Question

An advertisement board is in the form of an isosceles triangle with perimeter 36m and each of the equal sides are 13m . find the cost of the painting it at $17.50 per square meter​

Answer 1

$\huge\mathbb{\underline{\red{SOLUTION:-}}}$

$\bold{Given:-}\begin{cases}\sf{two\:Sides\:of\: isosceles\:triangle=13m}\\ \sf{perimeter=36m}\end{cases}$

$\huge\sf{\red{To\:Find:-}}$

Cost of painting the advertisement board

$\bf\underline{\underline{\blue{According\:To\: Question:-}}}$

• First find the third side of triangle

$\boxed{\sf{\star{Perimeter\:of\: triangle=sum\:of\:all\:sides}}}$

• let the third side be x

$\sf\implies 36m=13m+13m+x\\ \\ \sf\implies 36m=26m+x\\ \\ \sf\implies 36m-26m=x\\ \\ \sf\implies 10m=x$

• The third side of triangle is 10m
• Now the area of triangle
• By heron's Formula of triangle

$\boxed{\sf{\blue{Area\:of\: triangle=\sqrt{s(s-a)(s-b)(s-c)}}}}$

• Where s=semi perimeter

• $\large\tt s=\frac{a+b+c}{2}$

• a ,b and c are the sides of triangle

$\tt\implies s=\frac{13+13+10}{2}\\ \\ \tt\implies s=\cancel{\frac{36}{2}}\\ \\ \tt\implies{\blue{semi\: perimeter=18m}}$

• Now the area of triangle

$\sf\implies Area\:of\: triangle=\sqrt{18(18-13)(18-13)(18-10)}\\ \\ \sf\implies Area\:of\: triangle=\sqrt{18\times 5\times 5\times 8}\\ \\ \sf\implies Area\:of\: triangle=\sqrt{3\times 3\times 2\times 5\times 5\times 2\times 2\times 2}\\ \\ \sf\implies Area\:of\: triangle=3\times 2\times 5\times 2\\ \\ \sf\implies Area\:of\: triangle=6\times 10\\ \\ \sf\implies Area\:of\: triangle=60m{}^{2}$

• The area of triangular advertisement board is 60 square metre unit

• Now the cost of its painting

$\sf\implies Cost\:of\: painting=60\times 17.50\\ \\ \sf\implies Cost\:of\: painting=1050$

$\boxed{\star{\sf{\blue{Cost\:of\: painting\:the\: advertisement\:board=1050 .}}}}$

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