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CAT 2019 Slot 1 — QA Question 1

Basics of TrianglesEasy

Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

Answer & solution

  • A

    5 : 6

  • B

    4 : 5

  • C

    3 : 4

  • 2 : 3

Solution

Easy

A regular hexagon is cut from an equilateral triangle by trimming the three corners. By symmetry the cut corners must themselves be small equilateral triangles, and the hexagon side equals the corner side. Compare areas using the formula 34(side)2\dfrac{\sqrt3}{4}(\text{side})^2.

H T
1

Set the side lengths. Let the hexagon side be xx. Each cut corner is an equilateral triangle of side xx, and they sit at the three corners of TT. So each side of TT is made of three such lengths.

side of T=x+x+x=3x(corner + hexagon edge + corner)\begin{aligned} &\text{side of }T = x + x + x = 3x \quad\text{(corner + hexagon edge + corner)} \end{aligned}
2

Area of the hexagon. A regular hexagon of side xx splits into 66 equilateral triangles of side xx.

[H]=6⋅34x2=634x2(6 unit triangles)\begin{aligned} &[H] = 6\cdot\frac{\sqrt3}{4}x^2 = \frac{6\sqrt3}{4}x^2 \quad\text{(6 unit triangles)} \end{aligned}
3

Area of the big triangle. Side 3x3x.

[T]=34(3x)2=934x2(from step 1)\begin{aligned} &[T] = \frac{\sqrt3}{4}(3x)^2 = \frac{9\sqrt3}{4}x^2 \quad\text{(from step 1)} \end{aligned}
4

Take the ratio. Divide step 2 by step 3.

[H][T]=63/493/4=69=23\begin{aligned} &\frac{[H]}{[T]} = \frac{6\sqrt3/4}{9\sqrt3/4} = \frac{6}{9} = \frac{2}{3} \end{aligned}
[H]:[T]=2:3[H]:[T] = 2:3

Linear scale factor of TT to the unit triangle is 33, so [T]=9[T]=9 units. The hexagon is 66 of those units, giving 6:9=2:36:9=2:3 directly.

CAT 2019 Slot 1 QA Q1: Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area — Solution | TheCATExam