Question : The perimeter of a triangle is 54 m and its sides are in the ratio of 5 : 6 : 7. The area of the triangle is:
Option 1: $18\ \text{m}^2$
Option 2: $54\sqrt6\ \text{m}^2$
Option 3: $27\sqrt2\ \text{m}^2$
Option 4: $25\ \text{m}^2$
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Correct Answer: $54\sqrt6\ \text{m}^2$
Solution :
Given: The perimeter of a triangle is 54 m and its sides are in the ratio of 5 : 6 : 7.
Let the sides be 5$x$, 6$x$, and 7$x$, respectively.
According to the question,
5$x$ + 6$x$ + 7$x$ = 54
⇒ 18$x$ = 54
⇒ $x$ = 3
So, the sides are (5 × 3), (6 × 3) and (7 × 3) i.e., 15 m, 18 m and 21 m respectively.
Now, the semi perimeter = $\frac{15+18+21}{2}$ = 27 cm
Therefore, the area of the triangle
= $\sqrt{27(27-15)(27-18)(27-21)}$
= $\sqrt{27×12×9×6}$
= $54\sqrt{6}\ \text{m}^2$
Hence, the correct answer is $54\sqrt{6}\ \text{m}^2$.
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