Solve it from class 10
answer given => 98 cm²
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❤️HEY MATE ❤️
Area of shaded region = area of semicircle of diameter BC - {area of quadrant of radius AB /AC - area of ∆ABC }
So, area of semicircle of diameter BC = 1/2 πr²
= 1/2 × 22/7 × 7√2 × 7√2 [ ∵ BC is hypotenuse of right angle ∆ABC , here AB = BC = 14 so, BC = 14√2 = 2 × radius ⇒ radius = 7√2 ]
= 11 × 7 × 2 = 154 cm²
area of quadrant of radius AB/AC = 1/4 πr²
= 1/4 × 22/7 × 14 × 14
= 22 × 7 = 154 cm²
area of ∆ABC = 1/2 height × base
= 1/2 × 14 × 14 = 98 cm²
Now, area of shaded region = 154cm² - { 154cm² - 98cm²} = 98cm²
Hence, area of shaded region = 98 cm²
I HOPE IT HELPS YOU!
Area of shaded region = area of semicircle of diameter BC - {area of quadrant of radius AB /AC - area of ∆ABC }
So, area of semicircle of diameter BC = 1/2 πr²
= 1/2 × 22/7 × 7√2 × 7√2 [ ∵ BC is hypotenuse of right angle ∆ABC , here AB = BC = 14 so, BC = 14√2 = 2 × radius ⇒ radius = 7√2 ]
= 11 × 7 × 2 = 154 cm²
area of quadrant of radius AB/AC = 1/4 πr²
= 1/4 × 22/7 × 14 × 14
= 22 × 7 = 154 cm²
area of ∆ABC = 1/2 height × base
= 1/2 × 14 × 14 = 98 cm²
Now, area of shaded region = 154cm² - { 154cm² - 98cm²} = 98cm²
Hence, area of shaded region = 98 cm²
I HOPE IT HELPS YOU!
AsifAhamed4:
hehe
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