### Factor:

A number that will divide into another number exactly.

The factors of 6 are 1, 2 and 3.

Skip to content# Factor – F – BreathMath’s math glossary

### Factor:

# Expansion

# Factorization – F – BreathMath’s math glossary

## Factorization:

# Equilateral triangle – E – BreathMath’s math glossary

# Equilateral Triangle:

# Empty set – E – BreathMath’s math glossary

## Empty set:

# Discount – D – breathmath’s math glossary

## Discount

# Diagonal – D – Breathamth’s Math Glossary

### Diagonal:

# Distributive Property – D – Breathmath’s Math Glossary

### Distributive Property

# 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!

Category: Mathematics

A number that will divide into another number exactly.

The factors of 6 are 1, 2 and 3.

A quantity expressed as a sum of a series of terms i.e., an expression written as sum of its series of terms.

For example *x (3y + z)= 3xy + 3z *. Here this *3xy + 3z *is expansion of x* (3y + z). *

Another example, (x + 3)(x +2) = x^2 + 2x + 3x + 6

The process of changing algebraic or numerical expressions from a sum of terms into a product i.e.,

6 = 2 x 3

A triangle whose sides are all the same length.

A set with no elements is called empty

set. It is called null set and is denoted by {}

The difference between the issue price of a stock and its nominal value when the issue price is less than the nominal value.

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.

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.

- 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.