No bio :(Find the Area of the Triangular field of sides 55 m, 60 m, and 65 m. Find the cost of laying the grass in the triangular field at the rate of Rs 8 per m2.. Created by ammuqueen68. Math

Answer:➺ Sides of triangular field is 55 m, 60 m, and 65 m. ➺ The cost of laying the grass in the triangular field is Rs 8 per m².➺ The cost of laying the grass in the triangular field. Here ★ a = 55 m★ b = 60 m★ c = 65 m Firstly, Finding the semi perimeter of Triangle Substuting the values ∴ The Semi perimeter of Triangle is 90 m. Now, Finding the area of Triangle by heron's formulaSubstuting the values ∴ The Area of Triangle is 1537.04 m². Now, Calculating the cost of laying the grass in the triangular field.Substuting the values ∴ The cost of laying the grass in the triangular field at the rate of Rs 8 per m² is Rs.12296.32. Formulas of area↠ If there is any difficulty viewing this answer in app, kindly see this answer at website Brainly.in for clear steps and understanding.▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

No bio :(Find the Area of the Triangular field of sides 55 m, 60 m, and 65 m. Fi...

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Answer:

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

➺ Sides of triangular field is 55 m, 60 m, and 65 m.
➺ The cost of laying the grass in the triangular field is Rs 8 per m².

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

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

➺ The cost of laying the grass in the triangular field.

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

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

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

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

$\bigstar{\underline{\boxed{\bf{\red{Cost = Area \: of\: Triangular \: field \times rate \: of \: laying \: per \: {m}^{2} }}}}}$

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

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

$\red\bigstar$ Here

★ a = 55 m
★ b = 60 m
★ c = 65 m

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

$\red\bigstar$ Firstly, Finding the semi perimeter of Triangle

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

Substuting the values

${\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2}\bigg(55 + 60 + 65\bigg)}}}$

${\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2}\bigg( \: 180 \: \bigg)}}}$

${\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{1}{2} \times 180}}}$

${\dashrightarrow{\sf{ Semi \ perimeter =\dfrac{180}{2}}}}$

${\dashrightarrow{\sf{ Semi \ perimeter = \cancel\dfrac{180}{2}}}}$

${\dashrightarrow{\sf{ Semi \ perimeter =90 \: m}}}$

${\bigstar{\underline{\boxed{\bf{\green{Semi \ perimeter =90 \: m}}}}}}$

∴ The Semi perimeter of Triangle is 90 m.

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

$\red\bigstar$ Now, Finding the area of Triangle by heron's formula

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

Substuting the values

${\dashrightarrow{\sf{ Area \ of \ triangle=\sqrt{90(90 - 55)(90 - 60)(90 - 65)}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=\sqrt{90(35)(30)(25)}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=\sqrt{90 \times 35 \times 30 \times 25}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=\sqrt{3 \times 3 \times 10\times 7 \times 5 \times 3 \times 10 \times 5 \times 5}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=\sqrt{ \underbrace{ \: 3 \times 3 \: }\times \underbrace{ \: 10\times 10 \: }\times \underbrace{ \: 5 \times 5 \: } \times 3 \times 5 \times 7}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=3 \times 10 \times 5\sqrt{3 \times 5 \times 7}}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=150\sqrt{105} \: {m}^{2} }}}$

${\dashrightarrow{\sf{ Area \ of \ triangle=150 \times 10.2469}}}$

${\dashrightarrow{\sf{ Area \ of \ triangle= 1537.04 \: {m}^{2} }}}$

$\bigstar{\underline{\boxed{\bf{\green{ Area \ of \ triangle= 1537.04 \: {m}^{2} }}}}}$

∴ The Area of Triangle is 1537.04 m².

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

$\red\bigstar$ Now, Calculating the cost of laying the grass in the triangular field.

${\dashrightarrow{\bf{Cost = Area \: of\: Triangular \: field \times rate \: of \: laying \: per \: {m}^{2} }}}$

Substuting the values

${\dashrightarrow{\sf{Cost = 1537.04 \times 8 }}}$

${\dashrightarrow{\sf{Cost=Rs.12296.32}} }$

$\bigstar\underline{\boxed{\bf{\green{Cost=Rs.12296.32}}}}$

∴ The cost of laying the grass in the triangular field at the rate of Rs 8 per m² is Rs.12296.32.

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

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

$\red\bigstar$ Formulas of area

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

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

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

↠ If there is any difficulty viewing this answer in app, kindly see this answer at website Brainly.in for clear steps and understanding.

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MissieSquircles

$\huge \mathcal \colorbox{deep}{{ \color{h} \huge {Given:-}}}$

The Triangular field sides i.e 55 m, 60 m, and 65 m
The Triangular field sides i.e 55 m, 60 m, and 65 mThe triangular field the rate of Rs 8 per m2.

$\huge \mathcal \colorbox{deep}{{ \color{h} \huge {To Find:-}}}$

Area of the Triangular field.
Area of the Triangular field. The cost of laying the grass in the triangular field.

$\small\mathcal \colorbox{deep}{{ \color{h} \huge {Formula required:-}}}$

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \frac{1}{2}( a \: + b + c ) \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area \: of \:triangle = \sqrt{s(s -a )(s - b)(s -c )} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm:➥ cost \: = area \: of \: field \times rate \: per \: {m}^{2} \rm \end{gathered} \end{gathered} }$

$\huge \mathcal \colorbox{deep}{{ \color{h} \huge {solution:-}}}$

$\begin{gathered}\frak{sides \: of \: triangle}\begin{cases}\sf{\;\;\;a \: = \bf{55\;m}}\\\sf{\;\;\; b= \bf{60\;m }}\\\sf{\;\;\; c= \bf{65\;m }}\end{cases}\end{gathered}$

● Now, first we will find semi perimeter.

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \frac{1}{2}( a \: + b + c ) \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \frac{1}{2}( 55 \: + 60 + 65 ) \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \frac{1}{2}( 180 ) \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \frac{1}{2} \times 180 \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ semi \: perimeter = \cancel\frac{180}{2} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm \red{ :➥ semi \: perimeter = 90m }\rm \end{gathered} \end{gathered} }$

● Now, we will find area of triangle field.

${\begin{gathered}\begin{gathered}\rm :➥ area \: of \:triangle = \sqrt{s(s -a )(s - b)(s -c )} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{90(90 -55 )(90 - 60)(90 -65 )} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{90(35 )(30)(25 )} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{90 \times 35 \times 30 \times 25 } \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{3 \times 3 \times 10 \times 3 \times 10 \times 5 \times 5 \times 7 \times 5 } \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{3 \times 10 \times 5 \sqrt{3 \times 5 \times 7} } \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = \sqrt{150 \sqrt{105 {m}^{2} } } \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm :➥ area = 150 \times 10.2469 \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered} \red{\rm :➥ area = 1537.04 {m}^{2} } \rm \end{gathered} \end{gathered} }$

● Now, we will find the cost of laying grass in the triangle field.

Cost of laying grass in 1m² = Rs. 8
Cost of laying grass in 1537.04m² = ?

${\begin{gathered}\begin{gathered}\rm:➥ cost \: = area \: of \: field \times rate \: per \: {m}^{2} \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm:➥ cost \: = 1537.04 \times 8 \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm:➥ cost \: = 12296.32 \rm \end{gathered} \end{gathered} }$

${\begin{gathered}\begin{gathered}\rm ∴ \red{cost \: = rs .12296.32} \rm \end{gathered} \end{gathered} }$

∴ The cost of laying the grass in the triangular field at the rate of Rs 8 per m² is Rs.12296.32.

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} {\boxed{\begin{array}{cc} {\sf{Area \: of \: parallelogram = Base × Altitude }} \\ \\ \sf{ Area \: of \: rectangle = Length × breadth}\\ \\ \sf{ Area \: of \: square = Side × Side } \\ \\ \sf{Area \: of \: cylinder's \: surface = 2πr(h + r)} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

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