# Hexagon Questions for CAT PDF

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## Hexagon Questions for CAT

Download important CAT Hexagon Questions PDF based on previously asked questions in CAT exam. Practice Hexagon questions with solutions for CAT exam.

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Question 1:

a)  3, 3

b) 4, 2

c) 2, 4

d) 4, 4

Question 2: In the figure below, ABCDEF is a regular hexagon and $\angle{AOF}$ = 90° . FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?

a) 1/12

b) 1/6

c) 1/24

d) 1/18

Question 3: ABCDEF is a regular hexagon and PQR is an equilateral triangle of side a. The area of the shaded portion is X and CD : PQ : : 2 : 1. Find the area of the circle circumscribing the hexagon in terms of X.

a) $\frac{16 \pi}{23 \sqrt{3}}$ X

b) $\frac{42 \pi}{5 \sqrt{3}}$ X

c) $\frac{2 \pi}{3 \sqrt{3}}$ X

d) $2\sqrt{3 \pi}$ X

Question 4: Let PQRSTU be a regular hexagon. The ratio of the area of the triangle PRT to that of the hexagon PQRSTU is

a) 0.3

b) 0.5

c) 1

d) None of the above

Question 5: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

a) $3\sqrt{2}$

b) $3$

c) $4$

d) $\sqrt{3}$

When the hexagon is divided into number of similar triangle AOF we get 12 such triangles . Hence required ratio of area is 1/12.

Let ‘a’ be the length of side of the hexagon. Therefore, the area of hexagon = $6*\dfrac{\sqrt{3}}{4}\times a^2$.

The area of triangle PQR = $\dfrac{\sqrt{3}}{4}\times (a/2)^2$

Therefore, the are of the shaded region = $6*\dfrac{\sqrt{3}}{4}\times a^2$ – $\dfrac{\sqrt{3}}{4}\times (a/2)^2$ = $\dfrac{23\sqrt{3}a^2}{16} = X$

$\Rightarrow$ $a^2 = \dfrac{16X}{23\sqrt{3}}$

The area of the circle = $\pi*a^2$

$\Rightarrow$ $\dfrac{16 \pi}{23 \sqrt{3}}$ X. Hence, option A is the correct answer.

It’s given that PQRSTU is a regular hexagon and O is the center of the hexagon.

If we fold $\triangle$TSR ,$\triangle$PQR ,$\triangle$TUP along lines TR, PR, PT respectively then vertices S,Q,U will overlap each other exactly at center of the hexagon.

Hence we can say that Area of hexagon PQRSTU =  2($\triangle$PRT)

So Area of $\triangle$PRT = 0.5(Area of hexagon PQRSTU)

The length of the diagonals of a regular hexagon with side s are $\sqrt{3}s$.
$\sqrt{3}s$ = $\sqrt{3}$ cms
Hence area of the square = $\sqrt{3}^2$ = 3 sq cm