Question

.....
Calculate the area of the design region in Fig . 12 .34 common between the two quadrants of circles of the radius 8cm each .

Answer 1

Answer:

name the virtices of the square as A,B,C,D anticlockwise from bottom left corner respectively

area of the shaded region = area of triangle ABD + are of triangle DBC

Step-by-step explanation:

= 2* ( area of the sector - are of triangle ABD)

solve the rest using the formula and derive the answer

the answer is    256/7 .

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Answer 2

Answer :

We have to find the area of the designed region.

Hence,

We can say that ,

Area of the design region = Area of Quadrants - Area of the square

Let's find the area of the Quadrant 1

$\sf{Area \: of \: Quadrant \: 1 = \frac{ \theta}{360} \times \pi r {}^{2}}$

$\sf{ = \frac{90}{360} \times \frac{22}{7} \times 8 \times 8}$

$= \frac{22}{7} \times 8 \times 2$

$\sf{ = \frac{352}{7} \: cm {}^{2} }$

Now,

Area of Quadrant 1 = Area of Quadrant 2 = 352/7 cm^2

Let's find area of square

Area of square = 8 × 8

= 64 sq. cm

Now,

AREA OF THE DESIGN REGION =

$\sf{= ( \frac{352}{7} + \frac{352}{7}) - 64}$

$\sf{ = \frac{704}{7} - 64}$

$\sf{ = \frac{704}{7} - \frac{448}{7}}$

$\sf{ = \frac{704 - 448}{7}}$

$\sf{ = \frac{256}{7}}$


= 36.5 sq cm.

Hence,

Area of the design region is 36.5 cm^2.

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