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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?
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.