Question

find the cost of turfing a triangular field at the rate of rs5/m^2 having lengths of its sides as 40m , 70m , 90 m.​

Answer 1

Given : Find the cost of turfing a triangular field at the rate of ₹5m⁻² having lengths of its sides as 40m , 70m , 90 m.

Need To Find : The cost of turfing a triangular field.

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★ Cᴏɴᴄᴇᴘᴛ :

According to the question, here we are given that the length of the 1ˢᵗ side is 40m, 2ⁿᵈ side is 70m, 3ʳᵈ is 90m. The cost of turfing each m² is ₹5. So, first we need to get the side of the triangle by adding all the sides and then dividing it by 2 and then get the area of the triangle using Heron's Formula and then multiply the area with 5 to get the cost.

★ Sᴏʟᴜᴛɪᴏɴ :

Finding Side

➻ S = (40m + 70m + 90m)/2

➻ S = 200m/2

➻ S = 100m

Now,

❍ Heron's Formula : √s(s - a)(s - b)(s - c) ❍

Here, let's assume

• a as the length i.e. 40m

• b as the breadth i.e. 70m

• c as the height i.e. 90m

Now, putting the required values we get

➬ √s(s - a)(s - b)(s - c)

➬ √100(100 - 40)(100 - 70)(100 - 90)

➬ √100 × 60 × 30 × 10

➬ √1800000

➬ 1341.64m²

Hence, the area is 1341.64m²

Finding Cost

Cost = Area × Rate

Cost = 1341.64 × ₹5

Cost = ₹6708.2

The cost of turfing a triangular field is ₹6708.2

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Answer 2

Answer:

Gɪᴠᴇɴ :

• ➛ The sides of a triangular field are 40 m, 70 m and 90 m.
• ➛ The rate of levelling the field is Rs.5 per meter square.

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

Tᴏ Fɪɴᴅ :

• ➛ Semi perimeter of triangle.
• ➛ Area of triangle.
• ➛ Cost of turfing a triangular field

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

Usɪɴɢ Fᴏʀᴍᴜʟᴀs :

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{ Semi \: perimeter = \dfrac{1}{2} \big(a + b + c \big)}}}}}}}$

$\longrightarrow{\small{\underline{\boxed{\pmb{\sf{ Area=\sqrt{s(s - a)(s - b)(s - c)}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Cost = Area \: of\: field \times rate \: of \: turfing \: per \: {m}^{2}}}}}}}}$

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

Sᴏʟᴜᴛɪᴏɴ :

☼ Firstly finding the semi perimeter of triangle by substituting the given values in the formula :-

$\begin{gathered}\begin{gathered}\qquad{\dashrightarrow{\small{\sf{ S = \dfrac{1}{2} \bigg(a + b + c \bigg)}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ S = \dfrac{1}{2} \bigg(40 + 70 + 90 \bigg)}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ S = \dfrac{1}{2} \bigg( \: 200 \: \bigg)}}}} \\ \\ \quad{\dashrightarrow{ \small{\sf{ S = \dfrac{1}{2} \times 200}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ S = \dfrac{200}{2}}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ S = \cancel{\dfrac{200}{2}}}}}} \\ \\ \qquad{\dashrightarrow{\small{\sf{ S = 100 \: m}}}} \\ \\ \qquad\small\bigstar{\underline{\boxed{\sf{\pink{Semi \: Perimeter = 100 \: m}}}}}\end{gathered}\end{gathered}$

∴ The semi perimeter of triangle is 100 m.

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☼ Now, finding area of triangle by substituting the values in heron's formula :-

$\begin{gathered}\begin{gathered} \qquad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}=\sqrt{s(s - a)(s - b)(s - c)}}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}=\sqrt{100(100- 40)(100 - 70)(100- 90)}}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}=\sqrt{100(60)(30)(10)}}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}=\sqrt{100 \times 60 \times 30 \times 10}}}}}\\ \\ \quad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}=\sqrt{1800000}}}}} \\ \\ \quad{\dashrightarrow{\small{\sf{ Area_{(\triangle)}\approx{1341.64}}}}} \\ \\ \qquad\small{\bigstar{\underline{\boxed{\sf{\pink{ Area \: of \: triangle \approx{1341.64 \: m}}}}}}}\end{gathered}\end{gathered}$

∴ The area of triangle is 1341.64 m.

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☼ Now, finding the cost of levelling the field at the rate of Rs 1.20 per meter square :-

$\begin{gathered}\begin{gathered} \quad{\dashrightarrow{\small{\sf{Cost = Area \: of\: field \times rate \: of \: levelling \: per \: {m}^{2}}}}} \\ \\ {\dashrightarrow{\small{\sf{Cost =1341.64\times 5}}}} \\ \\ {\dashrightarrow{\small{\sf{Cost = \dfrac{134164}{100} \times 5}}}} \\ \\ {\dashrightarrow{\small{\sf{Cost =\dfrac{134164 \times 5}{100}}}}} \\ \\ {\dashrightarrow{\small{\sf{Cost = \dfrac{670820}{100}}}}} \\ \\ {\dashrightarrow{\small{\sf{Cost = \cancel{\dfrac{670820}{100}}}}}}\\ \\{\dashrightarrow{\small{\sf{Cost = Rs.6708.20}}}} \\ \\ \quad\small\bigstar{\underline{\boxed{\sf{\pink{Cost = Rs.6708.20}}}}} \end{gathered}\end{gathered}$

∴ The cost of turfing a triangular field is Rs.6708.20.

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

Exᴛʀᴀ Iɴғᴏʀᴍᴀᴛɪᴏɴ :

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

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

Rᴇʟᴀᴛᴇᴅ Qᴜᴇsᴛɪᴏɴ :

The sides of a triangular field are 33 m, 44 m and 55 m. the cost of levelling the field at the rate of Rs 1.20 per meter square is :

https://brainly.in/question/47457753

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