In Figure, ABC is a right-angled triangle, `/_B=90^0,A B=28 c m` and `B C=21 c mdot` With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

Explanation

Given: `/_B=90^0 AB=28cm,BC=21cm `

Applying pythagoras theorem in `/_\ABC`

`(AC)^2 = ( (AB)^2 + (BC)^2)`


`:. AC = sqrt( (28)^2 + (21)^2) = 35`

Area of shaded region= area of `/_\ABC` + area of semicircle having diameter AC - area of quarter circle having radius BC

`:.`Area of shaded region= `(1/2 xx BC xx AB) + (1/2 xx pi xx ((AC)/2)^2) - (1/4 xx pi xx (BC)^2)`

`:.`Area of shaded region= `(1/2 xx 21 xx 28) + (1/2 xx (22)/7 xx ((35)/2)^2) - (1/4 xx (22)/7 xx (21)^2)`

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