Form a rectangular region ABCD , a right triangle AED with AE= 9 cm and DE = 12cm is cut off. On other end taking BC as diameter , a semi - circle is added on the region. Find the area of the shaded region. (Use pi= 3.14)

Complete question  : Form a rectangular region ABCD with AB  =  20 cm , a right triangle AED with AE= 9 cm and DE = 12cm is cut off. On other end taking BC as diameter , a semi - circle is added on the region. Find the area of the shaded region. (Use pi= 3.14) ,

Solution :  As ABCD is a rectangular region , SO AB  =  CD and BC  =  AD

And our diagram from given information is , As :

Now we Apply Pythagoras theorem in $∆$ AED , and get

AD² =  AE² + DE²

AD² = 9² + 12²

AD² = 81 + 144

AD² = 225

AD  =  15 cm

So,

BC  =  15 cm   ( As we know AD  =  BC )

So,

Radius of semicircle  =  $\frac{BC}{2} = \frac{15}{2} = 7.5 cm$
And
We know Area of triangle  = $\frac{1}{2}\times Base \times height$ , So
Area of $∆$ AED = $\frac{1}{2}\times 9 \times 12 = 9 \times 6 = 54 c{m}^2$
And Area of rectangle  =  Length $\times$ Breadth , So
Area of rectangular part ABCD  = 20 $\times$ 15  =  300 cm²
And
We know Area of semicircle  = $\frac{\pi r^2}{2}$ , So
Area of semicircle  = $\frac{{\displaystyle 3.14\times 7.5 \times 7.5}}{2} = \frac{{\displaystyle 176.625}}{2} = \frac{618.75}{7} = 88.3125 c{m}^2$
So,
Area of shaded region  =  Area of rectangular part  + Area of semicircle  -  Area of $∆$ AED
Area of shaded region  = 300 + 88.3125  - 54  =  334.3125 cm²                                                ( Ans )