Question

in figure APB and AQP are semicircles and AO=OB. if the perimeter of the figure is 40 cm, find the area of the shaded region

Answer 1

A correct figure of question is shown in attachment,
also you did mistake in typing you should write AQO in place of AQP .

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

Answer:

Step-by-step explanation:

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