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Question 10

Both parallel sides of a trapezium are increased by 10% while its height kept constant. By what percent does its area increase?

Solution

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The area is $$\frac{1}{2}(p + q)h$$, and each of $$p$$ and $$q$$ becomes $$1.1$$ times its value while $$h$$ is unchanged. So the new area is $$\frac{1}{2}(1.1p + 1.1q)h = 1.1 \times \frac{1}{2}(p + q)h$$, an increase of 10%.

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