question_answer 4)

                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?

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