If the perimeter of an isosceles triangle is 32 cm and the ratio of the equal side to its base is 3: 2, then area

Q: If the perimeter of an isosceles triangle is 32 cm and the ratio of the equal side to its base is $3: 2$, then area of the triangle is

(d) $32 \sqrt{2} cm^2$ Let $a cm$ be the base and each equal side be $b cm$. Then, $a+2 b=32$$\quad$[Given] ...(i) It is given that $\frac{b}{a}=\frac{3}{2} \Rightarrow 2 b=3 a \Rightarrow b=\frac{3 a}{2}$. Putting $b=\frac{3 a}{2}$ in (i), we obtain $\begin{array}{ll}& a+3 a=32 \Rightarrow a=8 cm \\ \therefore \quad & b=12 cm\end{array}$ Let $A$ be the area of the triangle. Then, $A=\frac{a}{4} \sqrt{4 b^2-a^2}=\frac{8}{4} \sqrt{4 \times 12^2-8^2} cm^2=2 \sqrt{512} cm^2=32 \sqrt{2} cm^2$

M.C.Q

CBSE - English Medium

If the perimeter of an isosceles triangle is 32 cm and the ratio of the equal side to its base is $3: 2$, then area of the triangle is

A$16 \sqrt{2} cm^2$
B$20 \sqrt{2} cm^2$
C$30 \sqrt{2} cm^2$
D$32 \sqrt{2} cm^2$

STD 9
Maths
Herons Formula
RD Sharma

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