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Question 61

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

Solution

Let the side of the hexagon be a.

The area of the whole hexagon is going to be $$\frac{3\sqrt{3}a^2}{2}$$

We know that the longest diagonal of a hexagon is 2 times the side of the hexagon.

PQ = $$\frac{\left(BC+AD\right)}{2}= \frac{\left(a+2a\right)}{2} =1.5a$$

The area of the trapezium = Average of parllel sides * Height = $$\frac{\left(BC+PQ\right)}{2}\cdot h=\frac{5}{4}a\cdot h$$

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