Question

Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area.​

Answer 1

Question :-

• Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area.

Given :-

• Sides of a triangle are in the ratio of 12:17:25.
• And perimeter of triangle 540cm.

To find :-

• find its area.

Solution :-

Let the sides be 12x 17x and 25x.

➺ 12x + 17x + 25x = 540

➺ 29x + 25x = 540

➺ 54x = 540

➺ x = 10

Sides of the triangle are 120cm,170cm and 250cm.

Area of the ∆ = √ s (s – a) (s – b) (s – c)

➺ √270 (270 – 120) (270 – 170) (270 – 250)

➺ √270 × 150 × 100 × 20

➺ √81000000

➺ 9000cm².

Hope it helps you.

Answer 2

Answer:

$\begin{gathered}{\underline{\underline{\maltese{\textsf{\textbf{\:Given :}}}}}}\end{gathered}$

• ↠ Sides of a triangle are in the ratio = 12:17:25
• ↠ Perimeter of triangle = 540 cm

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

$\begin{gathered}{\underline{\underline{\maltese{\textsf{\textbf{\:To Find :}}}}}}\end{gathered}$

• ↠ Area of triangle

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

$\begin{gathered}{\underline{\underline{\maltese{\textsf{\textbf{\:Using Formulae :}}}}}}\end{gathered}$

$\bigstar\red{\underline{\boxed{\bf{Perimeter \: of \: triangle =a + b + c}}}}$

$\bigstar\red{\underline{\boxed{\bf{ Semi \ perimeter =\dfrac{1}{2}\bigg(a \: + \: b \: + \: c \bigg)}}}}$

$\bigstar\red{\underline{\boxed{\bf{ Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)}}}}}$

$\green\bigstar$ Where

• ↠ s = semi perimeter
• ↠ a = first side of triangle
• ↠ b = second side of triangle
• ↠ c = third side of triangle

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

$\begin{gathered}{\underline{\underline{\maltese{\textsf{\textbf{\:Solution :}}}}}}\end{gathered}$

$\green\bigstar$ Let the angles of triangle be

• ➱ a = 12x
• ➱ b = 17x
• ➱ c = 25x

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

$\rule{200}2$

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

$\green\bigstar$ Firstly, Finding the angles of triangle.

$\begin{gathered}{\dashrightarrow{\bf{Perimeter \: of \: triangle =a + b + c}}}\end{gathered}$

• Substuting the values

$\begin{gathered}{\dashrightarrow{\sf{Perimeter \: of \: \triangle =a + b + c}}} \\ \\ \qquad{\dashrightarrow{\sf{540 \: cm=12x+ 17x + 25x}}} \\ \\ {\dashrightarrow{\sf{540 \: cm =54x}}} \\ \\ {\dashrightarrow{\sf{x = \dfrac{540}{54}}}} \\ \\ {\dashrightarrow{\sf{x = \cancel{\dfrac{540}{54}}}}} \\ \\ {\dashrightarrow{\sf{x = 10}}} \\ \\ \bigstar{\underline{\boxed{\textsf{\textbf{\pink{x = 10}}}}}} \\ \end{gathered}$

$\green\bigstar$ Thus

• ➱ First side of △ = 12 × 10 = 120 cm
• ➱ Second side of △ = 17 × 10 = 170 cm
• ➱ Third side of △ = 25 × 10 = 250 cm

∴ The sides of triangle are 120 cm, 170cm and 250 cm.

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

$\rule{200}2$

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

$\green\bigstar$ Now, Calculating the semi perimeter

$\begin{gathered}{\dashrightarrow{\bf{ Semi \ perimeter =\dfrac{1}{2}\bigg(a \: + \: b \: + \: c \bigg)}}} \end{gathered}$

• Substuting the values

$\begin{gathered}{\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2}\bigg(120\: + \: 170\: + \: 250 \bigg)}}} \\ \\ {\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2}\bigg(540 \bigg)}}} \\ \\ {\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2} \times 540}}} \\ \\ \qquad{\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{\cancel{2}} \times \cancel{540}}}} \\ \\ {\dashrightarrow{\sf{ Semi \ perimeter =1 \times 270}}} \\ \\ {\dashrightarrow{\sf{ Semi \ perimeter =270 \: cm}}} \\ \\ \bigstar{\underline{\boxed{\textsf{\textbf{\pink{Semi perimeter = 270 cm}}}}}}\end{gathered}$

∴ The semi perimeter of triangle is 270 cm.

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

$\rule{200}2$

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

$\green\bigstar$ Now, Finding the area of triangle by heron's formula

$\begin{gathered}{\dashrightarrow{\bf{ Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)}}}} \end{gathered}$

• Substuting the values

$\begin{gathered}{\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{s(s-a)(s-b)(s-c)}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{270(270-120)(270-170)(270-250)}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{270(150)(100)(20)}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{270 \times 150 \times 100 \times 20}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{81000000}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=\sqrt{9000 \times 9000}}}} \\ \\ {\dashrightarrow{\sf{ Area \ of \ \triangle=9000 \: {cm}^{2}}}} \\ \\ {\dashrightarrow{\underline{\boxed{\textsf{\textbf{\pink{Area of triangle = 9000 square cm}}}}}}}\end{gathered}$

∴ The area of triangle is 9000 cm².

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

$\begin{gathered}{\underline{\underline{\maltese{\textsf{\textbf{\:Learn More :}}}}}}\end{gathered}$

$\green\bigstar$ Formulas of area

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

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