### Denominator:

The denominator is the bottom number of a fraction.

**The denominator represents the total number of parts i.e.,**

Skip to content# Denominator – D – Breathmath’s Math Glossary

### Denominator:

# Diagonal – D – Breathamth’s Math Glossary

### Diagonal:

# Diagonal Matrix – D – Breathmath’s Math Glossary

### Diagonal Matrix:

# Distributive Property – D – Breathmath’s Math Glossary

### Distributive Property

# Diameter – D – Breathmath’s Math Glossary

### Diameter:

### Diameter of the Circle:

# Circles Exercise 10.8 Solution – Class 10

## Circles Exercise 10.8 – Questions:

### I(A).

### (B)

**Circles Exercise 10.8 – Solution:**

# Circles Exercise 10.7 Solution – Class 10

## Circles Exercise 10.7 – Question

**Circles Exercise 10.7 – Solution**

# Circles Exercise 10.6 Solution – Chapter Circles – Class 10

## Circles Exercise 10.6 – Questions:

**Circles Exercise 10.6 solution:**

# Circles Exercise 10.5 – Class 10

## Circles Exercise 10.5 – Questions:

## Circles Exercise 10.5 – Solutions:

# Circles Exercise 10.4 – Chapter Circles – Class 10

Mathematics! Mathematics! Mathematics! All you just have to do is studying here!

The denominator is the bottom number of a fraction.

**The denominator represents the total number of parts i.e.,**

A line segment joining two non-adjacent vertices of a polygon.

Ex: A square has two diagonals.

Here in square ABCD, AC and BD are its diagonals.

Similarly, a hexagon has 3 diagonals.

A matrix having non-zero elements only in the diagonal running from the upper left to the lower right.

The **Distributive Property** is an algebra property which is used to multiply a single term and two or more terms inside a set of parentheses.

A straight line passing from side to side through the centre of a body or figure, especially a circle or sphere.

Any straight line segment that passes through the center of the circle and whose endpoints lie on the circle.

Diameter is the longest chord of the circle. Diameter is the twice of the radius of the circle.

- Draw two congruent circles of radii 3 cm, having their centres 10 cm apart, Draw a direct common tangent
- Draw two direct common tangents to two congruent circles of radii 3.5 cm and whose distance between them is 3 cm
- Draw two direct common tangent to two externally toucan circles of radii 4.5 cm
- Draw a pair of direct common tangents to two circles of radii 2.5 cm whose centres are at 4 cm apart.

- Construct a direct common tangent to two circles of radii 5 cm and 2 cm whose centres are 3 cm apart.
- Draw a direct common tangent to two internally touching circles of radii 4.5 cm and 2.5 cm
- Construct a direct common tangent to two internally touching circles of radii 4cm and 2 cm whose centres are 8 cm apart. Measure and verify the length of the tangent.
- Two circles of radii 5.5 cm and 3.5 cm touch each other externally. Draw a direct common tangent and measure its length.
- Draw direct common tangents to two circles of radii 5 cm and 3 cm having their centres 5 cm apart.
- Two circles of radii 6 cm and 3 cm at a distance of 1 cm, Draw a direct common tangent, measure and verify its length.

**I(A). **

**Draw two congruent circles of radii 3 cm, having their centres 10 cm apart, Draw a direct common tangent**

Solution:

**Draw two direct common tangents to two congruent circles of radii 3.5 cm and whose distance between them is 3 cm**

Solution:

**Draw two direct common tangent to two externally toucan circles of radii 4.5 cm**

Solution:

**Draw a pair of direct common tangents to two circles of radii 2.5 cm whose centres are at 4 cm apart.**

Solution:

** (B) **

**Construct a direct common tangent to two circles of radii 5 cm and 2 cm whose centres are 3 cm apart.**

Solution:

**Draw a direct common tangent to two internally touching circles of radii 4.5 cm and 2.5 cm**

Solution:

**Construct a direct common tangent to two internally touching circles of radii 4cm and 2 cm whose centres are 8 cm apart. Measure and verify the length of the tangent.**

Solution:

**Two circles of radii 5.5 cm and 3.5 cm touch each other externally. Draw a direct common tangent and measure its length.**

Solution:

**Draw direct common tangents to two circles of radii 5 cm and 3 cm having their centres 5 cm apart.**

Solution:

**Two circles of radii 6 cm and 3 cm at a distance of 1 cm, Draw a direct common tangent, measure and verify its length.**

Solution:

- Draw two circles of radii 5 cm and 2 c touching externally.
- Construct two circles of radii 4.5 cm and 2.5 cm whose centres are at 7 cm apart.
- Draw two circles of radii 4 cm and 2.5 cm touching internally. Measure and verify the distance between their centres.
- Distance between the centres of two circles touching internally is 2 cm. If the radius of one of the circles is 4.8 cm, find the radius of the other side and hence draw the touching circles.

**Draw two circles of radii 5 cm and 2 c touching externally.**

Solution:

**Construct two circles of radii 4.5 cm and 2.5 cm whose centres are at 7 cm apart.**

Solution:

**Draw two circles of radii 4 cm and 2.5 cm touching internally. Measure and verify the distance between their centres.**

Solution:

**Distance between the centres of two circles touching internally is 2 cm. If the radius of one of the circles is 4.8 cm, find the radius of the other side and hence draw the touching circles.**

Solution:

I. Numerical problems on touching circles.

- Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.
- Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ∆ABC.
- In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.

II. Riders based on touching circles.

- A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
- Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.
- In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC

**Numerical problems on touching circles.**

**1.Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.**

Solution:

Let radius of the circles be AP = x , BQ = y and CR = z

AB = AP + BP = x + y = 7 cm

BC = BQ + CQ = y + z = 8 cm

AC = CR + AR = z + x = 9 cm

The perimeter of ∆ABC , AB + BC + CA = 7 + 8 + 9 = 24

AP + BP + BQ + CQ + CR + AR = 24

2x + 2y + 2z 24

x + y + z = 12

7 + z = 12⇒ z = 12 – 7 = 5 cm

x + 8 = 12 ⇒ x = 12 – 8 = 4 cm

y + 9 = 12 ⇒ y = 12 – 9 = 3 cm

**Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of****∆****ABC.**

Solution:

The perimeter of ∆ABC = AB + BC + AC

AB = AM – BM = 8 – 3 = 5 cm

BC = BQ + CQ = 3 + 2 = 5 cm

AC = AN – CN = 8 – 2 = 6 cm

The perimeter of ∆ABC = AB + BC + AC = 5 + 5 + 6 = 16 cm

**In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.**

Solution:

In ∆OPC, ∠PCO = 90˚

PC^{2} = OP^{2} – OC^{2}

x^{2} = (OQ – PQ)^{2} – (AC – OA)^{2}

[Since OP = OQ – PQ, OC = AC – AO]

x^{2} = (5 – x)^{2}+(6 – 5)^{2} [Since OQ = OA = 5]

x^{2} = 25 – 10x + x^{2} – 1

10x = 24

x = 2.4 cm

**Riders based on touching circles.**

**1.A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.**

Solution:

∠AOP = ∠BOQ [Vertically opposite angles]

∠APO = ∠AOP [ AO = AP radius of the circle]

∠BQO = ∠BOQ

∠APO = ∠BQO [alternate angles]

Therefore, AP||BQ

**Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.**

Solution:

∠BXP = ∠PYC [Alternate angles AB||CD]

∠BPX = ∠PBX [ XB = XP radii]

∠BPX + ∠PBX + ∠BXP = 180˚ ………….(1)

∠CPY = ∠PCY [YP = YC radii]

∠CPY + ∠PCY + ∠PYC = 180˚

2∠CPY + ∠PCY = 180˚…………(2)

From (1) and (2),

2∠BPX + ∠BXP = 2∠CPY + ∠PYC

2∠BPX = 2∠CPY

∠BPX = ∠CPY

**In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC**

Solution:

∠ADB = 90˚

∠ACO = 90˚

In ∆ADB and ∆AOC,

∠ADB = ∠ACO = 90˚

∠A = ∠A

∆ADB ∼ ∆AOC

^{BD}/_{OC} = ^{AB}/_{AO}

^{BD}/_{OC} = ^{2. AO}/_{AO} [AB = 2.OA]

^{BD}/_{OC} = 2

BD = 2.OC

**In the given figure AB = 8 cm, M is the midpoint of AB. A circle with centre O touches all three semicircles as shown. Prove that the radius of this circle is shown. Prove that the radius of this circle is**^{1}/_{6}AB

Solution:

In ∆OPC, ∠POC = 90˚

OC^{2} = OM^{2} + MC^{2} [By Pythagoras Theorem]

(CP + OP)^{2} = (MR – OR)^{2} + MC^{2}

(2 + x)^{2} = (4 – x)^{2} + 2^{2}

4 + 4x + x^{2} = 16 – 8x + x^{2} + 4

4 + 4x = 16 – 8x + 4

12x = 16

x = ^{16}/_{12} = ^{8}/_{6}

x = ^{1}/_{6} x 8

x = ^{1}/_{6} x AB [AB = 8 cm]

- 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.
- Construct a pair of tangents to a circle of radius 3.5 cm from a point 3.5 cm away from the circle.
- Construct a tangent to a circle of radius 5.5 cm from a point 3.5 cm away from it.
- Draw a pair of perpendicular tangents of length 5 cm to a circle.
- Construct tangents to two concentric circles of radii 2 cm and 4 cm from a point 8 cm away from the centre.

**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:

Lent of the tangents, t = √(d^{2} – r^{2})

Length of the tangents, t = √(10^{2} – 6^{2}) = √(100 – 36) = √64 = 8 cm

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

Solution:

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

Solution:

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

Solution:

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

Solution:

**1.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

**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:

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˚

AP^{2} + OP^{2} = OA^{2} [ Pythagoras Theorem]

AP^{2} + 5^{2} = 13^{2}

AP^{2} + 25 = 169

AP^{2} = 169 – 25 = 144

AP = 12 cm

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

**In 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

- In 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]

- In 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:

AP = 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

- Tangents 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

- In 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˚

- A 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

- If 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

- In 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.