Question

The sides of a triangular field are 33m,44m and 55 m.The cost of leveling the field at the rate of ₹1.20 per m sq is

Answer 1

Answer: Rs. 871.2

Step-by-step explanation:

Given sides of triangle a=33 m; b= 44 m;c= 55 m

Applying Heron's formula

$s=\dfrac{a+b+c}{2} \\\\\text{ Area of }\triangle =\sqrt{s(s-a)(s-b)(s-c)}$

Therefore

$s=\dfrac{33+44+55}{2} = 66$

Now

$\text{ Area of }\triangle =\sqrt{66(66-33)(66-44)(66-55)}= \sqrt{66\times 33\times 22\times 11} \\\\=\sqrt{11\times 2\times 3\times 11\times 3 \times 11\times 2\times 11} \\\\=11\times 11\times 2\times 3 = 726 m^{2}$

Cost of leveling the field $1m^{2}$= Rs . 1.20

Cost of leveling the field $726m^{2}$=  Rs . 1.20x 726=  Rs. 871.2

Hence, total cost of leveling the field is  Rs. 871.2

Answer 2

Given:

The Sides of a triangular field are 33m, 44m, and 55 m.

To find:

The cost of leveling the field at the rate of ₹1.20 per meter square.

Solution:

$On Applying Heron's formula, we get: s=\frac{a+b+c}{2} Area of \triangle=\sqrt{s(s-a)(s-b)(s-c)}$

$Therefore, s=\frac{33+44+55}{2}=66$

$Now,\\Area of \triangle=\sqrt{66(66-33)(66-44)(66-55)}=\sqrt{66 \times 33 \times 22 \times 11} \\ =\sqrt{11 \times 2 \times 3 \times 11 \times 3 \times 11 \times 2 \times 11} \\ =11 \times 11 \times 2 \times 3=726 \mathrm{~m}^{2} \\Cost of leveling the field 1 \mathrm{~m}^{2}= Rs .1 .20 \\Cost of leveling the field 726 \mathrm{~m}^{2}= Rs. 1.20 \times 726\\= Rs. 871.2$

Therefore, the cost of leveling the field at the rate of ₹1.20 per meter square is Rs. 871.2.