SSC CHSL 16 Apr 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 71 of 100

Marks: +1, -0

In

△ A B C, ∠ B=90^{∘}, A D

and

C E

are the medians drawn from

A

and

C

, respectively. If

A C=10 {cm}

and

A D

=√{55} {cm}

, then the length of

C E

is

$2 √{15} {cm}$
$√{70} {cm}$
$√{66} {cm}$
$5 √{3} {cm}$

Solution:

Given,

Since,

A D

and

C E

are medians.

∴ \;\; B D=C D

and

A E=B E

Let

B D=C D=x {cm}

In

△ A B D, A D^{2}=A B^{2}+B D^{2}

55=A B^{2}+x^{2}

⇒ \;\; A B^{2}=55-x^{2}

. . . (i)
In

△ A B C, A C^{2}=A B^{2}+B C^{2}

100=A B^{2}+4 x^{2}

⇒ \;\; A B^{2}=100-4 x^{2}

. . . (ii)
From Eqs. (i) and (ii),

A B^{2}=100-4 × 15

A B=√{100-60}=√{40}

A E=B E=\;{√{40}}/{2}

Now, in

∆ B E C

C E^{2}=B E^{2}+B C^{2}

=(\;{√{40}}/{2})^{2}+(2 × √{15})^{2}

C E^{2} \;\; =\;{40}/{4}+60=70

∴ \;\; C E \;\; =√{70} {cm}

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