Question

a rectangular field of 80 m long and 50 broadcast has two roads 10 m constructed are parallel to the side of rectangle each cutting together at right angles through the mid part of the field find​

Answer 1

Answer:

GIVEN :-

Length of Rectangle = 80 m.

Breadth of Rectangle = 50 m.

Width of road = 10 m.

TO FIND :-

the area covered by road .

the area of the field with out road .

the cost to construct the road at rate of Rs. 50 /m².

SOLUTION :-

As we know that the area of rectangle is given by,

\begin{gathered}\\ : \implies \displaystyle \sf \: Area_{(field)} = length \times breadth \\ \\ \\\end{gathered}

:⟹Area

(field)

=length×breadth

\begin{gathered}: \implies \displaystyle \sf \: Area_{(field)} = 80 \: m \times 50 \: m \\ \\ \\\end{gathered}

:⟹Area

(field)

=80m×50m

\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \: \bold{ Area_{(field)} = 4000 \: m ^{2} }}} \\ \\\end{gathered}

:⟹

Area

(field)

=4000m

2

Now we are finding the area of the road which is parallel to the length of the rectangular field,

\begin{gathered}\\ : \implies \displaystyle \sf \: Area_{(road)} = length \: \times width \\ \\ \\\end{gathered}

:⟹Area

(road)

=length×width

\begin{gathered}: \implies \displaystyle \sf \: Area_{(road)} = 80 \: m \times 10 \: m \\ \\ \\\end{gathered}

:⟹Area

(road)

=80m×10m

\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \: \bold{ Area_{(road)} = 800 \: m ^{2} }}} \\ \\\end{gathered}

:⟹

Area

(road)

=800m

2

Now we are finding the area of the road which is parallel to the breadth of the rectangular field,

\begin{gathered}\\ : \implies \displaystyle \sf \: Area_{(road)} = breadth \times width \\ \\ \\\end{gathered}

:⟹Area

(road)

=breadth×width

\begin{gathered}: \implies \displaystyle \sf \: Area_{(road)} = 50 \: m \times 10 \: m \\ \\ \\\end{gathered}

:⟹Area

(road)

=50m×10m

\begin{gathered}: \implies \underline { \boxed{\displaystyle \sf \: \bold{ Area_{(road)} = 500 \: m ^{2} }}} \\ \\\end{gathered}

:⟹

Area

(road)

=500m

2

Now , If two roads are intersecting each other at a point . So that point is repeated two times i.e one road is coming which is parallel to the length and the other one which is parallel to the breadth . So , we will find the area of common portion now,

\begin{gathered}\\ : \implies \displaystyle \sf \: Area_{(common)} = width \times width \\ \\ \\\end{gathered}

:⟹Area

(common)

=width×width

\begin{gathered}: \implies \displaystyle \sf \: Area_{(common)} = 10 \: m \times 10 \: m \\ \\ \\\end{gathered}

:⟹Area

(common)

=10m×10m

\begin{gathered}: \implies \underline{ \boxed{ \displaystyle \sf \: \bold{ Area_{(common)} = 100 \: m ^{2} }}} \\ \\\end{gathered}

:⟹

Area

(common)

=100m

2

Now we are finding the area covered by the roads which is equal to the area covered by Both the roads,

\begin{gathered}\\ : \implies \displaystyle \sf \: Area_{(both \: road)} = 800 + 500 - (area_{(common)}) \\ \\ \\\end{gathered}

:⟹Area

(bothroad)

=800+500−(area

(common)

)

\begin{gathered}: \implies \displaystyle \sf \: Area_{(both \: road)} =1300 - 100 \: m ^{2} \\ \\ \\\end{gathered}

:⟹Area

(bothroad)

=1300−100m

2

\begin{gathered}: \implies \underline { \boxed{ \displaystyle \sf \: \bold{ Area_{(both \: road)} =1200 \: m ^{2} }}} \\ \\\end{gathered}

:⟹

Area

(bothroad)

=1200m

2

Now we are finding the cost to construct the road at rate of Rs. 50 /m².

\begin{gathered}\\ : \implies \displaystyle \sf \: cost_{(construction)} = area \times rate \\ \\ \\\end{gathered}

:⟹cost

(construction)

=area×rate

\begin{gathered}: \implies \displaystyle \sf \: cost_{(construction)} = 1200 \times 50 \\ \\ \\\end{gathered}

:⟹cost

(construction)

=1200×50

\begin{gathered}: \implies \underline{ \boxed{ \displaystyle \sf \: \bold{ cost_{(construction)} = \: rs. \: 60000 \: }}} \\\end{gathered}

:⟹

cost

(construction)

=rs.60000