The sides of a triangle are 7 cm, 9 cm and 14cm. Its area is:

Q: The sides of a triangle are $7\ cm, 9\ cm$ and $14cm$. Its area is:

Let $a = 7\ cm, b = 9\ cm, c = 14\ cm$ Semi-perimeter = $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7+9+14}{2}=15\text{cm}$ $s - a = 15 -7 = 8\ cm, s - b = 15 - 9 = 6\ cm$ and $s - c = 15 - 14 = 1\ cm$ Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ $=\sqrt{15\times8\times6\times1}$ $=\sqrt{5\times3\times4\times2\times3\times2}$ $=12\sqrt{5}\text{cm}^2$ Hence, correct option is $(a)$.

M.C.Q

GSEB - English Medium

The sides of a triangle are $7\ cm, 9\ cm$ and $14cm$. Its area is:

A$12\sqrt{5}\text{cm}^2$
B$12\sqrt{3}\text{cm}^2$
C$24\sqrt{5}\text{cm}^2$
D$63\text{cm}^2$

STD 9
Maths
Herons Formula
RD Sharma

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