ML Aggarwal Circle Exe-15 Class 8 ICSE Ch-15 Maths Solutions. We Provide Step by Step Answer of Exe-15 Questions for Circle as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Circle Exe-15 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-15 | Circle |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-15 Questions |

Edition | 2023-2024 |

**Circle Exe-15**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-264

**Question 1. Draw a circle with centre O and radius 2-5 cm. Draw two radii OA and OB such that ∠ AOB = 60°. Measure the length of the chord AB.**

**Answer:**

- Draw a circle with centre O and radius = 2·5 cm.
- Join OA where A is any point on the circle.

- Draw ∠AOB = 60°.
- Join AB and on measuring we get AB = 2·5 cm.

**Question 2. Draw a circle of radius 3·2 cm. Draw a chord AB of this circle such that AB = 5 cm. Shade the minor segment of the circle.**

**Answer:**

- Draw a circle with centre C and radius 3·2 cm.
- Take a point A on the circle.
- With A as centre and radius = 5 cm,

draw an arc to meet the circle at B - Join AB and shade the minor segment.

**Question 3. Find the length of the tangent drawn to a circle of radius 3 cm, from a point at a distance 5 cm from the centre.**

**Answer:**

Draw a circle with centre C and radius CT = 3 cm.

Let PT be the tangent drawn from

point P to a circle with centre C.

CP = 5 cm

CT = 3 cm (given radius)

∠CTP = 90°

∵ Radius is ⊥ to tangent

From ∆CPT,

by Pythagoras theorem, we get

CP^{2} = PT^{2} + CT^{2}

(5)^{2} = PT^{2} + 32

PT^{2} = 25 – 9 = 16

PT = √16 = 4

Hence, length of tangent = 4 cm

**Question 4. In the adjoining figure, PT is a tangent to the circle with centre C. Given CP = 20 cm and PT = 16 cm, find the radius of the circle.**

**Answer:**

We know that, radius is always ⊥ to longest.

i. e., CT ⊥ PT

∴ ∆CPT is right ∠d ∆

Where CP = hypotenuse

In rt. ∆CPT, by Pythagoras theorem,

CP^{2} = PT^{2} + CT^{2}

CT^{2} = CP^{2} – PT^{2 }= 20^{2} – 16^{2} = 400 – 256 = 144

CT = √144 = 12 cm

Hence, radius of circle = 12 cm

**Question 5. In each of the following figure, O is the centre of the circle. Find the size of each lettered angle :**

**Answer:**

##### (i) In the figure, AB is the diameter

and O is the centre of the circle ∠ CAB = 32°,

∠ ABD = 50° , ∠C = 90° (Angle in the semicircle)

By ∠ sum property of ∆

In ∆ABC, ∠C + ∠CAB + ∠ABC = 180°

⇒ 90° + ∠CAB + x = 180°

⇒ 32° + x = 180° – 90°

⇒ x = 90° – 32°

⇒ x = 58°

Similarly in right ∆ ADB

∠ADB = 90°

By ∠ sum property of ∆

∠ABD + ∠D + ∠BAD = 180°

⇒ 50° + 90° + ∠BAD = 180°

⇒ ∠y + 140° = 180°

⇒ ∠y = 180° – 140° = 40°

⇒ ∠y = 40°

##### (ii) In the figure,

AC in the diameter of circle with centre O

∠DAC = 37°, AD || BC

∵ AD || BC

∠ACB = ∠DAC (Alternate angles)

∴ x = 37°

In ∆ABC, ∠B = 90° (Angle in a semicircle)

∴ By ∠sum property of ∆

∠x + ∠y + ∠B = 180°

⇒ 37° + ∠y + 90° = 180°

⇒ y = 180° – 127° = 53°

##### (iii) In the figure,

AC is the diameter of the circle with centre O.

BA = BC

∴ ∠BAC = ∠BCA (∠s of isosceles ∆)

But ∠ABC = 90° (Angle in a semicircle)

In ∆ABC

(By ∠ sum property of ∆)

∠BAC + ∠ABC + ∠BCA = 180°

⇒ ∠BAC + ∠BCA = 180°- 90°

⇒ x + x = 90°

⇒ 2x = 90°

∴ x = 45°

##### (iv) In the figure,

AC is the diameter of the centre with centre O,

∠ACD = 122°

∵ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ ∠ACB + 122° = 180°

⇒ ∠ACB + 180° – 122° = 58°

In ∆ABC, ∠ABC = 90° (Angle in a semicircle)

∴ By angle sum prop, of ∆

∠ABC + ∠BCA + ∠ACB = 180°

90° + 58° + x = 180°

x = 180°- 148° = 32°

##### (v) In the figures,

AC is the diameter of the circle with centre O,

OD || CB and ∠CAB = 40°

In ∆ABC,

∠B = 90°, (Angle in a semicircle)

By ∠ sum prop, of ∆

∠BCA + ∠ABC + ∠BAC = 180°

∠BCA + ∠CAB + 90° = 180°

∴ ∠BCA + ∠CAB = 90°

⇒ x + 40° = 90° ⇒ x = 90°- 40°= 50°

∴ x = 50°

∵ OD || CB

so ∴ ∠AOD = ∠BCA (corresponding angles)

∠AOD = x = 50°

But ∠AOD + ∠DOC = 180° (Linear pair)

⇒ 50°+ y = 180° ⇒ y = 180°- 50°= 130°

Hence x = 50° and y = 130°

##### (vi) In the figure,

AC is the diameter of the circle with centre O

BA = BC = CD

In ∆ABC,

∠ABC = 90° (Angle in a semicircle)

By ∠ sum prop, of ∆

∠BAC + ∠BCA + ∠ABC = 180°

∠BAC + ∠BCA + 90° = 180°

∴ ∠BAC + ∠BCA = 90°

But BA = BC (given)

∴ ∠BAC = ∠BCA = x

∴ x + x = 90°

2x = 90°

∴ x = 45°

In ∆BCD,

BC = CD

∴ ∠CBD = ∠CDB = y

and ext. ∠ACB = Sum of interior opposite angles

∠CBD + ∠CDB

x = y + y = 2y

∴ 2y = 45°

y = 45°/2 = 22.5° or 22(1/2)°

##### (vii) In the figure,

AB is the diameter of circle with centre O.

ST is the tangent at B

∠ASB = 65°

In ∆ABS

∵ TS is the tangent and OB is the radius

OB ⊥ ST or ∠ABS = 90°

But in ∆ASB

∠BAC + ∠ASB + ∠ABS = 180° (Angles of a triangle)

x + 65° + 90° = 180°

⇒ x° + 155° = 180° ⇒ x = 180° – 155° = 25°

Hence x = 25°

##### (viii)In the figure,

AB is the diameter of the circle with centre

O. ST is the tangent to the circle at B.

AB = BS

∴ ST is the tangent and OB is the radius

∴ OB ⊥ ST or ∠OBS = 90°

∴ In ∆ABS,

∠BAS + ∠BSA + ∠ABS = 180°

[By ∠ sum property of ∆]

⇒ ∠BAS + ∠BSA + 90° = 180°

∠BAS + ∠BSA = 90° ⇒ x + y = 90°

∵ AB = BS

∴ x = y

hence ∴ x = y = 90°/2 = 45°

##### (ix) In the figure,

RS is the diameter of the circle with centre O.

SR is produced to Q. QT is tangent to the circle at P

OP is joined.

∠Q = 36°

QPT is tangent and OP is the radius of the circle

∴. OP ⊥ QT

∠OPQ = 90°

∴ Now in ∆OPQ

By ∠ sum prop, of ∆

∠OQP + ∠POQ + ∠OPQ = 180°

∠OQP + ∠POQ + 90° = 180°

∴ ∠OQP + ∠POQ = 90°

⇒ 36° + x = 90° ⇒ x = 90° – 36° = 54°

In ∆OPS, OP = OS (Radii of the circle)

∴ ∠OPS = ∠OSP = y

and Ext. ∠POQ = ∠OPS + ∠OSP

= y + y = 2y

⇒ 2y = x = 54°

**Circle Exe-15**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-265

**Question 6. In each of the following figures, O is the j centre of the circle. Find the values of x and y.**

**Answer**

**(i) O is the centre of the circle**

AB = 15 cm, BC = 8cm

∠ABC = 90° (Angles in a semicircle)

By Pythagoras Theorem,

AC^{2} = AB^{2} + AC^{2}

= (15)^{2} + (8)^{2} = 225 + 64

= 289 = (17)^{2}

∴ AC = 17 cm

∴ x = 17 cm

and ∴ y = 1/2

(∵ AC is the diameter and AO is the radius of the circle)

= 1/2 × 17 = 17/2 cm = 8.5 cm

**(ii) O is the centre of the circle.**

PT and PS are the tangents to the circle from P.

OS and OT are the radii of the circle

∴ ∠OSP = ∠OTP = 90°

OS = OT = 5 cm, PT = PS = 12 cm

Now in right ∆OPT (By Pythagoras Theorem)

OP^{2} = OT^{2} + PT^{2} = (5)^{2} + (12)^{2}

= 25 + 144 = 169 = (13)^{2}

∴ OP = 13 cm ⇒ x = 13cm

∵ PS = PT = 12 cm

∴ y = 12 cm

**(iii) O is the centre of the circle OT _{1} is the radius,**

From P, PT

_{1}and PT

_{2}are the tangents.

OT

_{1}= 24 cm PT

_{1}= 18cm

∵ OT, is the radius and PT

_{1}is the tangent

∴ OT

_{1}⊥ PT

_{1}

Now in right ∆OPT, (By Pythagoras Theorem)

OP

^{2}= OT

_{1}

^{2}+ PT

_{1}

^{2}= (24)

^{2}+ (18)

^{2}

= 576 + 324 = 900 = (30)

^{2}

∴ OP = 30

⇒ x = 30 cm

∵ PT

_{1}and PT

_{2}are the tangents from P

∴ PT

_{2}= PT

_{1}= 18 cm

⇒ y = 18 cm

— End of Circle** Exe-15 **Class 8 ICSE Maths Solutions :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

Thanks