Corners are cut from an equilateral triangle to produce a regular convex hexagon as shown in the figure above. The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is

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GENERAL APTITUDE

ELECTRONICE & COMMUNICATION 2021 GATE EXAM 

Question:

Corners are cut from an equilateral triangle to produce a regular convex hexagon as shown in the figure above. The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is

(A) 4 : 5 

(B) 5 : 6 

(C) 3 : 4 

(D) 2 : 3 

ANSWER : D

Given that, corners are cut from an equilateral triangle to produce a regular convex hexagon. The smaller triangles cut are equilateral triangles (If we cut off corners to create a regular hexagon, then each angle of the hexagon is 

120∘${120}^{\circ }$, and so each angle of every removed triangle is 60∘${60}^{\circ }$ making these triangles equilateral)

Let the original triangle side be 3a.$3a.$

The area of regular convex hexagon H1=33√2s2,$H_1=\frac{3\sqrt{3}}{2}{s}^2,$ the area of equilateral triangle T1=3√4s2,$T_1=\frac{\sqrt{3}}{4}{s}^2,$ where s$s$ is the side length. 

Now, H1=33√2a2,A1=3√4(3a)2=93√4a2$H_1=\frac{3\sqrt{3}}{2}{a}^2,A_1=\frac{\sqrt{3}}{4}(3a{)}^2=\frac{9\sqrt{3}}{4}{a}^2$

The required ratio =H1A1=33√2a293√4a2=23.$={\displaystyle \frac{H_1}{A_1}}={\displaystyle \frac{\frac{3\sqrt{3}}{2}{a}^2}{\frac{9\sqrt{3}}{4}{a}^2}}=\frac{2}{3}.$

∴$\therefore$ The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is 2:3.$2:3.$