ML Aggarwal Solutions for Class 8 Maths Chapter 15 Circle provide accurate answers to the textbook questions, which are designed by a set of expert faculty at BYJU’S. Students who want to excel in exams are suggested to follow ML Aggarwal Solutions as a reference guide. Practising these solutions help them to understand complex concepts effortlessly, and it also enhances the analytical and logical thinking abilities of 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. The construction of the circle is the main concept explained in this chapter. 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
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 ∠AOB = 600. Measure the length of the chord AB.
Solution:
1. Draw a circle, taking the centre as O and radius equal to 2.5 cm
2. Join OA, where A is any point on the circle
3. Draw ∠AOB equal to 600
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 the centre as O and radius = 3.2 cm
2. Take the point A on the circle
3. Taking A as the 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 the 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)
∠CTP = 900 (since the radius is perpendicular to tangent)
From △CPT,
CP2 = PT2 + CT2 (By Pythagoras theorem)
(5)2 = PT2 + (3)2
We get,
PT2 = 25 – 9
PT2 = 16
PT = √16
We get,
PT = 4
Therefore, the length of the 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,
The radius is always perpendicular to the tangent
i.e, CT ⊥ PT
Therefore,
△CPT is a right angled triangle, where CP = hypotenuse
In right-angled triangle,
By Pythagoras theorem, we get,
CP2 = PT2 + CT2
CT2 = CP2 – PT2
CT2 = (20)2 – (16)2
We get,
CT2 = 400 – 256
CT2 = 144
CT = √144
We get,
CT = 12 cm
Therefore, the radius of the circle = 12 cm
5. In each of the following figures, 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 ∠CAB = 320
∠ABD = 500
∠C = 900 (angles in the semicircle)
By the angle sum property of the triangle, we get,
∠C + ∠CAB + ∠ABC = 1800
900 + ∠CAB + ∠x = 1800
900 + 320 + ∠x = 1800
320 + ∠x = 1800 – 900
We get,
∠x = 900 – 320
∠x = 580
Similarly,
In the right-angled triangle ADB,
By the angle sum property of the triangle, we get,
∠ABD + ∠D + ∠BAD = 1800
500 + 900 + ∠BAD = 1800
500 + 900 + ∠y = 1800
∠y = 1800 – 1400
We get,
∠y = 400
(ii) In the figure,
AC is the diameter of the circle with centre O
∠DAC = 370
AD || BC
∠ACB = ∠DAC (Alternate angles)
Hence,
x = 370
In △ABC,
∠B = 900 (Angle in a semicircle)
By the angle sum property of the triangle, we get,
∠x + ∠y + ∠B = 1800
370 + ∠y + 900 = 1800
∠y = 1800 – 1270
We get,
∠y = 530
(iii) In the figure,
AC is the diameter of the circle with centre as O
BA = BC
Hence,
∠BAC = ∠BCA (angles of isosceles triangle)
But ∠ABC = 900 (angles in a semicircle)
In triangle ABC,
By the angle sum property of the triangle, we get,
∠BAC + ∠ABC + ∠BCA = 1800
∠BAC + ∠BCA = 1800 – 900
∠x + ∠x = 900
∠2x =900
We get,
∠x = 450
(iv) In the figure,
AC is the diameter of the circle, with the centre as O,
∠ACD = 1220
∠ACB + ∠ACD = 1800 (Linear pair)
∠ACB + 1220 = 1800
∠ACB = 1800 – 1220
We get,
∠ACB = 580
In △ABC,
∠ABC = 900 (Angles in a semicircle)
By the angle sum property of the triangle, we get,
∠ABC + ∠BCA + ∠ACB = 1800
900 + x + 580 = 1800
x = 1800 – 1480
We get,
x = 320
(v) In the figure,
AC is the diameter of the circle, with the centre as O,
OD || CB and ∠CAB = 400
In △ABC,
∠B = 900 (Angle in a semicircle)
By the angle sum property of the triangle, we get,
∠BCA + ∠ABC + ∠BAC = 1800
∠BCA + ∠BAC + 900 = 1800
∠BCA + ∠BAC = 1800 – 900
∠BCA + ∠BAC = 900
x + 400 = 900
x = 900 – 400
We get,
x = 500
∵ OD || CB
Hence,
∠AOD = ∠BCA (corresponding angles)
∠AOD = 500
But ∠AOD + ∠DOC = 1800 (Linear pair)
500 + y = 1800
y = 1800 – 500
We get,
y = 1300
Therefore, x = 500 and y = 1300
(vi) In the figure,
AC is the diameter of the circle with centre as O
BA = BC = CD
In △ABC,
∠ABC = 900 (Angle in a semicircle)
By the angle sum property of the triangle, we get,
∠BAC + ∠BCA + ∠ABC = 1800
∠BAC + ∠BCA + 900 = 1800
∠BAC + ∠BCA = 900
But given that, BA = BC
Therefore, ∠BAC = ∠BCA = x
x + x = 900
2x = 900
x = 450
In △BCD,
BC = CD
Hence,
∠CBD = ∠CDB = y and
Exterior ∠ACB = Sum of interior opposite angles
∠ACB = ∠CBD + ∠CDB
x = y + y
Therefore,
2y = x = 450
y = 450 / 2
y = 22.50 or
y = (
)0
(vii) In the figure,
AB is the diameter of the circle with centre O
ST is the tangent at point B
∠ASB = 650
In △ABS
∵ TS is the tangent, and OB is the radius
OB is perpendicular to ST or
∠ABS = 900
But in △ASB,
∠BAC + ∠ASB + ∠ABS = 1800
x + 650 + 900 = 1800
x + 1550 = 1800
x = 1800 – 1550
We get,
x = 250
Therefore, x = 250
(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 ⊥ ST or ∠OBS = 900
In △ABS,
∠BAS + ∠BSA + ∠ABS = 1800
By angle sum property of the triangle
∠BAS + ∠BSA + 900 = 1800
∠BAS + ∠BSA = 900
x + y = 900
∵ AB = BS
Hence,
x = y
Therefore, x = y = 900 / 2 = 450
(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
∠Q = 360
QT is the tangent, and OP is the radius of the circle
Hence,
OP is perpendicular to QT
∠OPQ = 900
In △OPQ,
By the angle sum property of the triangle, we get,
∠OQP + ∠POQ + ∠OPQ = 1800
∠OQP + ∠POQ + 900 = 1800
Hence,
∠OQP + ∠POQ = 900
360 + x = 900
x = 900 – 360
We get,
x = 540
In △OPS,
OP = OS (Radii of the circle)
Hence,
∠OPS = ∠OSP = y and
Exterior angle ∠POQ = ∠OPS + ∠OSP
x = y + y
x = 2y = 540
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
∠ABC = 900 (Angles in a semicircle)
By Pythagoras Theorem,
AC2 = AB2 + BC2
AC2 = (15)2 + (8)2
AC2 = 225 + 64
AC2 = 289
AC2 = (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 × 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, ∠OSP = ∠OTP = 900
OS = OT = 5 cm and
PT = PS = 12 cm
Now,
In the right angled triangle OTP,
By Pythagoras Theorem
OP2 = OT2 + PT2
= (5)2 + (12)2
= 25 + 144
= 169
We get,
OP2 = (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
OT1 is the radius of the circle
PT1 and PT2 are the tangents of the circle from point P
OT1 = 24 cm and PT1 = 18 cm
Here,
OT1 is the radius and PT1 is the tangent,
Hence,
OT1 ⊥ PT1
Now,
In the right angled triangle OPT,
By Pythagoras Theorem
OP2 = OT12 + PT12
OP2 = (24)2 + (18)2
OP2 = 576 + 324
We get,
OP2 = 900
OP2 = (30)2
OP = 30 cm
x = 30 cm
Since PT1 and PT2 are the tangents from point P
PT1 = PT2 = 18 cm
i.e, y = 18 cm
Comments