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Answer :
We have length of base as \(\frac{4}3\)
Let the length of the equal sides by each of x.
Perimeter of the isosceles triangle=sum of the base +sum of two equal sides=\(4\frac{2}{15}\)
\(\qquad = (\frac{4}3+x+x)cm =4\frac{2}{15}\)
∴ \(2x+\frac{4}3=4\frac{2}{15}\)
\(\Rightarrow 2x+\frac{4}3=\frac{62}{15}\)
\(\Rightarrow 2x=\frac{62}{15}-\frac{4}3\quad \)[Transposing \(\frac{4}3\) from (+) to (-)]
\(\Rightarrow 2x=\frac{62-20}{15}\)
\(\Rightarrow 2x=\frac{42}{15}\)
\(\Rightarrow x=\frac{42}{15}÷2\quad \)[Transposing 2 from (x) to ÷]
\(\Rightarrow x=\frac{42}{15}×\frac{1}2\)
\(\Rightarrow x=\frac{7}{5}=1\frac{2}5\)
Hence, the length of each equal side is =\(1\frac{2}5\)