Question

find the area of a triangle whose sides are 91cm ,98cm and 105cm is length. find the height of corrosponding to the largest side?

Answer 1

$\huge\mathtt{\blue{Answer:-}}$

$\sf\implies 4116cm{}^{2}$

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

$\bold{Given}\begin{cases}\sf{Side=91cm,98cm,105cm}\end{cases}$

• We find the area of this triangle by heron's Formula
• Heron's formula of triangle

$\sf\implies \sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf where \:s=\frac{a+b+c}{2}$

$\sf\underline{\red{According\:to\: question:-}}$

$\sf\implies s=\frac{91+98+105}{2}\\ \\ \sf\implies s=\frac{\cancel{294}}{\cancel2}=147\\ \\ \sf\implies s=147cm$

$\sf Area=\sqrt{147(147-91)(147-98)(147-105)}\\ \\ \sf\longrightarrow Area=\sqrt{147×56×49×42}\\ \\ \sf\longrightarrow Area=\sqrt{7×7×3×7×2×2×2×7×7×7×3×2}\\ \\ \sf\longrightarrow Area\:of\: triangle= 7×7×7×3×2×2\\ \\ \sf\longrightarrow Area\:of\: triangle=4116cm{}^{2}$

The area of triangle=$\sf 4116cm{}^{2}$

The area of triangle=$\tt\frac{1}{2}×Base\times height$

$\bold{Given}\begin{cases}\sf{Area=4116}\\ \sf{longest\:side=105cm}\end{cases}$

we have to find the height of triangle

$\sf Area\:of\: triangle=\frac{1}{2}\times base\times Height\\ \\ \sf\leadsto 4116=\frac{1}{2}\times 105\times height\\ \\ \sf\leadsto 4116×2=105\times height \\ \\ \sf\leadsto 8232=105\times height\\ \\ \sf\leadsto \frac{\cancel{8232}}{\cancel{105}}=Height\\ \\ \sf\implies Height=78.4cm$

●The height of triangle=78.4cm

$\boxed{\bold{Area=4116cm{}^{2}\: height=78.4cm}}$

#BAL

#Answerwithquality