Short Answer TypeIn a circle of radius 21 cm, an arc subtends an angle 60° at the centre. Find
(i) the length of the arc
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.
Two circles touch externally. The sum of their areas is 130π sq. cm and the distance between their centres is 14 cm. Find the radii of the circle.
Let r and R be the radii of the circles having Centres O and O' respectively, then
OO' = R + r
⇒ 14 = r + R
⇒ r + R = 14 ....(i)
Fig. 12.60.
It is also given that,
Sum of the area of two circles = 130π cm2
⇒ πR2 + πr2 = 130 π
⇒ π (R2 + r2) = 130 π
⇒ R2 + r2 = 130 ...(ii)
We know that,
(R + r)2 = R2 + r2 + 2Rr ...(iii)
Putting the values of (i) and (ii) in (iii), we get
(14)2 = 130 + 2Rr
⇒ 196 = 130 + 2Rr
⇒ 66 = 2Rr
⇒ 2Rr = 66
⇒ Rr = 33
Now, (R - r)2 = R2 + r2 - 2 Rr
= 130 - 2 (33)
[From (ii) R2 + r2 = 130]
= 130 - 66 = 64
⇒ R - r = 8 ....(iv)
Solving (i) and (iv), we get
R = 11 cm
and r = 3 cm.
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