Question

if triangle aob is isosceles with ab equal to ob and a semicircle is drawn on the side of oa then find the area of triangle and area of semicircle angle aob is theta.

Answer 1

As given we have a isosceles ∆ ABC , where ∠B  = 90°  ,

So, Here line AB  = BC  =  x  , As we know if ∠B  = 90° , then line AC is hypotenuse of triangle that can't be equal to any other side of triangle .

So we apply Pythagoras theorem in ∆ ABC , As :

AC2 =  AB2 +  BC2 

AC2 =  x 2 + x 2

AC2 =  2 x 2

AC  = 2√ x

And as given ∆  ACP and ∆  ABQ are similar triangle than we know ratio of their Area As :

Area of ∆ ABQArea of ∆ ACP  = ( Corresponding side)2( Corresponding side)2

So we get

Area of ∆ ABQArea of ∆ ACP  = ( AB)2( AC)2

Area of ∆ ABQArea of ∆ ACP  = ( x)2( 2√ x)2

Area of ∆ ABQArea of ∆ ACP  =  x2 2 x2

Area of ∆ ABQArea of ∆ ACP  =  1 2
So,

Area of ∆ ABQ  :  Area of ∆ ACP  =  1 :  2  (Ans  )