Question

for 25 points.....please solve it fast....it's really urgent....

Answer 1

Solution :- area of shaded region = area of semicircle APB + area of semicircle AQO [ as shown in attachment ]

Given perimeter of figure = 40 cm
Let radius of semicircle APB is R then, radius of AQO is R/2
Now, perimeter of structure = arc length of APB + arc length of AQO + length of OB
40 = πR + π(R/2) + R
40 = R( 3π/2 + 1) = R(3 × 3.14/2 + 1) = R(1.57 × 3 + 1)
40 = R( 5.71) ⇒R = 40/5.71 cm
Hence , radius of APB = 40/5.71 cm
Raidus of AQO = 20/5.71 cm

Now, area of shaded region = area of APB + area of AQO [ SEMICIRCLE]
= 1/2 × 3.14 × (40/5.71)² + 1/2 × 3.14 × (20/5.71)²
= 1/2 × 3.14 × 400/(5.71)² [ 4 + 1 ]
= 3140/5.71 × 5.71 cm = 96.30 cm²

Answer 2

perimeter = AQO + APB + OB

= pie ( R + r) + OB

OB = r

R = r/2

So pie (3 r/2) + r = 40

5.71 r= 40

r = 7

Now area = pie/2. ( r^2 + r^2/4)

= pie /2. ( 5 r^2/4)

= 96.16

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