Question

The side of a triangle field are 2.33 m 44 and 55 the cost of living the field rate of Rs 1.20 per metre square is is

Answer 1

Answer:

Area of triangle=

s(s−a)(s−b)(s−c)

Here, s=

2

33+44+55

=66cm

Therefore area =

66(66−33)(66−44)(66−55)

=

66(33)(22)(11)

=726cm

2

Now area =726cm

2

and,base=44cm

Therefore height =

base

area(2)

=33cm

Answer 2

Given :

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

The rate of levelling the field is Rs 1.20 per meter square.

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

To Find :

The cost of levelling the field.

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

Using Formulas :

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

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

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

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

Solution :

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

$\begin{gathered}\qquad\small{\longrightarrow{\sf{ S = \dfrac{1}{2} \bigg(a + b + c \bigg)}}} \\ \\ \quad\small{\longrightarrow{\sf{ S = \dfrac{1}{2} \bigg(33 + 44 + 55 \bigg)}}} \\ \\ \quad\small{\longrightarrow{\sf{ S = \dfrac{1}{2} \bigg(132\bigg)}}} \\ \\ \quad\small{\longrightarrow{\sf{ S = \dfrac{1}{2} \times 132}}} \\ \\ \quad\small{\longrightarrow{\sf{ S = \dfrac{132}{2}}}} \\ \\ \quad\small{\longrightarrow{\sf{ S = \cancel{\dfrac{132}{2}}}}} \\ \\ \qquad\small{\longrightarrow{\sf{ S = 66 \: m}}} \\ \\ \qquad\small\bigstar{\underline{\boxed{\sf{\pink{Semi \: Perimeter = 66 \: m}}}}}\end{gathered}$

∴ The semi perimeter of triangle is 66 m.

$\rule{300}{1.5}$

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

$\begin{gathered} \qquad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{s(s - a)(s - b)(s - c)}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{66(66- 33)(66 - 44)(66 - 55)}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{66(33)(22)(11)}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{66 \times 33 \times 22 \times 11}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{2 \times 3 \times 11 \times 3 \times 11 \times 2 \times 11 \times 11}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{2 \times 2 \times 3 \times 3 \times 11 \times 11\times 11 \times 11}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}=\sqrt{ \underbrace{2 \times 2} \times \underbrace{3 \times 3} \times \underbrace{11 \times 11}\times \underbrace{11 \times 11}}}}} \\ \\ \quad\small{\longrightarrow{\sf{ Area_{(\triangle)}={2 \times 3 \times 11 \times 11}}}} \\ \\ \small{\longrightarrow{\sf{ Area_{(\triangle)}={726 \: m}}}} \\ \\ \small {\bigstar{\underline{\boxed{\sf{\pink{ Area \: of \: triangle={726 \: m}}}}}}}\end{gathered}$

∴ The area of triangle is 726 m.

$\rule{300}{1.5}$

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

$\begin{gathered} \quad\small{\longrightarrow{\sf{Cost = Area \: of\: field \times rate \: of \: levelling \: per \: {m}^{2}}}} \\ \\ \small{\longrightarrow{\sf{Cost = 726 \times 1.20}}} \\ \\ \small{\longrightarrow{\sf{Cost = 726 \times \dfrac{120}{100}}}} \\ \\ \small{\longrightarrow{\sf{Cost = 726 \times \dfrac{120}{100}}}} \\ \\ \small{\longrightarrow{\sf{Cost = \dfrac{87120}{100}}}} \\ \\ \small{\longrightarrow{\sf{Cost = \cancel{\dfrac{87120}{100}}}}} \\ \\ \small{\longrightarrow{\sf{Cost = Rs.871.20}}} \\ \\ \quad\small\bigstar{\underline{\boxed{\sf{\pink{Cost = Rs.871.20}}}}} \end{gathered}$

∴ The cost of levelling the field is Rs.871.20.