Find the area enclosed by each of the following

Solution:

(i) The given figure can be divided into a rectangle and a trapezium as shown below:

From the above firgure:

Area of the complete figure $=($ Area of square $\mathrm{ABCF})+($ Area of trapezium $\mathrm{CDEF})$

$=(\mathrm{AB} \times \mathrm{BC})+\left[\frac{1}{2} \times(\mathrm{FC}+\mathrm{ED}) \times(\right.$ Distance between $\mathrm{FC}$ and $\left.\mathrm{ED})\right]$

$=(18 \times 18)+\left[\frac{1}{2} \times(18+7) \times(8)\right]$

$=324+100$

$=424 \mathrm{~cm}^{2}$

(ii) The given figure can be divided in the following manner:

From the above figure:

$\mathrm{AB}=\mathrm{AC}-\mathrm{BC}=28-20=8 \mathrm{~cm}$

So that area of the complete figure $=($ area of rectangle $\mathrm{BCDE})+($ area of trapezium $\mathrm{ABEF})$

$=(\mathrm{BC} \times \mathrm{CD})+\left[\frac{1}{2} \times(\mathrm{BE}+\mathrm{AF}) \times(\mathrm{AB})\right]$

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

$=300+84$

$=384 \mathrm{~cm}^{2}$

(iii) The given figure can be divided in the following manner:

From the above figure:

$\mathrm{EF}=\mathrm{AB}=6 \mathrm{~cm}$

Now, using the Pythagoras theorem in the right angle triangle CDE:

$5^{2}=4^{2}+\mathrm{CE}^{2}$

$\mathrm{CE}^{2}=25-16=9$

$\mathrm{CE}=\sqrt{9}=3 \mathrm{~cm}$

And, $\mathrm{GD}=\mathrm{GH}+\mathrm{HC}+\mathrm{CD}=4+6+4=14 \mathrm{~cm}$

$\therefore$ Area of the complete figure $=($ Area of rectangle $\mathrm{ABCH})+($ Area of trapezium GDEF $)$

$=(\mathrm{AB} \times \mathrm{BC})+\left[\frac{1}{2} \times(\mathrm{GD}+\mathrm{EF}) \times(\mathrm{CE})\right]$

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

$=24+30$

$=54 \mathrm{~cm}^{2}$