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:
