ML Aggarwal Solutions for Class 8 Maths Chapter 15 Circle has accurate answers designed by a set of expert faculty at BYJUâ€™S. Students who want to excel in exams, are suggested to follow ML Aggarwal Solutions as a reference guide. Practising these solutions helps them to understand the complex concepts effortlessly. This also enhances analytical and logical thinking abilities among students, which are important from the exam perspective. To speed up the problem solving skills, ML Aggarwal Solutions for Class 8 Maths Chapter 15 Circle PDF, links are available with a free download option.

Chapter 15 explains all the fundamental concepts related to Circle. Construction of the Circle is the main concept discussed here. Circle is one of the significant chapters, as it continues in higher classes as well. Students can analyse their exam preparation by cross checking their answers while solving textbook problems.

## ML Aggarwal Solutions for Class 8 Maths Chapter 15 Circle Download PDF

## Access ML Aggarwal Solutions for Class 8 Maths Chapter 15 Circle

**1. Draw a circle with centre O and radius 2.5 cm. Draw two radii OA and OB such that âˆ AOB = 60 ^{0}. Measure the length of the chord AB.**

**Solution:**

1. Draw a circle, taking centre as O and radius equal to 2.5 cm

2. Join OA, where A is any point on the circle

3. Draw âˆ AOB equal to 60^{0}

4. Now, join AB and on measuring we get, AB = 2.5 cm

**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.**

**Solution:**

1. Draw a circle, taking centre as O and radius = 3.2 cm

2. Take a point A on the circle

3. Taking A as centre and radius = 5 cm, draw an arc to meet the circle at point B

4. Now, join AB and shade the minor segment of the circle

**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.**

**Solution:**

Draw a circle, taking C as centre and radius CT = 3 cm

Let PT be the tangent, drawn from point P to a circle with centre C

Let CP = 5 cm

CT = 3 cm (given)

âˆ CTP = 90^{0} (since radius is perpendicular to tangent)

From â–³CPT,

CP^{2} = PT^{2} + CT^{2} (By Pythagoras theorem)

(5)^{2} = PT^{2} + (3)^{2}

We get,

PT^{2} = 25 â€“ 9

PT^{2} = 16

PT = **âˆš**16

We get,

PT = 4

Therefore, length of tangent = 4 cm

**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.**

**Solution:**

We know that,

Radius is always perpendicular to tangent

i.e, CT âŠ¥ PT

Therefore,

â–³CPT is a right angled triangle, where CP = hypotenuse

In right angled triangle,

By Pythagoras theorem, we get,

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

CT^{2} = CP^{2} â€“ PT^{2}

CT^{2} = (20)^{2} â€“ (16)^{2}

We get,

CT^{2} = 400 â€“ 256

CT^{2} = 144

CT = **âˆš**144

We get,

CT = 12 cm

Therefore, radius of circle = 12 cm

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

**Solution:**

(i) In the given figure,

AB is the diameter and O is the centre of the circle

Given âˆ CAB = 32^{0}

âˆ ABD = 50^{0}

âˆ C = 90^{0} (angles in the semicircle)

By angle sum property of triangle, we get,

âˆ C + âˆ CAB + âˆ ABC = 180^{0}

90^{0} + âˆ CAB + âˆ x = 180^{0}

90^{0} + 32^{0} + âˆ x = 180^{0}

32^{0} + âˆ x = 180^{0} â€“ 90^{0}

We get,

âˆ x = 90^{0} – 32^{0}

âˆ x = 58^{0}

Similarly,

In right angled triangle ADB,

By angle sum property of triangle, we get,

âˆ ABD + âˆ D + âˆ BAD = 180^{0}

50^{0} + 90^{0} + âˆ BAD = 180^{0}

50^{0} + 90^{0} + âˆ y = 180^{0}

âˆ y = 180^{0} – 140^{0}

We get,

âˆ y = 40^{0}

(ii) In the figure,

AC is the diameter of circle with centre O

âˆ DAC = 37^{0}

AD || BC

âˆ ACB = âˆ DAC (Alternate angles)

Hence,

x = 37^{0}

In â–³ABC,

âˆ B = 90^{0} (Angle in a semicircle)

By angle sum property of triangle, we get,

âˆ x + âˆ y + âˆ B = 180^{0}

37^{0} + âˆ y + 90^{0} = 180^{0}

âˆ y = 180^{0} – 127^{0}

We get,

âˆ y = 53^{0}

(iii) In the figure,

AC is the diameter of the circle with center as O

BA = BC

Hence,

âˆ BAC = âˆ BCA (angles of isosceles triangle)

But âˆ ABC = 90^{0} (angles in a semicircle)

In triangle ABC,

By angle sum property of triangle, we get,

âˆ BAC + âˆ ABC + âˆ BCA = 180^{0}

âˆ BAC + âˆ BCA = 180^{0} – 90^{0}

âˆ x + âˆ x = 90^{0}

âˆ 2x =90^{0}

We get,

âˆ x = 45^{0}

(iv) In the figure,

AC is the diameter of the circle, with centre as O,

âˆ ACD = 122^{0}

âˆ ACB + âˆ ACD = 180^{0} (Linear pair)

âˆ ACB + 122^{0} = 180^{0}

âˆ ACB = 180^{0} – 122^{0}

We get,

âˆ ACB = 58^{0}

In â–³ABC,

âˆ ABC = 90^{0} (Angles in a semicircle)

By angle sum property of triangle, we get,

âˆ ABC + âˆ BCA + âˆ ACB = 180^{0}

90^{0} + x + 58^{0} = 180^{0}

x = 180^{0} – 148^{0}

We get,

x = 32^{0}

(v) In the figure,

AC is the diameter of the circle, with centre as O,

OD || CB and âˆ CAB = 40^{0}

In â–³ABC,

âˆ B = 90^{0} (Angle in a semicircle)

By angle sum property of triangle, we get,

âˆ BCA + âˆ ABC + âˆ BAC = 180^{0}

âˆ BCA + âˆ BAC + 90^{0} = 180^{0}

âˆ BCA + âˆ BAC = 180^{0} â€“ 90^{0}

âˆ BCA + âˆ BAC = 90^{0}

x + 40^{0} = 90^{0}

x = 90^{0} – 40^{0}

We get,

x = 50^{0}

âˆµÂ OD || CB

Hence,

âˆ AOD = âˆ BCA (corresponding angles)

âˆ AOD = 50^{0}

But âˆ AOD + âˆ DOC = 180^{0} (Linear pair)

50^{0} + y = 180^{0}

y = 180^{0} â€“ 50^{0}

We get,

y = 130^{0}

Therefore, x = 50^{0} and y = 130^{0}

(vi) In the figure,

AC is the diameter of the circle with centre as O

BA = BC = CD

In â–³ABC,

âˆ ABC = 90^{0} (Angle in a semicircle)

By angle sum property of triangle, we get,

âˆ BAC + âˆ BCA + âˆ ABC = 180^{0}

âˆ BAC + âˆ BCA + 90^{0} = 180^{0}

âˆ BAC + âˆ BCA = 90^{0}

But given that, BA = BC

Therefore, âˆ BAC = âˆ BCA = x

x + x = 90^{0}

2x = 90^{0}

x = 45^{0}

In â–³BCD,

BC = CD

Hence,

âˆ CBD = âˆ CDB = y and

Exterior âˆ ACB = Sum of interior opposite angles

âˆ ACB = âˆ CBD + âˆ CDB

x = y + y

Therefore,

2y = x = 45^{0}

y = 45^{0} / 2

y = 22.5^{0} or

y = (

)^{0}

(vii) In the figure,

AB is the diameter of circle with centre O

ST is the tangent at point B

âˆ ASB = 65^{0}

In â–³ABS

âˆµÂ TS is the tangent and OB is the radius

OB is perpendicular to ST or

âˆ ABS = 90^{0}

But in â–³ASB,

âˆ BAC + âˆ ASB + âˆ ABS = 180^{0}

x + 65^{0} + 90^{0} = 180^{0}

x + 155^{0} = 180^{0}

x = 180^{0} – 155^{0}

We get,

x = 25^{0}

Therefore, x = 25^{0}

(viii) In the figure,

AB is the diameter of the circle with centre O

ST is the tangent to the circle at point B

AB = BS

Hence,

ST is the tangent and OB is the radius

OB âŠ¥ ST or âˆ OBS = 90^{0}

In â–³ABS,

âˆ BAS + âˆ BSA + âˆ ABS = 180^{0}

By angle sum property of triangle

âˆ BAS + âˆ BSA + 90^{0} = 180^{0}

âˆ BAS + âˆ BSA = 90^{0}

x + y = 90^{0}

âˆµÂ AB = BS

Hence,

x = y

Therefore, x = y = 90^{0} / 2 = 45^{0}

(ix) In the figure,

RS is the diameter of the circle with centre as O

SR is produced to Q

QT is the tangent to the circle at point P

OP is joined

âˆ Q = 36^{0}

QT is the tangent and OP is the radius of the circle

Hence,

OP is perpendicular to QT

âˆ OPQ = 90^{0}

In â–³OPQ,

By angle sum property of triangle, we get,

âˆ OQP + âˆ POQ + âˆ OPQ = 180^{0}

âˆ OQP + âˆ POQ + 90^{0} = 180^{0}

Hence,

âˆ OQP + âˆ POQ = 90^{0}

36^{0} + x = 90^{0}

x = 90^{0} – 36^{0}

We get,

x = 54^{0}

In â–³OPS,

OP = OS (Radii of the circle)

Hence,

âˆ OPS = âˆ OSP = y and

Exterior angle âˆ POQ = âˆ OPS + âˆ OSP

x = y + y

x = 2y = 54^{0}

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

**y.**

**Solution:**

(i) Given

O is the centre of the circle

AB = 15 cm,

BC = 8 cm

âˆ ABC = 90^{0} (Angles in a semicircle)

By Pythagoras Theorem,

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

AC^{2} = (15)^{2} + (8)^{2}

AC^{2} = 225 + 64

AC^{2} = 289

AC^{2} = (17)^{2}

AC = 17 cm

x = 17 cm

y = 1 / 2

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

= 1 / 2 Ã— 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 point P

OS and OT are the radii of the circle

Hence, âˆ OSP = âˆ OTP = 90^{0}

OS = OT = 5 cm and

PT = PS = 12 cm

Now,

In right angled triangle OTP,

By Pythagoras Theorem

OP^{2} = OT^{2 }+ PT^{2}

= (5)^{2} + (12)^{2}

= 25 + 144

= 169

We get,

OP^{2} = (13)^{2}

Therefore, OP = 13 cm

i.e, x = 13 cm

Since PS = PT = 12 cm

Therefore,

y = 12 cm

(iii) O is the centre of the circle

OT_{1} is the radius of the circle

PT_{1} and PT_{2} are the tangents of the circle from point P

OT_{1} = 24 cm and PT_{1} = 18 cm

Here,

OT_{1} is the radius and PT_{1} is the tangent,

Hence,

OT_{1} âŠ¥ PT_{1}

Now,

In right angled triangle OPT,

By Pythagoras Theorem

OP^{2} = OT_{1}^{2} + PT_{1}^{2}

OP^{2} = (24)^{2} + (18)^{2}

OP^{2} = 576 + 324

We get,

OP^{2}_{ }= 900

OP^{2} = (30)^{2}

OP = 30 cm

x = 30 cm

Since, PT_{1} and PT_{2} are the tangents from point P

Therefore,

PT_{1 }= PT_{2} = 18 cm

i.e, y = 18 cm