Question

Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a rectangle.

Answer 1

$\text{The given figure is:}$



$$\begin{gathered} \text{In the given figure, we have a rectangle of length 50 cm and width 10 cm, and two similar trapeziums with parallel sides as 30 cm and 10 cm at both ends.} \\ \text{Suppose x is the perpendicular distance between the parallel sides in both the trapeziums.} \\ \text{We have:} \\ \text{Total length of the given figure= Length of the rectangle+ 2×Perpendicular distance}\mathrm{between}\mathrm{the}\mathrm{parallel}\mathrm{sides}\mathrm{in}\mathrm{both}\mathrm{the}\text{trapeziums} \\ 70=50+2\times \text{x} \\ \text{2×x=70-50=20} \\ \text{x=}\frac{20}{2}\text{=10 cm} \\ \text{Now, area of the complete figure=(area of the rectangle with sides 50 cm and 10 cm)+2×(area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm)} \\ \text{=(50×10)+2×[}\frac{1}{2}\text{×(30+10)×(10)]} \\ =500+2\times \left[200\right] \\ =900{\text{cm}}^2 \end{gathered}$$