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

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Given:

The base of an isosceles triangle $=\frac{4}{3}\ cm$

The perimeter of the triangle $=4 \frac{2}{15}\ cm = \frac{(4\times 15+2)}{15}\ cm = \frac{(60+2)}{15}\ cm = \frac{62}{15}\ cm$.

To do:

We have to find the length of either of the remaining equal sides.

Solution :

Let the length of the equal sides be $y\ cm$ each.

We know that,

The perimeter of a triangle is the sum of the three sides of the triangle.

Therefore,

$\frac{4}{3} + y + y = \frac{62}{15}$

$\frac{4}{3} + 2y = \frac{62}{15}$

$2y=\frac{62}{15} - \frac{4}{3}$

$2y = \frac{(62-4\times 5)}{15}$         (LCM of 15 and 3 is 15)

$2y = \frac{(62-20)}{15}$

$2y = \frac{42}{15}$

$y = \frac{(2\times 21)}{(15\times 2)}$

$y = \frac{21}{15}$

$y = \frac{7}{5}$

Therefore, the length of either of the remaining sides is $\frac{7}{5}$cm.

Updated on 10-Oct-2022 13:47:38