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Question
Find the area of Fig. as the sum of the areas of two trapezium and a rectangle.
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Solution
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.
Suppose x is the perpendicular distance between the parallel sides in both the trapeziums.
We have:
Total length of the given figure = Length of the rectangle + 2 x Perpendicular distancebetween the parallel sides in both the trapeziums
\[70 = 50 + 2 \times x\]
\[2\times x=70-50=20\]
\[x=\frac{20}{2}=10 cm\]
Now, area of the complete figure = (area of the rectangle with sides 50 cm and 10 cm) + 2 x (area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm
\[=(50\times10)+2\times[\frac{1}{2}\times(30+10)\times(10)]\]
\[ = 500 + 2 \times [200]\]
= 900 cm2
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