# 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)

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 $\u2206$ AED , and get

AD

^{2}= AE

^{2}+ DE

^{2}

AD

^{2}= 9

^{2}+ 12

^{2}

AD

^{2}= 81 + 144

AD

^{2}= 225

AD = 15 cm

So,

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

So,

Radius of semicircle = $\frac{BC}{2}=\frac{15}{2}=7.5cm$

And

We know Area of triangle = $\frac{1}{2}\times Base\times height$ , So

Area of $\u2206$ AED = $\frac{1}{2}\times 9\times 12=9\times 6=54c{m}^{2}$

And Area of rectangle = Length $\times $ Breadth , So

Area of rectangular part ABCD = 20 $\times $ 15 = 300 cm

^{2}

And

We know Area of semicircle = $\frac{\pi {r}^{2}}{2}$ , So

Area of semicircle = $\frac{{\displaystyle 3.14\times 7.5\times 7.5}}{2}=\frac{{\displaystyle 176.625}}{2}=\frac{618.75}{7}=88.3125c{m}^{2}$

So,

Area of shaded region = Area of rectangular part + Area of semicircle - Area of $\u2206$ AED

Area of shaded region = 300 + 88.3125 - 54 =

**334.3125 cm**

^{2}( Ans )
**
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