Answer:
Let the length of either of the remaining equal sides be \[x\]. Base \[=\frac{4}{3}\,cm\] \[\therefore \] Perimeter of the triangle \[=\left( x+x+\frac{4}{3} \right)cm\] \[=\left( 2x+\frac{4}{3} \right)cm\] According to the question, \[2x+\frac{4}{3}\,=4\frac{2}{15}\] \[\Rightarrow \] \[2x+\frac{4}{3}=\frac{62}{15}\] \[\Rightarrow \] \[2x=\frac{62}{15}-\frac{4}{3}\] | Transposing \[\frac{4}{3}\] to RHS \[\Rightarrow \] \[2x=\frac{62-20}{15}\] \[\Rightarrow \] \[2x=\frac{42}{15}\] \[\Rightarrow \] \[x=\frac{42}{15\times 2}\] |Dividing both sides by 2 \[\Rightarrow \] \[x=\frac{21}{15}\] \[\Rightarrow \] \[x=\frac{21\div 3}{15\div 3}\] |Dividing the numerator and denominator by 3 \[\Rightarrow \] \[x=\frac{7}{5}\] \[\Rightarrow \] \[=1\frac{2}{5}\] Hence, the length of either of the remaining equal sides is \[1\frac{2}{5}\,cm\]. Check: Perimeter \[=\left( 1\frac{2}{5}+1\frac{2}{5}+\frac{4}{3} \right)\,cm\] \[=\left( \frac{7}{5}+\frac{7}{5}+\frac{4}{3} \right)\,cm\] \[=\frac{21+21+20}{15}\,cm\] \[=\frac{62}{15}\,cm\] \[=4\,\frac{2}{15}\,cm\]. |as desired
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