Test Index

Properties of Triangles

Section: Mathematics

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Question : 18 of 50

Marks: +1, -0

In a triangle

A B C

\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}, a=2, c=5

b

is an integer, then the area (in sq. units) of triangle ABC is

[AP EAPCET 21 May 2025 S2]

$\frac{\sqrt{297}}{4}$
$\frac{\sqrt{231}}{4}$
$\frac{\sqrt{385}}{4}$
$\frac{\sqrt{185}}{4}$

Solution:

\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}} a=2 c=5

\cos A = 1-2 \sin^2 \frac{A}{2} = 1-2 \cdot \frac{1}{16} \times \frac{3}{5} = \frac{37}{40}

\text{ Now, } \cos A = \frac{b^2+c^2-a^2}{2 \times b \times c}

\Rightarrow \frac{37}{40} = \frac{b^2+25-4}{2 \times b \times 5}

\Rightarrow \frac{37}{4} = \frac{b^2+21}{b}

\Rightarrow 4b^2-37b+84=0

\therefore (4b-21)(b-4)=0

\therefore b=4

because given

b

is an integer

\because \sin A = \sqrt{1-\cos^2 A} = \sqrt{1- \left(\frac{37}{40}\right)^2}

\therefore \text{ Area of triangle } = \frac{1}{2} b \times c \times \sin A

= \frac{1}{2} \times 4 \times 5 \sqrt{1- \left(\frac{37}{40}\right)^2}

= 10 \sqrt{\frac{77}{40} \times \frac{3}{40}} = \frac{1}{4} \sqrt{231}

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