Circle Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-15. We provide step by step Solutions of Exercise / lesson-15 Circle Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-15 Objective Type Questions (including Mental Maths Multiple Choice Questions ) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

## Circle Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-15

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Exercise-15,

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

### Circle Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-15

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.

1. Draw a circle with centre O and radius = 2·5 cm.
2. Join OA where A is any point on the circle.
3. Draw ∠AOB = 60°.
4. 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.

1. Draw a circle with centre C and radius 3·2 cm.
2. Take a point A on the circle.
3. With A as centre and radius = 5 cm,
draw an arc to meet the circle at B
4. 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.

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
CP2 = PT2 + CT2
(5)2 = PT2 + 32
PT2 = 25 – 9 = 16
PT = $\sqrt{16}$ = 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.

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,
CP2 = PT2 + CT2
CT2 = CP2 – PT= 202 – 162 = 400 – 256 = 144
CT = $\sqrt{144}$ = 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 :

##### (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°
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
∠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 = $\frac{45^{\circ}}{2}$ = 22.5° or $22 \frac{1}{2}^{\circ}$

##### (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 = $\frac{90^{\circ}}{2}$ = 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°

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

(i) O is the centre of the circle
AB = 15 cm, BC = 8cm

∠ABC = 90° (Angles in a semicircle)
By Pythagoras Theorem,
AC2 = AB2 + AC2
= (15)2 + (8)2 = 225 + 64
= 289 = (17)2
∴ AC = 17 cm
∴ x = 17 cm
and  ∴ y = $\frac{1}{2}$
(∵ AC is the diameter and AO is the radius of the circle)
$\frac{1}{2}$ × 17 = $\frac{17}{2}$ 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)
OP2 = OT2 + PT2 = (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 OT1 is the radius,
From P, PT1 and PT2 are the tangents.
OT1 = 24 cm PT1 = 18cm

∵ OT, is the radius and PT1 is the tangent
∴ OT1 ⊥ PT1
Now in right ∆OPT, (By Pythagoras Theorem)
OP2 = OT12 + PT12 = (24)2 + (18)2
= 576 + 324 = 900 = (30)2
∴ OP = 30
⇒ x = 30 cm
∵ PT1 and PT2 are the tangents from P
∴ PT2 = PT1 = 18 cm
⇒ y = 18 cm

### Objective Type Questions,Circle Class-8 ML Aggarwal ICSE Mathematics Solutions

#### Mental Maths

Question 1.
Fill in the blanks:
(i) A chord of a circle is a line segment with its end points ……………
(ii) A diameter of a circle is a chord that …………… the centre of circle.
(iii) A line meets a circle at most in …………… points.
(iv) One-half of the whole arc of a circle is called a …………… of the circle.
(v) The angle subtended by an arc of a circle at the centre of the circle is called the …………… by the arc.
(vi) A line which meets a circle in one and only one point is called a …………… to the circle.
(vii) The tangent at any point of a circle and the radius through that point are …………… to each other.
(viii)From a point outside the circle, …………… tangents can be drawn to the circle.
(ix) The measure of an angle in a semicircle is ……………

(i) A chord of a circle is a line segment with its endpoints on the circle.
(ii) A diameter of a circle is a chord that passes through the centre of the circle.
(iii) A line meets a circle almost in two points.
(iv) One-half of the whole arc of a circle is called a semicircle of the circle.
(v) The angle subtended by an arc of a circle at the centre
of the circle is called the angle subtended by the arc.
(vi) A line which meets a circle in one and only
one point is called a tangent to the circle.
(vii) The tangent at any point of a circle and the radius
through that point are perpendicular to each other.
(viii)From a point outside the circle, two tangents can be drawn to the circle.
(ix) The measure of an angle in a semicircle is right angle

#### Question 2.

State whether the following statements are true (T) or false (F):
(i) A line segment with its end-points lying on a circle is called a radius of the circle.
(ii) Diameter is the longest chord of the circle.
(iii) The end-points of a diameter of a circle divide the circle into two parts; each part is called a semicircle.
(iv) A diameter of a circle divides the circular region into two parts; each part is called a semicircular region.
(v) The diameters of a circle are concurrent. The centre of the circle is the point common to all diameters.
(vi) Every circle has unique centre and it lies inside the circle.
(vii) Every circle has unique diameter.
(viii)From a given point in the exterior of a circle, two tangents can be drawn to it and these two tangents are equal in length.

(i) A line segment with its end-points lying on a
circle is called a radius of the circle. False
Correct:
It is called a chord of the circle.
(ii) Diameter is the longest chord of the circle. True
(iii) The end-points of a diameter of a circle divide the circle
into two parts; each part is called a semicircle. True
(iv) A diameter of a circle divides the circular region
into two parts; each part is called a semicircular region. True
(v) The diameters of a circle are concurrent.
The centre of the circle is the point common to all diameters. True
(vi) Every circle has unique centre and it lies inside the circle. True
(vii) Every circle has unique diameter. False
Correct:
It has infinite number of diameters.
(viii) From a given point in the exterior of a circle,
two tangents can be drawn to it and these two tangents are equal in length. True

### MCQs

Multiple Choice Questions Circle Class-8 ML Aggarwal ICSE Mathematics Solutions

Choose the correct answer from the given four options (3 to 6):

Question 3.
If P and Q are any two points on a circle, then the line segment PQ is called a
(b) diameter of the circle
(c) chord of the circle
(d) secant of the circle

P and Q are two points on a circle.

Then line segment PQ is called a chord of the circle. (c)

Question 4.
If P is a point in the interior of a circle with centre O and radius r, then
(a) OP = r
(b) OP > r
(c) OP > r
(d) OP < r

P is a point in the interior of a circle with centre O, r is the radius
∴ OP < r (d)

Question 5.
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 20 cm

AB = 12 cm, BC = 16 cm
AC is the diagonal of ∆ABC
and AC is the diameter of the circle (∵ ∠B = 90°)

Question 6.
In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°

In circle with centre O, AB is its diameter.
∴ ∠C = 90° (Angle in a semi-circle)
By ∠ sum property of ∆
∠A + ∠B +∠C = 180°
∠A + ∠B + 90°= 180°
∴ ∠A + ∠B = 90°
∵ CA = CB
∴ ∠A = ∠B = $\frac{90^{\circ}}{2}$ = 45°
∴ ∠CAB = 45° (b)

Circle Class-8 ML Aggarwal ICSE Mathematics Solutions

Question 1.

Draw a circle of radius 2·7 cm. Draw a chord PQ of length 4 cm of this circle. Shade the major segment of this circle.

(i) Draw a circle of radius = 2·7 cm.
(ii) Take a point P anywhere on the circle.
(iii) With P as centre and 4 cm as radius,
draw an arc which cuts the circle at Q.

(iv) Join PQ which is the required chord.

Question 2.
Draw a circle of radius 3·2 cm and in it make a sector of angle.
(i) 30°
(ii) 135°
(iii) $2 \frac{2}{3}$ right angles
Draw separate diagrams and shade the sectors.

(i) 30°
Steps :
(a) Draw a circle with centre C and radius CB = 3.2 cm
(b) From C, make an angle of 30°.
(c) Shade the region enclosed in ABC.

(ii) 135°
Steps :
(a) Draw a circle with centre C and radius CB = 3.2 cm.
(b) From C, make an angle of 135°.
(c) Shade the region enclosed in ACB.

(iii) $2 \frac{2}{3}$ right angles
Steps :
(a) Draw a circle with centre C and radius 3.2 cm.
(b) From C, make an ∠, $2 \frac{2}{3}$ of right angle
$2 \frac{2}{3}$ × 90°
$\frac{8}{3}$ × 90° = 240°
(c) Shade the region enclosed in ACB.

$\frac{8}{3}$ × 90° = 240°

Question 3.
Draw a line segment PQ = 6·4 cm. Construct a circle on PQ as diameter. Take any point R on this circle and measure ∠PRQ.

(i) Draw a line segment PQ = 6·4 cm.
(ii) Draw ⊥ bisector of PQ.
(iii) With O as centre and OP or OQ as radius draw a circle
which passes through P as well as through Q.

(iv) Take point R on the circle.
(v) Join PR and QR.
(vi) Measure ∠PRQ, we get ∠PRQ = 90°

Question 4.
In the adjoining figure, the tangent to a circle of radius 6 cm from an external point P is of length 8 cm. Calculate the distance of the point P from the nearest point of the circumference.

C is the centre of the circle
PT is the tangent to the circle from P.
∴ CT ⊥ PT
CT = CR = 6 cm, PT = 8 cm
Now in right ∆CPT (By Pythagoras Theorem)
CP2 = PT2 + CT2 = (8)2 + (6)2
= 64 + 36 = 100 = (10)2
∴ CP = 10 cm
Now PR = CP – CR = 10 – 6 = 4 cm

Question 5.
In the given figure, O is the centre of the circle. If ∠ABP = 35° and ∠BAQ = 65°, find
(i) ∠PAB
(ii) ∠QBA

In the figure,
AB is the diameter of the circle with centre O
∠ABP = 35° and ∠BAQ = 65°
(i) ∠APB = 90° (Angle in a semicircle)
In ∆APB, By ∠sum property of ∆
∠PAB + ∠P + ∠ABP = 180°
∠PAB + 90° + ∠ABP = 180°
∴ ∠PAB + ∠ABP = 90°
⇒ ∠PAB + 35° = 90° ⇒ ∠PAB = 90° – 35° ∠PAB = 55°

(ii) Similarly ∠AQB = 90° (Angle in a semicircle)
In ∆AOB, By angle sum property of ∆
∠BAQ + ∠Q + ∠QBA = 180°
∠BAQ + ∠QBA + 90° = 180°
∴ ∠BAQ + ∠QBA = 90°
⇒ 65° + ∠QBA = 90°
⇒ ∠QBA = 90°- 65° = 25°
Hence ∠PAB = 55° and ∠QBA = 25°

— End of  Circle Class-8 ML Aggarwal Solutions :–