Question

In the given figure,∆PQR is an equilateral trisngle of side 8cm and P,Q,R are centres of circular discs ,each ofradius 4cm find the area of shaded region.
(Use π=3.14 and √3= 1.732)​

Answer 1

${\huge{\underline{\underline{\sf{\red{Solution:-}}}}}}$

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${\small{\underline{\underline{\sf{\green{Area\:of\:shaded\:region=Area\:of∆-\:Area\:of\:3sector\:region}}}}}}$

$⠀⠀ ⠀ ⠀\small\red{ = \frac{ \sqrt{3} }{4}(side)^{2} - 3( \frac{θ\pi{r}^{2} }{360}) }$

$⠀ ⠀ ⠀\small\red{ = \frac{1.73}{4} \times 8\times 8 - 3 \times \frac{1}{6} \times 3.14 \times 4 \times 4 }$

$⠀ ⠀ ⠀\small\red{ = 1.73 \times 16 - 3.14 \times 8}$

$⠀ ⠀ ⠀ \small\red{ = 27.712 - 3.14 \times 8}$

$⠀ ⠀ ⠀\small\red{ = 25.9cm^{2} }$

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Answer 2

ANSWER:-

Area of shaded region = Area of ∆ - area of 3 sector region.

=> √3/4(side)² -3(∅πr²)

=> 1.73/4 ×8× 8-3 × 1/6 × 3.14× 4× 4

=> 1.73× 16-3× 1/6× 3.14× 16

=> 27.712 - 3.14×8

=> 25.9cm² Ans.