SSC CHSL 12 Oct 2020 Shift 1 Paper

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Question : 56 of 100

Marks: +1, -0

M is the circumcentre of

\Δ\ABC

with circumradius

15 {\cm}

. Let

\BC=24 {\cm}

and ML is perpendicular to BC. Then the length of ML is

$9 {\cm}$
$10 {\cm}$
$12 {\cm}$
$8 {\cm}$

Solution:

M is circumcentre

\CM=

circum radius

=15 {\cm}

\LC={1}/{2} \BC={1}/{2} \×24=12 {\cm}

Now, in right

\Δ\MLC

,
By Pythagoras theorem,

\CM^{2}=\ML^{2}+\LC^{2}

(15)^{2}=\ML^{2}+(12)^{2}

225=\ML^{2}+144

\ML^{2}=225-144

\ML^{2}=81

\ML=9 {\cm}

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