Sainik School Entrance Exam 2020 Solved Paper

© examsiri.com

Question : 48 of 150

Marks: +1, -0

The sides of the triangle are

45 {\cm}, 60 {\cm}

and

75 {\cm}

. Find the length drawn to the longest side from its opposite vertex

$27 {\cm}$
$21 {\cm}$
$39 {\cm}$
$36 {\cm}$

Solution:

(d) Let a triangle PQR in which PQ = 45 cm

\PR=60 \cm

and

\QR = 75 \cm

from vertex

\P (\PS⊥\QR)

\QS=a ∴ \SR=75-a

Then,

\PS^2=\PQ^2-\QS^2=\PR^2-\SR^2

⇒ 45^2-a^2=60^2-(75-a)^2

⇒ 2025-a^2=3600-(75^2+a^2-150a)

⇒ 2025-a^2=3600-(5625+a^2-150a)

⇒ 2025-a^2=3600-5625-a^2+150a

⇒ 2025-a^2= -2025-a^2+150a

⇒ 150a=4050 ⇒ a = 4050/150 = 27

∴

\PS^2=\PQ^2-\QS^2=45^2-27^2

= 2025-729=1296

∴

\PS^2=1296=(36)^2 ⇒ \PS = 36 \cm

© examsiri.com

Go to Question:

Prev Question

Next Question