Question

the perimeter of a triangular field is 450m and its sides are inthe ratio 13 :12:5find the area of triangle​

Answer 1

➤ Answer

• Area of triangle = 6750 m².

➤ Given

• The perimeter of a triangular field is 450m and its sides are in the ratio 13 : 12 : 5.

➤ To Find

• The area of the triangle.

➤ Step By Step Explanation

Given that the perimeter of a triangular field is 450m and its sides are in the ratio 13 : 12 : 5.

We need to find the area.

Let's do it !!

First we need to find the unknown sides.

Let the sides be 13x, 12x and 5x.

➠ Equation

13x + 12x + 5x = 450

30x = 450

x = 450/30

x = 15

➠ Length of sides

13x = 13 × 15 = 195

12x = 12 × 15 = 180

5x = 5 × 15 = 75

Now, Area ⤵

➠ By using Herons formula

$\dag\underline{\boxed{\mathfrak{\pink{s = \cfrac{a + b + c}{2}}}}}$

$\dag\underline{\boxed{\mathfrak{\sqrt{{\red{(s) \times (s - a)(s - b)(s - c)}}}}}}$

➠ By substituting the values

$\longmapsto\tt{s=\cfrac{195+180+75}{2}}\\\\\longmapsto\tt{s=\cancel\cfrac{450}{2} } \\ \\\longmapsto\tt{s = 225}$

➠ Now ⤵

$\tt \sqrt{225 \times (225 - 195) (225 - 180) (225 - 75)} \\ \\\longmapsto\tt \sqrt{225\times30\times45\times150}\\\\\longmapsto\tt\sqrt{45562500} \\\\\longmapsto\bf{\green {Area_{(Triangle)} = 6750 {m}^{2}}}$

➢ Therefore, Area of triangle = 6750m²

_____________________

Answer 2

Answer:

${\large{\pmb{\underline{\underline{\frak\color{purple}{Given...}}}}}}$

• $\dashrightarrow$ Perimeter of a triangular field = 450 m
• $\dashrightarrow$ Sides of triangle in the ratio = 13:12:5

$\begin{gathered}\end{gathered}$

${\large{\pmb{\underline{\underline{\frak{\color{purple}{To \: Find...}}}}}}}$

• $\dashrightarrow$ Area of Triangle

$\begin{gathered}\end{gathered}$

${\large{\pmb{\underline{\underline{\frak{\color{purple}{Using \: Formulae ...}}}}}}}$

$\green\bigstar$ Semi Perimeter of Triangle

$\dag{\underline{\boxed{\sf{\red{Semi \: Perimeter = \dfrac{a + b + c}{2}}}}}}$

$\green\bigstar$ Area of Triangle by herom Formula

$\dag{\underline{\boxed{\sf{\red{Area \: of \: Triangle = {\sqrt{s(s - a)(s - b)(s - c)}}}}}}}$

$\begin{gathered}\end{gathered}$

${\large{\pmb{\underline{\underline{\frak{\color{purple}{Solution...}}}}}}}$

$\ddag \: \: {\underline{\frak{\blue{Let \: the \: sides \: of \: triangle ..}}}}$

• 13x
• 12x
• 5x

$\begin{gathered}\end{gathered}$

$\ddag \: \: {\underline{\frak{\blue{According \: to \: the \: question..}}}}$

$\begin{gathered} \dashrightarrow{\sf{13x + 12x + 5x = 450 \: m}} \\\\\dashrightarrow{\sf{30x = 450 \: m}} \\ \\\dashrightarrow{\sf{x = \frac{450}{30}}} \\ \\ \dashrightarrow{\sf{x=\cancel{\frac{450}{30}}}} \\ \\ \dashrightarrow{\sf{x = {15}}} \\ \\ \dag\underline{\boxed{\sf{\pink{x = {15}}}}} \end{gathered}$

$\begin{gathered}\end{gathered}$

$\ddag \: \: {\underline{\frak{\blue{Thus,The\: Length \: of \: Sides \: are..}}}}$

$\dashrightarrow{\sf{13x = 13 \times 15 = \bf\purple{195 \: m}}}$

$\dashrightarrow{\sf{12x = 12\times 15 = \bf\purple{180\: m}}}$

$\dashrightarrow{\sf{5x = 5\times 15 = \bf\purple{75\: m}}}$

$\begin{gathered}\end{gathered}$

$\ddag \: \: {\underline{\frak{\blue{Finding \: the \: semi \: perimeter \: of \: Triangle..}}}}$

$\quad{ : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{a + b + c}{2}}}}}$

• Substituting the values

$\begin{gathered} \begin{gathered}\quad{ : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{195 + 180 + 75}{2}}}}} \\ \\ { : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{450}{2}}}}} \\ \\ \qquad{ : \implies{\sf{Semi \: Perimeter = \bf{\cancel{\dfrac{450}{2}}}}}} \\ \\ \qquad{: \implies{\sf{Semi \: Perimeter = \bf{225}}}} \\ \\ \qquad{\dag{\underline{\boxed{\sf{\pink{Semi \: Perimeter = {225}}}}}}}\end{gathered}\end{gathered}$

$\begin{gathered}\end{gathered}$

$\ddag \: \: {\underline{\frak{\blue{Now \: finding \: the \: area \: of \: Triangle..}}}}$

$\quad{: \implies{\sf{Area \: of \: Triangle = \bf{\sqrt{s(s - a)(s - b)(s - c)}}}}}$

• Substituting the values

$\begin{gathered}\begin{gathered}{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{225(225 - 195)(225- 180)(225 - 75)}}}}} \\ \\{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{225(30)(45)(150)}}}}} \\ \\ \quad\qquad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{225 \times 30 \times 45 \times 150}}}}} \\ \\ \quad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{45562500}}}}} \\ \\ \qquad\qquad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{6750 \times 6750}}}}} \\ \\ \qquad{: \implies{\sf{Area \: of \: \triangle = \bf{6750 \: {m}^{2}}}}} \\ \\ { \dag{\underline{\boxed{\sf{\pink{Area \: of \: Triangle = {6750 \: {m}^{2}}}}}}}}\end{gathered}\end{gathered}$

${\therefore{\underline{\underline{\sf{\red{The \: Area \: of \: Triangle \: is \: 6750 \: {m}^{2} .}}}}}}$

$\begin{gathered}\end{gathered}$

${\large{\pmb{\underline{\underline{\frak\color{purple}{Learn \: More...}}}}}}$

$\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}$