Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:

#### 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 ABCF)+(Area of trapezium CDEF)

\[=(AB\times BC)+[\frac{1}{2}\times(FC+ED)\times(\text{ Distance between FC and ED })]\]

\[=(18\times18)+[\frac{1}{2}\times(18+7)\times(8)]\]

\[=324+100\]

\[ {=424 cm}^2\]

\[(ii)\]

The given figure can be divided in the following manner:\]

From the above figure:

AB = AC-BC=28-20=8 cm

So that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF)

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

\[=(20\times15)+[\frac{1}{2}\times(15+6)\times(8)]\]

\[=300+84\]

\[ {=384 cm}^2\]

The given figure can be divided in the following manner:

From the above figure:

EF = AB = 6 cm

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

\[ 5^2 {= 4}^2 {+CE}^2 \]

\[ {CE}^2 = 25-16=9\]

\[CE =\sqrt{9}= 3 cm\]

\[\text{ And, }GD=GH+HC+CD=4+6+4=14 cm\]

\[ \therefore\text{ Area of the complete figure }= (\text{ Area of rectangle ABCH })+(\text{ Area of trapezium GDEF })\]

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

\[=(6\times4)+[\frac{1}{2}\times(14+6)\times(3)]\]

\[=24+30\]

\[ {=54 cm}^2\]