Question

ABC is right angled at C. Can you locate its orthocentre without drawing any altitude?
If so, name it
​

Answer 1

Step-by-step explanation:

Given :

Ratio of lenhth and breadth of a rectangular field = 9 : 5

Area of the rectangular field = 14580 sq.m

To Find :

The cost of surrounding the field with a fence at the rate of 3.25 per m

Solution :

Area of a rectangle is given by ,

\begin{gathered} \\ \star \: {\boxed{\purple{\sf{Area_{(rectangle)} = length \times breadth}}}} \\ \\ \end{gathered}

⋆

Area

(rectangle)

=length×breadth

We are given that ratio of length and breadth of a rectangular field as 9 : 5. Let the ratio constant be "x". Then ,

Length = 9x

Breadth = 5x

We are given the area of rectangular field as 14580 sq.m.

Now ,

\begin{gathered} \\ : \implies \sf \: 14580 \: {m}^{2} = 9x \times 5x \\ \\ \end{gathered}

:⟹14580m

2

=9x×5x

\begin{gathered} \\ : \implies \sf \: 14580 \: {m}^{2} = 45 {x}^{2} \\ \\ \end{gathered}

:⟹14580m

2

=45x

2

\begin{gathered} \\ : \implies \sf \: {x}^{2} = \frac{14580 \: {m}^{2} }{45} \\ \\ \end{gathered}

:⟹x

2

=

45

14580m

2

\begin{gathered} \\ : \implies \sf \: {x}^{2} = 324 \: {m}^{2} \\ \\ \end{gathered}

:⟹x

2

=324m

2

\begin{gathered} \\ : \implies \sf \: x = \sqrt{324 \: {m}^{2} } \\ \\ \end{gathered}

:⟹x=

324m

2

\begin{gathered} \\ : \implies{\underline{\boxed {\red{\mathfrak{x = 18 \: m}}}}} \\ \\ \end{gathered}

:⟹

x=18m

Now ,

Length = 9x = 9(18) = 162 m

Breadth = 5x = 5(18) = 90 m

Perimeter of a rectangle is given by ,

\begin{gathered} \\ \star \: {\boxed{\purple{\sf{perimeter_{(rectangle)} = 2(length + breadth)}}}} \\ \\ \end{gathered}

⋆

perimeter

(rectangle)

=2(length+breadth)

Substituting the values we have ,

\begin{gathered} \\ : \implies \sf \: Perimeter_{(rectangular\:field)} = 2(162 + 90) \\ \\ \end{gathered}

:⟹Perimeter

(rectangularfield)

=2(162+90)

\begin{gathered} \\ : \implies \sf \:Perimeter_{(rectangular\:field)} = 2(252) \\ \\ \end{gathered}

:⟹Perimeter

(rectangularfield)

=2(252)

\begin{gathered} \\ : \implies{\underline{\boxed{\blue {\mathfrak{ Perimeter_{(rectangular \: field)} = 504 \: m}}}}} \: \bigstar\\ \\ \end{gathered}

:⟹

Perimeter

(rectangularfield)

=504m

★

We are given that cost of surrounding the field with fence per m as Rs. 3.25 .

Then , Cost of fencing the rectangular field per 504 m is

\begin{gathered} \\ \longrightarrow \sf \: 504 \times 3.25 = Rs.1638 \\ \\ \end{gathered}

⟶504×3.25=Rs.1638

Hence ,

The cost of surrounding the field with fence at the rate of Rs. 3.25 per m is Rs.1638