Question

$the \: base \: of \: an \: isosceles \: triangle \: is \: \frac{4}{3} cm.the \: perimeter \: of \: the \: triangle \: is \: 4 \times \frac{2}{15} cm.what \: is \: the \: length \: of \: either \: of \: the \: remaining \: equal \: sides$
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Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that the required isosceles triangle be ABC such that AB = AC and base is BC.

Let assume that AB = AC = x cm

Now, given that

$\rm \: Base\:of\:\triangle ABC, \: BC = \frac{4}{3} \: cm$

Further, given that

$\rm \: Perimeter\:of\:\triangle ABC \: = 4 \frac{2}{15} \: cm$

$\rm \: Perimeter\:of\:\triangle ABC \: = \frac{15 \times 4 + 2}{15} \: cm$

$\rm \: Perimeter\:of\:\triangle ABC \: = \frac{60 + 2}{15} \: cm$

$\rm \: Perimeter\:of\:\triangle ABC \: = \frac{62}{15} \: cm$

We know,

Perimeter of a triangle is defined as sum of its three sides.

So,

$\rm \: AB + AC + BC \: = \frac{62}{15} \:$

$\rm \: x + x + \frac{4}{3} = \frac{62}{15} \:$

$\rm \: 2x+ \frac{4}{3} = \frac{62}{15} \:$

$\rm \: 2x = \frac{62}{15} - \frac{4}{3} \:$

$\rm \: 2x = \frac{62 - 20}{15} \:$

$\rm \: 2x = \frac{42}{15} \:$

$\rm \: 2x = \frac{14}{5} \:$

$\rm\implies \:\rm \: x = \frac{7}{5} \: cm \:$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{AB = \dfrac{7}{5} \: cm} \\ \\ &\sf{AC = \dfrac{7}{5} \: cm} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Refer the given attachment