CAT 2025

Slot 2

QA

Question & 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 = $$\frac{\left(BC+PQ\right)}{2}\cdot h=\frac{5}{4}a\cdot h$$

The distance between BC and AD will be $$\sqrt{3}a$$, that is, the length of diagonal BF.

The distance between BC and AD will be $$\frac{\sqrt{3}a}{2}$$ 

And the height of the trapezium BCQP will be  $$\frac{\sqrt{3}a}{4}$$

Area of trapezium = $$\frac{5}{4}a\cdot \frac{\sqrt{3}a}{4}$$ = $$\frac{5\sqrt{3}a^2}{16}$$ 

Area of hexagon ABCDEF = $$\frac{3\sqrt{3}}{2}a^2$$

We have to find the ratio  = $$\frac{5\sqrt{3}a^2}{16}$$ : $$\frac{3\sqrt{3}}{2}a^2$$ = 5:24