

InterviewSolution
1. |
Find the area enclosed by each of the following figures [fig. (i) - (ii)] as the sum of the areas of a rectangle and a trapezium. |
Answer» Figure (i) Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (18 + 7) x 8 + 18 x 18 Area of given figure = 100 +324 = 424 Therefore area of given figure is 424 cm2 Figure (ii) Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (15 + 6) x 8 + 15 x 20 Area of given figure = 84 + 300 = 384 Therefore area of given figure is 384 cm2 Figure (iii) Using pythagorous theorem in the right angled triangle 52 = 42 + x2 x2 = 25 – 16 x2 = 9 x = 3 cm Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (14 + 6) x 3 + 4 x 6 Area of given figure = 30 + 24 = 54 Therefore area of given figure is 54 cm2 |
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