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CAT 2021 Slot 1 — QA Question 14

HexagonEasy

Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is

Answer & solution

  • A

    √14

  • B

    √12

  • √13

  • D

    √15

Solution

Easy

Place the hexagon on coordinates (or use the fact that opposite sides of a regular hexagon are 3×side\sqrt3\times\text{side} apart). Build a right triangle whose legs are this perpendicular distance and a horizontal offset, then apply the Pythagorean theorem to get ATAT.

A B C D E F T
1

Set coordinates. For a regular hexagon of side 22, the centre-to-vertex distance equals the side =2=2. Place the centre at the origin with A=(−1,3), B=(1,3)A=(-1,\sqrt3),\ B=(1,\sqrt3) on top and going clockwise C=(2,0), D=(1,−3)C=(2,0),\ D=(1,-\sqrt3).

2

Locate TT, the midpoint of CDCD.

T=(2+12,0−32)=(32,−32)\begin{aligned} &T=\left(\tfrac{2+1}{2},\tfrac{0-\sqrt3}{2}\right)=\left(\tfrac32,-\tfrac{\sqrt3}{2}\right) \end{aligned}
3

Apply the distance formula from A=(−1,3)A=(-1,\sqrt3) to T=(32,−32)T=\left(\tfrac32,-\tfrac{\sqrt3}{2}\right).

AT2=(32−(−1))2+(−32−3)2⇒ AT2=(52)2+(−332)2⇒ AT2=254+274=524=13\begin{aligned} &AT^2=\left(\tfrac32-(-1)\right)^2+\left(-\tfrac{\sqrt3}{2}-\sqrt3\right)^2\\ &\Rightarrow\ AT^2=\left(\tfrac52\right)^2+\left(-\tfrac{3\sqrt3}{2}\right)^2\\ &\Rightarrow\ AT^2=\tfrac{25}{4}+\tfrac{27}{4}=\tfrac{52}{4}=13 \end{aligned}
4

Take the square root.

AT=13 cm\begin{aligned} &AT=\sqrt{13}\ \text{cm} \end{aligned}
AT=13 cm (option c)AT=\sqrt{13}\ \text{cm (option c)}
Need a hint?

Drop a perpendicular from TT to line AFAF. The distance between the parallel sides AFAF and CDCD is 3×side=23\sqrt3\times\text{side}=2\sqrt3, and the foot lands 11 unit from AA along AFAF. Then AT2=(23)2+12=13AT^2=(2\sqrt3)^2+1^2=13.

CAT 2021 Slot 1 QA Q14: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the len — Solution | TheCATExam