Question

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:

Answer 1

$$\begin{gathered} \text{(i)} \\ \text{The given figure can be divided into a rectangle and a trapezium as shown below:} \end{gathered}$$

$$\begin{gathered} \text{From the above firgure:} \\ \text{Area of the complete figure = (Area of square ABCF)+(Area of trapezium CDEF)} \\ \text{=(AB×BC)+[}\frac{1}{2}\text{×(FC+ED)×(Distance between FC and ED)]} \\ \text{=(18×18)+[}\frac{1}{2}\text{×(18+7)×(8)]} \\ \text{=324+100} \\ {\text{=424 cm}}^2 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{The given figure can be divided in the following manner:} \end{gathered}$$

$$\begin{gathered} \text{From the above figure:} \\ \text{AB = AC-BC=28-20=8 cm} \\ \text{So that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF)} \\ \text{=(BC×CD)+[}\frac{1}{2}\text{×(BE+AF)×(AB)]} \\ \text{=(20×15)+[}\frac{1}{2}\text{×(15+6)×(8)]} \\ \text{=300+84} \\ {\text{=384 cm}}^2 \end{gathered}$$


$$\begin{gathered} \text{(iii)} \\ \text{The given figure can be divided in the following manner:} \end{gathered}$$

$$\begin{gathered} \text{From the above figure:} \\ \text{EF = AB = 6 cm} \\ \text{Now, using the Pythagoras theorem in the right angle triangle CDE:} \\ {\text{5}}^2{\text{= 4}}^2{\text{+CE}}^2 \\ {\text{CE}}^2\text{= 25-16=9} \\ \text{CE =}\sqrt{9}\text{= 3 cm} \\ \text{And, GD=GH+HC+CD=4+6+4=14 cm} \\ \therefore \text{Area of the complete figure=(Area of rectangle ABCH)+(Area of trapezium GDEF)} \\ \text{=(AB×BC)+[}\frac{1}{2}\text{×(GD+EF)×(CE)]} \\ \text{=(6×4)+[}\frac{1}{2}\text{×(14+6)×(3)]} \\ \text{=24+30} \\ {\text{=54 cm}}^2 \end{gathered}$$