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Find the area of a triangle whose sides are $5 \mathrm{~cm}, 12 \mathrm{~cm}$ and $13 \mathrm{~cm}$.

(a) $15 \mathrm{sq} . \mathrm{cm}$

(b) $30 \mathrm{sq} . \mathrm{cm}$

(c) $60 \mathrm{sq} \cdot \mathrm{cm}$

(d) $90 \mathrm{sq} \cdot \mathrm{cm}$

Solution

Here, $a=5 \mathrm{~cm}, b=12 \mathrm{~cm}$

$c=13 \mathrm{~cm}$

$\therefore \quad s=\frac{5+12+13}{2}=\frac{30}{2}=15 \mathrm{~cm}$

$\therefore$ Area of the given triangle

$$\begin{array}{l} =\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{15(15-5)(15-12)(15-13)} \\ =\sqrt{15 \times 10 \times 3 \times 2} \\ =\sqrt{3 \times 5 \times 5 \times 2 \times 3 \times 2} \\ =3 \times 5 \times 2=30 \text { sq. cm; Ans. } \end{array}$$

So, the correct option is (B)

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