Find the area of the Figur given in ncert book Exercise 12.2 Question Number 3 Page no. 206 Please illustrate the calculations.

$\mathrm{There} \mathrm{are} 5 \mathrm{areas}.$$\mathrm{Area} I : \mathrm{It} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle} \mathrm{with} \mathrm{equal} \mathrm{side} 5\mathrm{cm} \mathrm{and} \mathrm{base} 1\mathrm{cm}.$$\mathrm{Using} \mathrm{Heron}\text{'}s \mathrm{formula} , s=\frac{5+5+1}{2}=\frac{11}{2}$$s(s-a)(s-b)(s-c)=\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-5)(\frac{11}{2}-1)$$= \frac{11}{2}\frac{1}{2}\frac{1}{2}\frac{9}{2}=\frac{99}{16}$$\mathrm{Therefore} \mathrm{Area} I = \mathrm{square} \mathrm{root} \mathrm{of} \frac{99}{16}=\frac{3\sqrt{11}}{4}=2.5\left(\mathrm{approx}\right)$$$$\mathrm{Area} \mathrm{II} :$$\mathrm{It} \mathrm{is} a \mathrm{rectangle} \mathrm{with} \mathrm{sides} 6.5\mathrm{cm} \mathrm{and} 1\mathrm{cm}.$$\mathrm{Therefore} \mathrm{its} \mathrm{area} = 6.5\mathrm{x1}=\frac{13}{2}\mathrm{x1}=\frac{13}{2}=6.5\mathrm{sq} \mathrm{cm}$$\mathrm{Area} \mathrm{III} :$$\mathrm{This} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{trapezium} \mathrm{with} \mathrm{parallel} \mathrm{sides} 1\mathrm{cm} \mathrm{and} 2 \mathrm{cm}.$$\mathrm{Therefore} \mathrm{its} \mathrm{area} = \frac{1}{2}\mathrm{xheight} (\mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides})$$\mathrm{Find} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{trapezium} \mathrm{by} \mathrm{drawing} \mathrm{perpendicular} \mathrm{on} \mathrm{the} \mathrm{base} \mathrm{and} \mathrm{using} \mathrm{pythagoras} \mathrm{theorem}.$$\mathrm{In} \mathrm{the} \mathrm{right} -\mathrm{angled} \mathrm{triangle}, \mathrm{hyotenuse} = 1 \mathrm{cm} \mathrm{and} \mathrm{base} = 0.5 \mathrm{cm}$$\mathrm{therefore} (\mathrm{height}{)}^2 = 1^2-{\left(\frac{1}{2}\right)}^2= \frac{3}{4}.$$\mathrm{height} =h = \frac{\mathrm{sq} \mathrm{rt} 3}{2}$$\mathrm{Area} \mathrm{III} =\frac{1}{2}x\frac{\mathrm{sq} \mathrm{rt} 3}{2}x(1+2)=1.3\mathrm{sq} \mathrm{cm} \left(\mathrm{aprox}\right)$$$$\mathrm{Area} \mathrm{IV} \mathrm{and} V :$$\mathrm{These} \mathrm{are} \mathrm{two} \mathrm{right} \mathrm{angled} \mathrm{triangles} \mathrm{with} \mathrm{base} 6 \mathrm{cm} \mathrm{and} \mathrm{height} 1.5 \mathrm{cm}$$\mathrm{Therefore} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{triangle}=\frac{1}{2}\mathrm{x6x}\frac{3}{2}=\frac{9}{2}=4.5.$$\mathrm{The} \mathrm{total} \mathrm{Area} \mathrm{IV} \mathrm{and} V = 2x 4.5=9\mathrm{sq} \mathrm{cm}$$$$\mathrm{Thus} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{complete} \mathrm{figure} \mathrm{is} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{areas}$$=2.5+6.5+1.3+9 =19.3 \mathrm{sq} \mathrm{cm} \left(\mathrm{aprox}\right)$$$">