Question

In the given figure, AP=12 cm and PB=16 cm. Taking the value of pi as 3.14, the area of the shaded region is :

Answer 1

Step-by-step explanation:

first by Pythagoras

we can find AB

( because APB is a right angled triangle

because <APB is 90° because angle made from diameter of circle is always 90° .)

(AP)^2 + ( PB )^2 = (AB)^2

now put all values

(12)^2 + (16)^2 = (AB)^2

144 + 256 = (AB)^2

400 = (AB)^2

√400 = AB

20cm = AB

AB is diameter so AB/2 is radius .

AB/2 = 20/2 = 10cm

so now find area of both figures

area of semi - circle = Πr^2 /2 .....(Π is pie )

area of semi - circle = 3.14 * 10^2 /2

area of semi - circle = 314/2 = 107cm^2

now area of triangle = 1/2 * base * height

now put values

area of triangle = 1/2 * 12 * 16

area = 12 * 8

area = 96cm^2

now subtract the area of triangle from area of semi - circle to get the area of shaded region

107 cm^2 - 96cm^2 = 11cm^2

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Area\:of\:shaded\:region}}}$=532${cm}^{2}$

$\frak{ Given}\begin{cases}\textsf{Base\:of\: triangle\:=12\:cm}\\\textsf{Height\:of\: triangle\:=16\:cm}\end{cases}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline{\boxed{\sf Area\:of\:shaded\:region\:=\:Area\:of\:semicircle\:-\:Area\:of\: triangle}}$

$\underline{\bigstar\:\:\textsf{From Pythagoras theorem :}}$

In Right triangle ABP,

$(AB)= \sqrt{ {(AP)}^{2} + {(PB)}^{2} } \\ (AB) = \sqrt{ {16}^{2} + {12}^{2} } \\ (AB) = \sqrt{256 + 144} = \sqrt{400} \\ (AB) = 20cm$

$\underline{\bigstar\:\:\textsf{Area of triangle :}}$=$\frac{1}{2}×base×height$

$\implies$$\frac{1}{2}$×12×16

$\implies$96${cm}^{2}$

$\underline{\bigstar\:\:\textsf{Area of semicircle :}}$=$\frac{1}{2} \pi {r}^{2}$

Radius of semi circle = $\frac{20}{2}$ = 10 cm

$\implies$$\frac{1}{2}×3.14×{{10}^{2}}$

$\implies$$\frac{1}{2}×3.14×100$

$\implies{157{cm}^{2}}$

Now, area of shaded region = 157 - 96 = 61${cm}^{2}$

$\therefore$$\underline{\textsf{Area\:of\:shaded region is}}$$\bold{\underline{{61{cm}^{2}}}}.$