The base of an isosceles triangle …

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Sia ? 6 years, 8 months ago
The base of an isosceles triangle = {tex}\frac{4}{3}{/tex}cm
let two equal sides are = x
perimeter of the triangle = {tex}4\frac{2}{{15}}{/tex}cm
perimeter of the triangle = sum of all sides
{tex}62/15 = x +x+4/3{/tex}
or, {tex}62/15=2x+4/3{/tex}
or, {tex}62/15 =(6x+4) /3{/tex}
By crossmutliply,
or, {tex}186= 90x +60{/tex}
or, {tex}186 -60 =90x{/tex}
or, {tex}126= 90x{/tex}
or, {tex}126/90 = x{/tex}
or, {tex}7/5 = x{/tex}
{tex}1\frac{2}{5}cm{/tex} = {tex}x{/tex}
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