Question : The longest side of the obtuse triangle is 7 cm and the other two sides of the triangle are 4 cm and 5 cm. Find the area of the triangle.
Option 1: $1 \sqrt{3} \mathrm{~cm}^2$
Option 2: $6 \sqrt{3} \mathrm{~cm}^2$
Option 3: $3 \sqrt{2} \mathrm{~cm}^2$
Option 4: $4 \sqrt{6} \mathrm{~cm}^2$
Correct Answer: $4 \sqrt{6} \mathrm{~cm}^2$
Solution :
Given triangle is an obtuse triangle as $7^2 > 4^2 + 5^2$.
Semi perimeter of the triangle($s$) = $\frac{7+4+5}{2}$ = 8 cm
$\therefore$ Area of the triangle
= $\sqrt{8(8-7)(8-4)(8-5)}$
= $\sqrt{8×(1)×(4)×(3)}$
= $\sqrt{96}$
= $\sqrt{16×6}$
= $4\sqrt{6} \mathrm{~cm}^2$
Hence, the correct answer is $4 \sqrt{6} \mathrm{~cm}^2$.
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