 or x = –5/2

Thus, x = 4

Hence present age of son is 4 years and present age of  mother is 2x2 = 2(4)2 = 32.

  1. The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.

Solution:

A = 56 m2

56 = (x + 5)(x – 5)

56 = x2 – 5x + 5x – 25

56 = x2 – 25

56 + 25 = x2

81 = x2

x = √81 = 9

Therefore, base of the rectangle = x + 5 = 9 + 5 = 14

height of the rectangle = x – 5 = 9 – 5 = 4

  1. The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2. Find its base and height.

Solution:

Let the base of the triangle be x.

Then altitude of the triangle = x + 6

Area of the triangle = 108 cm2

Formula to find the area of triangle = 1/2 x base x height

108 = 1/2 x(x + 6)

216 = x(x + 6)

216 = x2 + 6x

x2 + 6x – 216 = 0

x2 + 18x – 12x – 216 = 0

x(x + 18) -12(x + 18) = 0

(x – 12)(x + 18) = 0

x – 12 = 0 or x + 18 = 0

x = 12

Let the base of the triangle be 12cm

Then height of the triangle 18cm

  1. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.

Solution:

. In rhombus ABCD, the diagonals AC and BD intersect at E. Therefore, angles at E are right angles. Hence, all the triangles formed by the diagonals AC and BD are right angled triangle.

In right angled triangle ABE, by Pythagoras theorem. We have

AB2 = AE2 + EB2

(x + 8)2 = x2 + (x + 7)2

x2 + 16x + 64 = x2 + x2 + 14x + 49

x2 + 16x + 64 – x2 – x2 – 14x – 49 = 0

-x2 + 2x + 15 = 0

-x2 + 5x – 3x + 15 = 0

x(-x + 5) +3(-x + 5) = 0

(-x + 5)(x + 3) = 0

x + 3 = 0 or –x +5 = 0

x = -3 or x = 5

The diagonal AC = AE + EC = 2(AE) = 2x = 10 cm

The diagonal BD = BE + ED = 2(BE)= 2(x+7) = 2(5 +7) = 2(12) = 24 cm

 

  1. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.

Solution:

Given, AB = BC = 2x + 1

In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.

Therefore triangle BCD is a right angled triangle.

By Pythagoras theorem, BC2 = BD2 + CD2

(2x + 1)2 = (2x – 1)2 + x2

4x2 + 4x  + 1 = 4x2 – 4x + 1 + x2

4x2 + 4x  + 1 – 4x2 + 4x – 1 – x2 = 0

-x2 + 8x = 0

x(-x + 8) = 0

-x + 8 = 0

x = 8

We know, AB = BC = 2x + 1 = 2(8) + 1 = 17cm

AC = AD + DC = 2(DC) = 2x = 2(8) = 16 cm

  1. A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.

Solution:

Let the speed of the stream be x km/hr. Then speed downstream = (15 + x) kn/hr

Speed upstream = (15 –x )km/hr

given, it took 4 hours 30 minutes to travel back to same place.

So, we have,

30/15 + x30/15 – x = 41/2

30*30/(15+x)(15-x) = 41/2

900/225 – x^2 = 9/2

900 *2 = (225 – x2)*9

1800 = 2025 – 9x2

9x2 = – 1800 + 2025

9x2 = 225

x2 = 225/9

x2 = 25

x = 5km/hr

  1. A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

Solution:

Let x be the cost price of the article.

The selling price of the article Rs. 24

x = 24 – x/x * 100

x2 = (24 – x)100

x2 = 2400 – 100 x

x2 + 100x – 2400 = 0

x2 + 120x – 20x – 2400 = 0

x(x + 120) -20(x + 120) = 0

(x + 120)(x – 20) = 0

x = -120 or x = 20

Therefore, cost price of the article is 20 Rs.

  1. Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?

Solution:

Let Shwetha take x – 6 days and Ankitha takes x days.

Shwetha one day work = 1/(x – 6) and Ankitha one day work = 1/x

Given, (A + B) can finish work in 4 day.

1/x + 1/(x – 6)  = 1/4

(x-6)+x/x(x-6) = 1/4

2x -6/x(x – 6) = 1/4

4(2x – 6) = x(x – 6)

8x – 24 = x2 – 6x

x2 – 6x – 8x + 24 = 0

x2 – 14x + 24 = 0

x2 – 12x – 2x + 24 = 0

x(x – 12)-2(x – 12) = 0

(x – 12)(x – 2) = 0

x = 12 days taken by Ankitha alone.

Quadratic Equations Exercise 9.10 – Class X

I.Solve the following equations graphically.

(i) x2 – 4x = 0

(ii) x2 + x – 12 = 0

(iii) x2 – x – 2 = 0

(iv) x2 – 5x + 6 = 0

II.

  1. Draw the graph of y = x2 and find the value of √3
  2. Draw the graph of y = 2x2 and find the value of √7
  3. Draw the graph of y = 1/2y2 and find the value of √10

Quadratic Equations – Exercise 9.10 – Solutions:

I.Solve the following equations graphically.

(i) x2 – 4x = 0

Solution:

Prepare the table of the values for the equation y = x2 – 4x

x-1012345
y50-3-4-305

Quadratic Equation Exercise 9.10

The parabola intersects the x – axis at (0, 0) and (4, 0)

Therefore, the roots of equation are 0 and 4.


 (ii) x2 + x – 12 = 0

Solution:

Prepare the table of the values for the equation  x2 + x – 12 = 0

x-4-3-2-101234
y06-10-12-12-10-608

Quadratic Equation Exercise 9.10
The parabola intersects the x – axis only at (3, 0) and (-4, 0)

Therefore, the roots of equation are 3 and -4.


(iii) x2 – x – 2 = 0

Solution:

Prepare the table of the values for the equation  x2 – x – 2 = 0

x-2-10123
y40-2-204

Quadratic Equation Exercise 9.10

The parabola intersects the x – axis at (-1, 0) and (2, 0)

Therefore, the roots of equation are -1 and 2.


 (iv) x2 – 5x + 6 = 0

Solution:

Prepare the table of the values for the equation y = x2 – 5x + 6

x012345
y620026

Quadratic Equation Exercise 9.10

The parabola intersects the x – axis at (2, 0) and (3, 0)

Therefore, the roots of equation are 2 and 3.

 


II.

  1. Draw the graph of y = x2 and find the value of √3

Solution:

y = x2

x0-11-22√3
y011443

Quadratic Equation Exercise 9.10

When x = √3, y = (√3)2 = 3

Draw a straight line y = 3 parallel to x-axis

The point on x – axis at which the perpendiculars meets are the values of √3 , x = ±1.7

Therefore, √3 = ±1.7

  1. Draw the graph of y = 2x2 and find the value of √7

Solution:

y = 2x2

x0-11-22-33√7
y02288181814

Quadratic Equation Exercise 9.10

When x = √7, y = 2(√7)2 = 14

Draw a straight line y = 14 parallel to x-axis.

The point on x – axis at which the perpendiculars meets are the values of √7 , x = ±2.6

Therefore, √7 = ±2.6

 

  1. Draw the graph of y = 1/2x2 and find the value of √10

Solution:

y = 1/2x2

x0-22-44√10
y022885

Quadratic Equation Exercise 9.10

When x = √10, y = 1/2(√10)2 = 5

Draw a straight line y = 5 parallel to x-axis.

The point on x – axis at which the perpendiculars meets are the values of √10 , x = ±3.1

Therefore, √10 = ±3.1

Quadratic Equation Exercise 9.8 – Class X

Quadratic Equations – Exercise 9.8 – Questions:

  1. Form the quadratic equation whose roots are
  2. i) 3, 5
  3. ii) 6, -5

iii) -3, 3/2

  1. iv) 2/3 , 3/2
  2. v) (2 + √3), (2 – √3)
  3. vi) (-3 + 2√5)(-3 – 2√5)

B.

  1. If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) 1/m + 1/n

(iii) m3n2 + n3m2

(iv) 1/n1/m

  1. IF a and b are the roots of the equation 3m2 = 6m + 5, find the value of

(i) a/b + b/a

(ii) (a + 2b)(2a + b)

  1. If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of

(i) (p + q)2 + 4pq

(ii) p3 + q3

  1. Form a quadratic equation whose roots are p/q and q/p
  2. Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root which is twice the other.
  3. Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.
  4. Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.
  5. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

Quadratic Equations – Exercise 9.8 – Solution:

  1. Form the quadratic equation whose roots are

i) 3, 5

Solution:

Let m and n be the roots

Then m = 3 and n = 5

Sum of the roots = m + n = 3 + 5 = 8

Product of the roots = mn = 3 x 5 = 15

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – 8x + 15 = 0

 

ii) 6, -5

Solution:

Let m and n be the roots

Then m = 6 and n = -5

Sum of the roots = m + n = 6 – 5 = 1

Product of the roots = mn = 6 x (-5) = -30

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – 1x + (-30) = 0

x2 – x – 30 = 0

 

iii) -3, 3/2

Solution:

Let m and n be the roots

Then m = -3 and n = 3/2

Sum of the roots = m + n = -3 + 3/2 = -6+3/2 = –3/2

Product of the roots = mn = -3 x 3/2 = -9/2

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (-3/2) x + (-9/2) = 0

x2 + 3/2x – 9/2 = 0

 

iv) 2/3 , 3/2

Solution:

Let m and n be the roots

Then m = 2/3 and n = 3/2

Sum of the roots = m + n = 2/3 + 3/2 = 4+9/6 = 13/6

Product of the roots = mn = 2/3 x 3/2 =1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (13/6) x + (1) = 0

x213/6x + 1 = 0

 

v) (2 + √3), (2 – √3)

Solution:

Let m and n be the roots

Then m = 2 + √3 and n = 2 – √3

Sum of the roots = m + n = (2 + √3)+( 2 – √3) = 4

Product of the roots = mn = (2 + √3)( 2 – √3) = 4 – 2√3 + 2√3 – 3 = 4 – 3 = 1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (4) x + (1) = 0

x2 – 4x + 1 = 0

 

vi) (-3 + 2√5)(-3 – 2√5)

Solution:

Let m and n be the roots

Then m = -3 + 2√5 and n = -3 – 2√5

Sum of the roots = m + n = (-3 + 2√5)+( -3 – 2√5) = -6

Product of the roots = mn = (-3 + 2√5)( -3 – 2√5) = 9 + 6√5 – 6√5 – 4×5 = 9 – 45 = – 36

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (-6) x + (-36) = 0

x2 + 6x – 36 = 0


B.

  1. If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) 1/m + 1/n

(iii) m3n2 + n3m2

(iv) 1/n1/m

Solution:

The given quadratic equation is x2 – 6x + 2 = 0 which is of the form ax2 + bx + c = 0 where a = 1, b = -6 and c = 2

Quadratic Equation Exercise 9.8

Therefore, m = 3 + √7 and n = 3 – √7

(i) (m + n)mn

= (3+ √7+ 3 – √7)( 3 + √7)( 3 – √7)

= 6 (3 + √7)( 3 – √7)

=6 (9 – 3√7 + 3√7 – 7)

= 6 x 2

= 12

 

(ii) 1/m + 1/n

= 1/(3 + √7) + 1/(3 – √7)

= (3 – √7)+(3 + √7)/(3 – √7)(3+√7)

= 6/2

= 3

 

(iii) m3n2 + n3m2

= (3 + √7)3( 3 – √7)2 + (3 – √7)3( 3 + √7)2

= (90 + 34√7)(16 – 6√7) + (90 – 34√7)(16 + 6√7)

= 12 + 4√7 + 12 – 4√7

= 24

 

(iv) 1/n1/m

= 1/(3 – √7)1/(3 + √7)

= (3 + √7)-(3 – √7)/(3 – √7)(3+√7)

= 2√7/2

= √7


  1. If a and b are the roots of the equation 3m2 = 6m + 5, find the value of

(i) a/b + b/a

(ii) (a + 2b)(2a + b)

Solution:

The given equation is 3m2 = 6m + 5, this can be written as 3m2 – 6m – 5 = 0, which is of the form mx2 + nx + p = 0 where m = 3 ; n = -6 and  p = -5

Quadratic Equation Exercise 9.8

Quadratic Equation Exercise 9.8


  1. If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of

(i) (p + q)2 + 4pq

(ii) p3 + q3

Solution:

Given p and q are the roots of the equation is 2a2 – 4a + 1 = 0, which is of the form ax2 + bx + c = 0 where a = 2, b = -4 , x = 1

Quadratic Equation Exercise 9.8

  1. Form a quadratic equation whose roots are p/q and q/p

Solution:

Let p/q and q/p be the roots of a quadratic equation

Sum of the roots = m + n = p/q + q/p = p^2+q^2/pq

Product of the roots = mn = p/q x q/p = 1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (p^2+q^2/pq) x + (1) = 0

pqx2 – (p2 + q2)x – pq = 0


  1. Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root equal to zero.

Solution:

The given quadratic equation is x2 + 4x + (k + 2) = 0 where a = 1, b = 4 , c = k+2

The sum of the roots m+n = –b/a = – 4/1 = -4

The product of the roots mn = c/a = k+2/1 = k + 2

If one root is m then other root is zero.

Thus, m = m and n = 0

Therefore, m + n = – 4 ⇒ m + 0 = -4 ⇒m = -4

mn = k + 2 ⇒ 0 = k + 2 ⇒ k = -2


  1. Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.

Solution:

The given quadratic equation is 2x2  – 3qx + 5q = 0 where a = 2, b = -3q , c = 5q

The sum of the roots m+n = –(-3q)/2 = 3q/2

The product of the roots mn = c/a = 5q/2

If one root is m then other root is 2m.

Thus, m = m and n = 2m

Therefore, m + 2m = 3q/2 ⇒ 3m = 3q/2 ⇒m = q/2

mn = 5q/2 ⇒ 2m2 = (5q/2)

⇒ 2(q/2)2 = 5q/2

⇒ 2(q^2/4) = 5q/2

q^2/4 = 5q/4

⇒q2 = 5q

⇒q = 5


  1. Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.

Solution:

The given quadratic equation is 4x2 – 8px + 9 = 0 where a = 4, b = -8p , c = 9

The sum of the roots m+n = –(-8p)/4 = 2p

The product of the roots mn = c/a = 9/4

If one root is m then other root is (m – 4).

Thus, m = m and n = (m – 4)

m+n = 2p⇒ m+m-4 = 2p⇒ 2m = 2p + 4⇒m = p + 2

mn = 9/4 ⇒m(m – 4) = 9/4

⇒m2 – 4m – 9/4 = 0

⇒(p+2)2 – 4(p+2) – 9/4 = 0

⇒p2 + 4p + 4 – 4p – 8 – 9/4 = 0

⇒p225/4 = 0

⇒p2 = 25/4

⇒p = ±5/2


  1. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

Solution:

The given quadratic equation is x2 + px + q = 0 where a = 1, b = p , c = q

The sum of the roots m+n = –(p)/1 = -p

The product of the roots mn = q/1 = q

If one root is m then other root is 3m.

Thus, m = m and n = 3m

m + n = -p ⇒ m+3m = -p

⇒4m = -p

⇒ m = -p/4

mn = q ⇒m(3m) = q

⇒3m2 = q

⇒3(-p/4)2 = q

⇒3p^2/16 = q

⇒3p2 = 16q


 

Quadratic Equations – Exercise 9.7 – Class X

Find the sum and product of the roots of the quadratic equation:

  1. x2 – 5x + 8 = 0
  2. 3a2 – 10a – 5 = 0
  3. 8m2 – m = 2
  4. 6k2 – 3 = 0
  5. pr2 = r – 5
  6. x2 + (ab)x + (a + b) = 0

Quadratic Equation – Exercise 9.7 – Solutions:

Find the sum and product of the roots of the quadratic equation:

  1. x2 – 5x + 8 = 0

Solution:

Let m and n are the roots of the quadratic equations x2 – 5x + 8 = 0 which is of the type ax2 + bx + c = 0, then, where a = 1 , b = -5 , c = 8

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -(-5)/1 = 5 and also, mn = c/a = 8/1 = 8

Thus, the sum of the roots of the quadratic  equation x2 – 5x + 8 = 0 is 5 and the product of the roots of the quadratic equation x2 – 5x + 8 = 0 is 8.


  1. 3a2 – 10a – 5 = 0

Solution:

Let m and n are the roots of the quadratic equations 3a2 – 10a – 5 = 0 which is of the type ax2 + bx + c = 0, then, where a = 3 , b = -10 , c = -5

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -(-10)/3 = 10/3 and also, mn = c/a = -5/3

Thus, the sum of the roots of the quadratic  equation 3a2 – 10a – 5 = 0 is 10/3 and the product of the roots of the quadratic equation 3a2 – 10a – 5 = 0 is -5/3.


  1. 8m2 – m = 2

Solution:

The given equation 8m2 – m = 2 can be written as 8m2 – m – 2 = 0

Let m and n are the roots of the quadratic equations 8m2 – m – 2 = 0 which is of the type ax2 + bx + c = 0, then, where a = 8 , b = -1 , c = -2

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -(-1)/8 = 1/8 and also, mn = c/a = -2/8 = –1/4

Thus, the sum of the roots of the quadratic  equation 8m2 – m – 2 = 0 is 1/8 and the product of the roots of the quadratic equation 8m2 – m – 2 = 0is –1/4.


  1. 6k2 – 3 = 0

Solution:

The given equation 6k2 – 3 = 0 can be written as 6k2 + 0.k – 3 = 0

Let m and n are the roots of the quadratic equations 6k2 – 3 = 0 which is of the type ax2 + bx + c = 0, then, where a = 6 , b = 0 , c = -3

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -(0)/6 = 0 and also, mn = c/a = -3/6 = –1/2

Thus, the sum of the roots of the quadratic  equation 6k2 – 3 = 0 is 0 and the product of the roots of the quadratic equation 6k2 – 3 = 0 is –1/2.


  1. pr2 = r – 5

Solution:

The given equation pr2 = r – 5 can be written as pr2 – r + 5 = 0

Let m and n are the roots of the quadratic equations pr2 – r + 5 = 0 which is of the type ax2 + bx + c = 0, then, where a = p , b = -1 , c = 5

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -(-1)/p = 1/p and also, mn = c/a = 5/p

Thus, the sum of the roots of the quadratic equation pr2 – r + 5 = 0 is 1/p and the product of the roots of the quadratic equation pr2 – r + 5 = 0 is 5/p.


  1. x2 + (ab)x + (a + b) = 0

Solution:

Let m and n are the roots of the quadratic equations x2 + (ab)x + (a + b) = 0 which is of the type ax2 + bx + c = 0, then, where a = 1 , b = (ab) , c = (a+b)

Formula to find the sum of the roots and product of the roots are m+n = -b/a and mn = c/a respectively.

Therefore, m + n = -b/a = -ab/1 = -ab and also, mn = c/a = (a+b)/1 = (a + b)

Thus, the sum of the roots of the quadratic equation x2 + (ab)x + (a + b) = 0 is (ab) and the product of the roots of the quadratic equation x2 + (ab)x + (a + b) = 0 is (a + b).


X – Table of Contents