Question

Find the area enclosed by each of the following figures $\left(i\right)--\left(iii\right)$ as the sum of the areas of a rectangle and a trapezium:

Answer 1

Figure $\left(i\right)$

From the figure we can write,

Area of figure $=$ Area of trapezium $+$ Area of rectangle.

$=\frac{1}{2}\times ($ Sum of lengths of parallel sides $)\times$ altitude $+$ Length $\times$ Breadth

$=\frac{1}{2}(18+7)\times 8+18\times 18$

$=\frac{1}{2}\times \left(25\right)\times 8+324$

$=100+324$

$=424$

$\therefore$ Area of figure is $424c{m}^2$

Figure $\left(ii\right)$

From the figure we can write,

Area of figure $=$ Area of trapezium $+$ Area of rectangle.

$=\frac{1}{2}\times ($ Sum of lengths of parallel sides $)\times$ altitude $+$ Length $\times$ Breadth

$=\frac{1}{2}(15+6)\times 8+15\times 20$

$=84+300$

$=384$

$\therefore$ Area of figure is $384c{m}^2$

Figure $\left(iii\right)$

Using Pythagoras theorem in the right angled triangle.

$\Rightarrow$ $5^2=4^2+x^2$

$\Rightarrow$ $x^2=25-16$

$\Rightarrow$ $x^2=9$

$\therefore$ $x=3cm$

From the figure we can write,

Area of figure $=$ Area of trapezium $+$ Area of rectangle.

$=\frac{1}{2}\times ($ Sum of lengths of parallel sides $)\times$ altitude $+$ Length $\times$ Breadth

$=\frac{1}{2}(14+6)\times 3+4\times 6$

$=30+24$

$=54$

$\therefore$ Area of figure is $54c{m}^2$.