VerifiedComplete step by step answer:
Now let’s name the rectangular path as ABCD shown in the fig 1,
With AB = 125m, BC = 65m, CD = 125m, DA = 65m.
Now we are told that a 3m wide path runs around and outside the rectangular path.
So, we can find out the dimension of new rectangle EFGH, they are
EF = (125+3+3)m = 131m
FG = (65+3+3)m = 71m
GH = (125+3+3)m = 131m
HE = (65+3+3)m = 71m
The path thickness is 3m, so we have added twice because it is surrounded in all the four sides of the rectangle (as can be seen in the above diagram).
Now we were asked to find the area of path which can be done by subtracting the area of rectangle ABCD from the area of rectangle EFGH.
At first we will find the area of rectangle EFGH using formula of area of rectangle which is $\text{length }\!\!\times\!\!\text{ breadth}\text{.}$
So, area of \[EFGH=\left( 131\times 71 \right){{m}^{2}}=9301{{m}^{2}}\]
Similarly, the area of rectangle ABCD is,
\[ar(ABCD)=\left( 125\times 65 \right){{m}^{2}}=8125{{m}^{2}}\]
The area of path will be:
area of (EFGH) – area of (ABCD)
$\Rightarrow 9310m^2-8125m^2$
$\Rightarrow 1176m^2 $
So, the area of path is $1176{{m}^{2}}.$
Note: Students should be careful while using the formula of which is $\text{length }\!\!\times\!\!\text{ breadth}\text{.}$They should also be careful about the calculations. Students generally make mistakes when finding the length and breadth of the new rectangle, they usually add the path just once, which is wrong. So drawing the diagram makes everything clear.
