Question

The base of an isosceles triangle is $\frac{4}{3}$ cm. The perimeter of the triangle is $4\frac{2}{15}$ cm. What is the length of either of the remaining equal sides?

• A. Length of either of the remaining equal sides is $1\frac{2}{5}$ cm
• B. Length of either of the remaining equal sides is $1\frac{1}{3}$ cm
• C. Length of either of the remaining equal sides is $2\frac{3}{4}$ cm
• D. Length of either of the remaining equal sides is $5\frac{1}{6}$ cm

Answer 1

Answer: (A)

Length of either of the remaining equal sides is $1\frac{2}{5}$ cm

Let each equal side of an isosceles triangle ABC be $x$ cm.

Then, its perimeter $=AB+BC+CA=(x+\frac{4}{3}+x)$ cm $=(2x+\frac{4}{3})$ cm

But it is given that the perimeter of the triangle $=4\frac{2}{15}$ cm $=\frac{62}{15}$ cm

Therefore, $2x+\frac{4}{3}=\frac{62}{15}$

$\Rightarrow 30x+20=62$ ...[On multiplying both sides by $15$]

$\Rightarrow 30x=62-20$

$\Rightarrow 30x=42$

$\Rightarrow x=\frac{42}{30}=\frac{7}{5}=1\frac{2}{5}$

Thus, each equal side of the isosceles traingle is of $1\frac{2}{5}$ cm.