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MAT Exam 2022 Paper

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Question : 115 of 200

Marks: +1, -0

In the given figure ABCD is a rectangle. AB is 3 m and BC is 2 m. It is given that CF is 1 m. AF intersects CD at E and it is known that BE is parallel to HG, GE is perpendicular to AB. Find the length of HG (in m).

$\frac{2\sqrt{5}}{3}$
$\sqrt{2}$
$\frac{4\sqrt{5}}{3}$
$\frac{2\sqrt{5}}{5}$

Solution:

Concept:

Use the basic proportionality theorem and similarity of triangles inside the rectangle.

Explanation:

In rectangle

ABCD

,

AB = 3

m and

BC = 2

m.

Given

CF = 1

m, so

BF = BC + CF = 2 + 1 = 3

m.

Since

AB \parallel EC

in

\triangle ABF

, we get:

\frac{EC}{AB} = \frac{CF}{BF}

Substituting the values:

\frac{EC}{3} = \frac{1}{3} \Rightarrow EC = 1 \text{ m}

Since

GE \perp AB

and

E

lies on

CD

, we have

GB = EC = 1

m.

Also,

GE = BC = 2

m.

In right

\triangle GBE

:

BE = \sqrt{GB^2 + GE^2} = \sqrt{1^2 + 2^2} = \sqrt{5} \text{ m}

Clearly,

AG = AB - GB = 3 - 1 = 2

m.

Given

HG \parallel BE

, triangles

AHG

and

AEB

are similar:

\triangle AHG \sim \triangle AEB

Therefore:

\frac{HG}{BE} = \frac{AG}{AB} = \frac{2}{3}

So:

HG = \frac{2}{3} \times \sqrt{5} = \frac{2\sqrt{5}}{3} \text{ m}

Answer:

HG = \frac{2\sqrt{5}}{3}

m

Correct option: A.

\frac{2\sqrt{5}}{3}

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