Math, asked by laxmichirag1984, 2 months ago

a triangle park has side 30m, 82m and 56m. a garden has to put a fence all around it and also plant grass inside it. how much area does he need to plant? find the cost of fencing it with the barbed wire at the rate of ₹22 per meter leaving a space of 6 meter wide for a gate on one side.​

Answers

Answered by MoodyCloud
53

Answer:

  • Area he need to plant is 504 m².
  • Cost of fencing it with wire is ₹3564.

Step-by-step explanation:

Given :-

  • Sides of triangular park are 30m, 82m and 56m.
  • Rate of wire is ₹22 per meter.

To find :-

  • Area he need to plant.
  • Cost of fencing it.

Solution :

Formulas to be used :

• Heron's formula :

Area of triangle = √s(s - a)(s - b)(s - c)

[Where, S is semi-perimeter and a, b and c are sides of triangle]

• Semi-perimeter = Perimeter/2

• Perimeter/Fence of triangle = Sum of all sides.

So,

⇒ S = 30 + 82 + 56/2

⇒ S = 168/2

⇒ S = 84

Semi-perimeter is 84 m.

Area of triangular park:

⇒ √84(84 - 30)(84 - 82)(84 - 56)

⇒ √84 × 54 × 2 × 28

⇒ √2 × 2 × 3 × 7 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 7

⇒ 2 × 2 × 2 × 3 × 3 × 7

504

Area he need to plant is 504 .

Now,

⇒ Fence = 30 + 84 + 56

⇒ Fence = 168

Fence of triangle is 168 m.

Gardener need to fence the triangular park with a wire at the rate of ₹22 per meter leaving space of 6m.

⇒ Cost of fencing = (168 - 6) × 22

⇒ Cost of fencing = 162 × 22

⇒ Cost of fencing = 3564

Cost of fencing it with wire is 3564.

Answered by MяMαgıcıαη
277

Understanding the question :

» Here we have a triangular park having sides 30 m , 82 m and 56 m. Park has to put a fence all around it and also plant grass inside it.

» We had to find out the area of park to be planted with grass and the cost of fencing at the rate of 22 per meter leaving a space of 6 m wide gate on one side.

Let's do it !!

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

\underbrace{\underline{\sf{Required\:solution\::}}}

Finding semi - perimeter of park :

\qquad:\mapsto\:\sf s = \dfrac{a + b + c}{2}

\qquad:\mapsto\:\sf s = \dfrac{30 + 82 + 56}{2}

\qquad:\mapsto\:\sf s = \dfrac{168}{2}

\qquad:\mapsto\:\sf s = \dfrac{\cancel{168}}{\cancel{2}}

\qquad:\mapsto\:\bold {s = \red{ 84\:m}}

Using heron's formula to find area of park :

\qquad:\implies\:\sf \sqrt{s\:(s - a)\:(s - b)\:(s - c)}

\qquad:\implies\:\sf \sqrt{84\:(84 - 30)\:(84 - 82)\:(84 - 56)}

\qquad:\implies\:\sf \sqrt{84\:(54)\:(2)\:(28)}

\qquad:\implies\:\sf \sqrt{84\:\times\:54\:\times\:2\:\times\:28}

\qquad:\implies\:\sf \sqrt{254016}

\qquad:\implies\:\bold \red {504\:m^2}

\small\underline{\boxed{\tt{The \:area \:of \:park\: to \:be \:planted \:with \:grass\:=\:\rm\purple{504\:m^2}}}}

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

➱ Semi - perimeter of park = 84 m

So,

➱ Perimeter of park = 84 × 2 = 168 m

Finding cost of fencing around the park leaving space for gate which is 6 m wide :

➱ Cost of fencing = 22 × (Perimeter - 6)

➱ Cost of fencing = 22 × (168 - 6)

➱ Cost of fencing = 22 × 162

➱ Cost of fencing = 3,564

\small\underline{\boxed{\tt{Cost\:of\:fencing\:around\:the\:park\:=\:\rm\purple{Rs.\:3,564 }}}}

Solution completed !!

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