Xavier Aptitude Test 2023 Paper

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Question : 58 of 102

Marks: +1, -0

A B C

is a triangle with

B C=5

.

D

is the foot of the perpendicular from

A

on

B C

.

E

is a point on

C D

such that

{\BE}=3

. The value of

A B^2-A E^2+6 C D

is:

5
10
14
18
21

Solution:

Given that,

A D ⟂ B C

\;B E=3

\;B C=5

Using Pythagoras' theorem,

\;A D^2+B D^2=A B^2

....(1)

\;A D^2+D E^2=A E^2

....(2)

\;A D^2+D C^2=A C^2

....(3)
(1) - (2) gives

\;B D^2+D E^2=A B^2-A E^2

\;x^2-(3-x)^2=A B^2-A E^2

\;A B^2-A E^2=6 x-9

\;A B^2-A E^2+6 C D=6 x-9+6(5-x) \;\; (C D=(5-x))

\;A B^2-A E^2+6 C D=6 x-9+30-6 x

\;A B^2-A E^2+6 C D=21

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