Question

Find the area of a triangular field whose side are 91m ,98m and 105m in length . Find the heigh corresponding to the longest side

Answer 1

hope this solution helps you

Answer 2

Given :-

Side (1) = 91 m

Side (2) = 98 m

Side (3) = 105 m

To Find :-

The height corresponding to the longest side.

Solution :-

Given that,

Consider that

a = 91 m

b = 98 m

c = 105 m

According to the question,

$\sf s = \dfrac{a+b+c} {2}$

Substituting their values, we get

$\sf s = \dfrac{91+98+105} {2}$

By division,

$\sf s=147$

We know by the formula,

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

By substituting the values,

$\sf Area=\sqrt{147(147-91)(147-98)(147-105)}$

So we get,

$\sf Area=\sqrt{147 \times 56 \times 49 \times 42}$

$\sf Area = 4116 \ m^{2}$

It is given as,

b = Longest side = 105 m

Consider h as the height corresponding to the longest side

We know that,

$\underline{\boxed{\sf Area \ of \ triangle = \dfrac{1}{2} \times b \times h}}$

By substituting the values

$\sf \sf Area \ of \ triangle = \dfrac{1}{2} b h=4116$

On further calculation.

$\sf 105 \times h = 4116 \times 2$

$\sf h=\dfrac{4116 \times 2}{105}$

$\sf h = 78.4 \ m$

Area = 4116²

Height = 78.4 m