Circle Class-8 ML Aggarwal ICSE Maths Solutions

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)

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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.

Answer

  1. Draw a circle with centre O and radius = 2·5 cm.
  2. Join OA where A is any point on the circle.
    ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Ex 15 1
  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.

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Answer

  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.
    ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Ex 15 2

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)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Ex 15 3
∠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.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Ex 15 4

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,
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 :
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Ex 15 5

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,

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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°

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