**The Circle – Maharashtra Board Class 8 Solutions for Mathematics**

MathematicsGeneral ScienceMaharashtra Board Solutions

**Exercise – 19**

**Solution 1:**

From the figure, we have

Seg OD is a radius.

Seg AB is a diameter.

Seg PQ is a chord.

The length of seg AB is twice that of seg OD.

**Solution 2:**

From the figure, we have

Centre of the circle: Point C

Radius of the circle: seg CL, seg CM and seg CD

Chord of the circle: seg TR and seg LM

Diameter of the circle: seg LM

**Solution 3:**

i. Seg TS is not a chord. (False)

ii. Seg KM is a chord. (True)

iii. Seg CK is a radius. (True)

iv. Seg KM is not a diameter. (False)

**Exercise – 20**

**Solution 1:**

Given: chord MN ≅ chord RT

Chord RT is 6 cm from the centre.

Congruent chords are equidistant from the centre.

∴ Distance of chord MN from the centre = 6 cm

**Solution 2:**

Seg OM ⊥ chord AB …(Given)

l(AM) = 1.5 cm

The perpendicular from the centre of a circle to the chord bisects the chord.

∴ l(AM) = l(BM) = 1.5 cm

l(AB) = 2 × l(AM) = 2 × 1.5 = 3 cm

Thus, l(BM) = 1.5 cm and l(AB) = 3 cm

**Solution 3:**

Given: m∠APB = 40° and chord AB ≅ chord CD.

Congruent chords make congruent angles at the centre.

∴ ∠APB ≅ ∠CPD

∴ m∠APB = m∠CPD

∴ m∠CPD = 40°

Thus, the measure of ∠CPD is 40°.

**Solution 4:**

Let O be the centre of the circle of radius 5 cm.

The distance of the chord PQ from the centre = 4 cm

i.e., l(OM) = 4 cm

In DOMP, Seg OM ⊥ seg PQ

By Pythagoras Theorem,

[l(OM)]2 + [l(PM)]2 = [l(OP)]2

∴ (4)2 + [l(PM)]2 = (5)2

∴ [l(PM)]2 = 25 – 16

∴ [l(PM)]2 = 9

∴ l(PM) = 3 cm

The perpendicular from the centre to the chord bisects the chord.

∴ l(PQ) = 2 × l(PM) = 2 × 3 = 6 cm

Thus, the length of the chord is 6 cm.

**Solution 5:**

**Solution 6:**