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H^2=√P^2+B^2
=√20^2+20^2
=√400+400
=√800
=√2×400
=20√2
formula of major sector
0/360πr^2-a^2
90/360×3.14×20√2-400
3.14/4×400×2-400
314×2/628-400
228cm^2
=√20^2+20^2
=√400+400
=√800
=√2×400
=20√2
formula of major sector
0/360πr^2-a^2
90/360×3.14×20√2-400
3.14/4×400×2-400
314×2/628-400
228cm^2
Answered by
1
firstly find the area of OABC
i.e 20*20
=400cm²
length of diagonal of OABC
=_/OA²+AB²
= 10√8 cm
then we will find area of quadrant
=3.14* 10√8*10√8 / 4
=3.14*800/4
=3.14*200
=628cm²
now the area of shaded region will be (628-400)cm²
=228cm²
i.e 20*20
=400cm²
length of diagonal of OABC
=_/OA²+AB²
= 10√8 cm
then we will find area of quadrant
=3.14* 10√8*10√8 / 4
=3.14*800/4
=3.14*200
=628cm²
now the area of shaded region will be (628-400)cm²
=228cm²
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