# Brilliant’s Composite Mathematics Class 8 Solutions Chapter 12

## Brilliant’s Composite Mathematics Class 8 Solutions Chapter 12 Circle

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 12, Circle. Here students can easily find step by step solutions of all the problems for Circle, Exercise 12.1, 12.2 and 12.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Circle Exercise 12.1 Solution

Question no – (1)

Solution : Since, ∆ AMO is a right angular triangle and ∠AMO is the right angle we have, AM2 + MO2 = AO2

Or, Am+ 32 = 52

Or, AM= 25 – 9 = 16

AM = √16 = 4 cm

AB = 2.4 = 8 cm.

Question no – (2)

Solution :

Required Figure, Here, ∆OCA is a right angular triangle and ∠OCA is the right angle. Here, OC = 6 cm, AC = 1/2 AB = 8cm. we have to find OA By Pythagoras theorem, OA2 = OC2 = AC2

Or, OA2 = 6+ 82 = 36 + 64 =100

OA = √100 = 10

The radius at the circle is 10 cm.

Question no – (3)

Solution :

Required figure, Here, OA = 5 cm, AC = 1/2 AB = 4 cm. ∆OAC is a right angular triangular

Triangle so we have OA2 = OC2 + AC2

Or, OC2 = OA2 – AC2

Or, OC= 52 – 32 = 25 – 16 = 9

OC =  √9 = 3 cm.

The distance from the center to the chord is 3 cm.

Question no – (4)

Solution :

Required figure, Here ∆ AOC is a right angular triangle in which AO is the hypogenous.

By Pythagoras theorem we have AO2 = OC2 + AC2

Or, AC= AO2 – OC2

Or, AC2 = 13– 122 = 169 – 144 = 25

AC = √25 = 5 cm.

Question no – (5)

Solution :

Required figure, ∆AMO is a right angular triangle, where AO is the hypotonus

AO2 = AM+ MO2

Or, MO2 = AO2 – AM2 = 52 – 42 = 25 – 16 = 9

MO = √9 = 3 cm.

Question no – (6)

Solution :

Required Figure, AM= OA– OM2 = 132 – 52 = 169 – 25

Or, AM2 = 144

AM = √144 = 12 cm.

Since, AB = 2AM

AB = 2 × 12 = 24 cm.

Question no – (7)

Solution :

The statement is True.

There are two different ways to think of it.

Firstly, starting from any point on a circle, if we draw a straight line in any direction, then we can observe that it touches at most one another point.

Secondly, if we cut any arc from a circle, then we can find that none of three points among them are colinear.

Question no – (8)

Solution :

Required Figure, Suppose, AB and CD are two parallel chords. Lets draw a radius OP such that OP || AB || CD.

Since, O is the centre of the circle and E and F are mid-points of AB and CD

∠AEO = 90°

And, OP || AE ⇒ ∠EOP = 90°……(i)

Also, OP || CD, thus, ∠POF = 90° …..(ii)

From (i) and (ii), we have ∠EOP + ∠POF = 180°

⇒ EOF is a straight line.

Question no – (11)

Solution :

Draw two biggest chords and notice, where do they intersect. That is the centre.

Question no – (12)

Solution :

Required Figure, AO = 1/2 AB = 3 cm

In AOM, we have AM² = AO² + OM²

Or, OM² = AM² – AO² = 5² – 3²

Or, OM² = 25 – 9 =  10

OM = √16 = 4 cm

In ANO, AN = AO + ON

Or, On = AN – AO

Or, ON = 4 – 3

= 16 – 9 =  7

ON = √16 = 2.6 cm

Distance between the centres,

=  OM + ON

= 4 + 2.6

= 6.6 cm

Circle Exercise 12.2 Solution

Question no – (1)

Solution :

Required Figure, Since, there a re 5 sides and the pentagon is a regular one,

Each of its angle are equal.

And, each of them is equal to 360°/58° =  72°

So, each inner angle  =  180° – 72° =  108° = ABC

Since, OB is a bisector of ∠ABC,

∠OBA = ∠OAB = 108/2 = 54°

Now, in △ABO, ∠AOB

= 180° – (54° + 54°)

= 180° – 108°

=  72°

Therefore, each side of a regular pentagon is 72°

Question no – (2)

Solution :

Required Figure, As we know square is a regular polygon of 4 sides.

So, each outer angle measures

= 360°/4

= 90°

Each inner angle

= 180° – 90°

= 90°

Now, AC is a bisector of ∠DAB

∠OAB = ∠OBA

= 90°/2 = 45°

∠AOB = 180° – (45° + 45°)

= 180° – 90°

= 90°

Therefore, each side will subtend 90° at the centre.

Question no – (3)

Solution :

 Angle 72° 36° 20° 45° 30° No of Sides 360°/72° = 5° 360°/36° = 10° 360°/20° = 20° 360°/45° = 8° 360°/30° = 12°

Question no – (4)

Solution :

Measure of each angle subtended at centre

= 360°/9

= 40°

Question no – (5)

Solution :

Measure of each angle subtend at centre,

= 360°/6

= 60°

Circle Exercise 12.3 Solution :

Question no – (1)

Solution :

Required Figure : Comparing △AOB and △AOC we get,

(i) OAB = OAC  [Since, OA = OB = OCC, △OAB, △OAC are isosceles]

(ii) ∠OAB = ∠OCA  [Since, OA = OB = OCC, △OAB, △OAC are isosceles]

(iii) OA common side

Therefore, △AOB ≅ △AOC

Question no – (2)

Solution :

Required Figure : Comparing △MSO and △NSO,

We get,

(i) ∠SNO = ∠SMO [both are right angle]

(ii) ∠OSM = ∠OSN [Given]

(iii) OS is common side.

Thus, △MSO ≅ △NSO ….(Proved)

Since, △MSO ≅ △ NSO, OM = ON …(Proved)

Question no – (3)

Solution :

Required Figure : Comparing △OPQ and OPR

We have,

(i) ∠OPQ = ∠OPQ [OP is the bisector]

(ii) ∠ORB = ∠OQR [Both are 90]

(iii) OP is common side

△OPQ ≅ △OPR

OR = OQ [Similar sides of equal triangles]

AB = CD [Equidistant chords of a circle]

Question no – (5)

Solution :

Required Figure : Comparing ∆BOL and ∆BOM, we have,

(i) ∠OLB = COMB

(ii) ∠LBO = ∠ OBM

(iii) OB is common side

∆DLB2 = ∆OMB

OL  = OM

AB = BC

Question no – (6)

Solution :

Required Figure : Comparing ∆ADO and ∆AED, we have,

(i) ∠DAO =  ∠EAO

(ii) ∠ODA =  ∠OEA

(iii) AO is the common side.

SO, ∆ADO and  ∆AEO are congruent.

OD = OE

AC = AB

Question no – (7)

Solution :

Required Figure : Comparing  ∆APO and  ∆DQO we have,

(i) OA = OD

(ii) AP = QD

(iii) ∠OPA = ∠OQD

∆APO2 = ∆DQO

OP = OQ

Question no – (8)

Solution :

Required Figure : Comparing ∆EFO and ∆EGO, we have,

(i) ∠OFE = ∠OGE

(ii) OF = OG

(iii) OE is common side

OFE= ∆OGE

Now, FE = GE

Or, FB + BE = GD + DE

Or, FB + BE = FB + PE

Or, BE = DE (Proved)

Previous Chapter Solution :

Updated: May 29, 2023 — 7:42 am