Math, asked by sanjoylaskar, 11 months ago

The sides of a triangular park are 5m, 7m, 8m respectively. Find the cost of leveling the park at the rate of RS 10.50 per square metre.please give the answer with steps..

Answers

Answered by Anonymous

Solution :-

Finding area of the triangular park

Lengths of the sides of a triangular park :

• a = 5 m

• b = 7 m

• c = 8 m

Semi perimeter (s) = (a + b + c)/2

s = (5 + 7 + 8)/2

s = 20/2

s = 10 m

$\rm Area \: of \: the \: traingular \: park = \sqrt{s(s - a)(s - b)(s - c)} \\$

$\\ \sf = \sqrt{10(10 - 5)(10 - 7)(10 - 8)} \\ \\ \sf = \sqrt{10(5)(3)(2)} \\ \\ \sf = \sqrt{50(6)} \\ \\ \sf = \sqrt{300} \\ \\ \sf = 17.32 \: m^2(approx.)$

Area of the triangular park = 17.32 m²

Finding the cost of leveling the park

Cost of leveling area 1 m² = Rs. 10.50

Cost of leveling area 17.32 m² = (10.50 * 17.32) = Rs. 181.86

Therefore the cost of leveling the triangular park is Rs. 181.86.

Answered by Anonymous

SOLUTION:-

Given:

The sides of a triangular park are 5m, 7m & 8m respectively.

So,

Let A=5m, B= 7m, C= 8m

Using Heron's Formula:

$S = \frac{A + B + C}{2} \\ \\ = > S= \frac{5m + 7m + 8m}{2} \\ \\ = > S = \frac{20m}{2} \\ \\ = > S = 10m$

Now,

Area of triangular park:

$A = \sqrt{S(S - A)(S - B)(S - C)} \\ \\ = > \sqrt{10(10 - 5)(10 - 7)(10 - 8)} \\ \\ = > \sqrt{10(5)(3)(2)} \\ \\ = > \sqrt{2 \times 5 \times 5 \times 3 \times 2} \\ \\ = > 2 \times 5 \sqrt{3} \\ \\ = > 10 \sqrt{3} \\ \\ = > 10 \times 1.732 \\ \\A = > 17.32 {m}^{2}$

So,

The cost of levelling the park at the rate of Rs.10.50 per m².

Therefore,

Total cost of levelling the park:

17.32m² × 10.50Rs.

=) Rs.181.86

Hope it helps ☺️

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