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In △ABC,∠A=90°
By Pythagoras theorem,
BC^2 =AB^2+AC^2
3^2+4^2
9+16=25
∴BC=5cm
A(shadded region)=A(△ ABC)+A (semicircle AB) + (semicircle AC) - A(semicircle BAC)
= 1/2×3×4+1/2×π(3/2)^2+1/2×π(4/2)^2-1/2×π(5/2)^2
= 6+6π/8+2π-25π/8
= 6+2π−2π
=6 cm^2
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