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: $4 \sqrt{6} \mathrm{~cm}^2$
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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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Question : If $\triangle \mathrm{ABC}$ is similar to $\triangle \mathrm{DEF}$ such that $\mathrm{BC}=3 \mathrm{~cm}, \mathrm{EF}=4 \mathrm{~cm}$ and the area of $\triangle \mathrm{ABC}=54 \mathrm{~cm}^2$, then the area of $\triangle \mathrm{DEF}$ is:
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