Question

The ratio of the sides of a triangular land is 5:12:13. If the area of the land be 900 sqm, find its perimeter [hint: use area = root over s×(s-a)×(s-b)×(s-c)]​

Answer 1

Given

• Ratio of Side's of triangle = 5:12:13
• Area of triangle = 900m²

To Find

• Perimeter of Triangle

Solutions

_____________________

$\bf \mathcal {Let \: the \: ratio \: of \: Side's \: of \: triangle \: be \: \bf\: 5x, \: 12x \: and \: 13x.}$

We have,

• a= 5x
• b=12x
• c= 13x

We know that,

$\boxed{Semi-Perimeter = \dfrac{a+b+c}{2} }$

${S = \dfrac{5x + 12x + 13x}{2}}$

${S = \dfrac{30x}{2}}$

${S = 15 x}$

Now,

• Let a, b and c be the sides of a triangle.
• Apply Heron's Formula of find the area of triangle.

${ \: \: \: \: \: \implies \bf Area \: of \: ∆ = \sqrt{S(S-a)(S-b)(S-c)}}$

${ ~~~~~\implies \sf900 = \sqrt{15x(15x -5 x)(15 - 12x)(15x - 13x)}}$

${ ~~~~~\implies \sf900 = \sqrt{15x \times 10x \times 3x \times 2x}}$

${ ~~~~~\implies \sf900 = \sqrt{ 900{x}^{4} }}$

${~~~~~\implies \sf {900}^{2} =({ \sqrt{ {30}^{2} \times( {x}^{2})^{2} }} )^{2} }$

${~~~~~\implies \sf \sqrt{810000} =30 {x}^{2} }$

${~~~~~\implies \sf 900 =30 {x}^{2} }$

${~~~~~\implies \sf {x}^{2} = 30 }$

${~~~~~\implies \sf x ≈5.477 }$

Now

• a=5.47722...×5=27.386127875258m
• b=5.47722...×12=65.726706900619m
• c=5.47722...×13=71.203932475671m

$\sf{Perimeter \: of \: ∆ = sum \: of \: all \: side's}$

$\sf{Perimeter \: of \: ∆ = 27.38m + 65.72m +71.20m}$

$\bf{Perimeter \: of \: ∆ = 164.3 meters}$

Figure

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

$\begin{gathered}\\\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Perimeter \: of \: rectangle = 2(l + b)}} \\ \\ \dashrightarrow \sf{Area \: of \: rectangle = length \: \times breadth }\\ \\ \dashrightarrow \sf{Perimeter \: of \: square = 4 \times side } \\ \\ \dashrightarrow \sf{Area \: of \: square =(side) ^{2} } \\ \\ \dashrightarrow \sf{Area \: of \: parallelogram = base \times height} \\ \\ \dashrightarrow \sf{Area \: of \: trapezium = \frac{1}{2}×sum \: of \: parallel \: side \: \times \: height }\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$