We have learnt that a plane figure, such as a rectangle, has a length and a breadth and is a two-dimensional shape. We have also learnt about the perimeter and area of closed figures like triangles, rectangles, squares, parallelograms. The amount of surface enclosed by a closed rectilinear figure is called its area.

Now think of the parks with crossroads and running tracks. Is it possible to find out the area of the crossroads and paths?? Yes, it’s possible to find the area of paths, roads and borders. In this article, we will learn to find the area of the crossroads, paths and borders.

Figures

Area and Perimeter

The area is the amount of space covered by a two-dimensional shape or surface. We measure the area in square units, \(\mathrm{cm}^{2}\) or \({{\rm{m}}^{\rm{2}}}{\rm{.}}\)

In the figure below, if one square block is \(1 \mathrm{~cm}\) long and \(1 \mathrm{~cm}\) wide, we can find the area of the shaded region just by counting the number of squares in the region, i.e., \(25\) squares.

So, the area is \(5 \mathrm{~cm} \times 5 \mathrm{~cm}=25 \mathrm{~cm}^{2}\)

The perimeter of a rectilinear figure is the distance around its boundary. Or we can say that the total length of the boundary of a closed shape or rectilinear figure is called its perimeter. So, the sum of the lengths of all sides of a rectilinear figure is its perimeter.

Area of Borders

We know to find the area of any rectilinear figure. Let us now understand the use of the area in measuring paths and verandahs with the help of an example.

Example: A rectangular field is \(42 \mathrm{~m}\) by \(30 \mathrm{~m}\). A \(2\;{\rm{m}}\) wide path is to be constructed along the sides outside the field. Find the area of the path.

Solution: In the given figure, the inner rectangle \(ABCD\) represents the given field, and the shaded region represents the path around it. So, we have,

Length of the outer rectangle \(EFGH = 42 + 2 + 2 = 46\,{\rm{m}}\)

The breadth of the outer rectangle \(EFGH=30+2+2=34 \mathrm{~m}\)

Area of the inner rectangle \(=42 \times 30=1260 \mathrm{~m}^{2}\)

Area of the outer rectangle \(=46 \times 34=1564 \mathrm{~m}^{2}\)

We can see that the outer rectangle \(EFGH\) includes the area of the inner rectangle \(ABCD\) and the area of the path.

Thus, area of the path = Area of the outer rectangle \(EFGH\) – Area of the rectangle \(ABCD\)

\(=1564-1260 \mathrm{~m}^{2}\)

\(=304 \mathrm{~m}^{2}\)

Hence, the area of the path is \(304 \mathrm{~m}^{2}\)

Area Between Two Rectangles and Two Concentric Circles

In our day-to-day life, we have observed that in parks or gardens, someplace is left around in the form of tracks, paths or in-between cross paths. We also come across some borders or frames, etc., attached to some objects. We need to find the areas of these paths or borders to find the cost of making them.

Let us look at a couple of examples to understand this better.

Example 1: The length and breadth of a rectangular park are \(65 \mathrm{~m}\) and \({\rm{45m}}\) respectively. Each \(5 \mathrm{~m}\) wide, two crossroads run at right angles through the centre and parallel to its sides. Find the cost of cementing the roads at the rate of \(₹25\) per \({{\rm{m}}^{\rm{2}}}\)

Solution: The area of the cemented roads is shown by the shaded portion. These roads are rectangles \(ABCD\) and \(EFGH\) of dimensions \({\rm{(65\,m \times 5\,m)}}\) and \({\rm{(45\,m \times 5\,m)}}\) respectively. 

We note that, while doing this, the area of square \(PQRS\) is taken twice, so we have to exclude it once.

Thus, the area of the roads \(=\) Area of rectangle \(ABCD+\) Area of rectangle \(EFGH-\) Area of square \(PQRS.\)

\({\rm{ = 65\,m \times 5\,m + 45\,m \times 5\,m – 5\,m \times 5\,m}}\)

\({\rm{ = 325\;}}{{\rm{m}}^{\rm{2}}}{\rm{ + 225\;}}{{\rm{m}}^{\rm{2}}}{\rm{ – 25\;}}{{\rm{m}}^{\rm{2}}}\)

\(=525 \mathrm{~m}^{2}\)

Therefore the cost of cementing the roads \(=₹ 25 \times 525= ₹13125\)

Example 2: The radius of the bigger circle is \(15 \mathrm{~cm}\). The radius of the smaller circle is \(8 \mathrm{~cm} .\) What is the area of the shaded region? (Take \(\pi=3.14)\)

Solution: Area of the bigger circle \(=\pi R^{2}\)

\(=3.14 \times 15 \times 15\)

\(=706.5 \mathrm{~cm}^{2}\)

Area of the smaller circle \(=\pi r^{2}\)

\(=3.14 \times 8 \times 8\)

\(=200.96 \mathrm{~cm}^{2}\)

Area of the shaded region \(=706.5-200.96\)

\(=505.54 \mathrm{~cm}^{2}\)

Solved Examples – Area of Paths, Roads and Borders

Q.1. A sheet of paper is \(60\,{\rm{cm}}\) by \(40\,{\rm{cm}}.\) A \(5\,{\rm{cm}}\) wide strip is cut from it all around. Find the area of the strip and the area of the remaining sheet.

Ans: In the above-given figure, the outer rectangle \(ABCD\) represents the sheet of the paper, and the shaded region represents the strip that is cut from it. The inner rectangle \(EFGH\) represents the sheet left after cutting the strip.
Area of the complete sheet \(=\) Area of the outer rectangle \(=60 \times 40=2400 \mathrm{~cm}^{2} .\)
Length of the inner rectangle \(EFGH=60-5-5=50 \mathrm{~cm}\)
The breadth of the inner rectangle \(EFGH=40-5-5=30 \mathrm{~cm}\)
Area of the sheet left after cutting the strip=Area of the inner rectangle \(=50 \times 30=1500 \mathrm{~cm}^{2}\).
Area of the strip \(=\) Area of the complete sheet \(-\) Area of the sheet left after cutting the strip
\(=2400-1500\)
\(=900 \mathrm{~cm}^{2}\)
Hence, the area of the strip is \(900 \mathrm{~cm}^{2}\) and the area of the remaining sheet is \(1500 \mathrm{~cm}^{2} .\)

Q.2. There are two concentric circular tracks of radii \(100\,{\rm{m}}\) and \(100\,{\rm{m}},\) respectively. Manu runs on the inner track and goes once around the track in \(1\) minute \(30\) seconds, while Dhruv runs on the outer track in \(1\) minute \(32\) seconds. Who runs faster?
Ans: Circumference of the inner track \(=(2 \pi \times 100) \mathrm{m}=200 \pi \,\mathrm{m}\)
Circumference of the outer track \(=(2 \pi \times 102) \mathrm{m}=204 \pi \,\mathrm{m}\)
Manu covers a distance equal to the circumference of the inner track in \(1\) minute \(30\) seconds, i.e., in \(\frac{3}{2}\) minutes.
So the distance travelled by Manu in \(\frac{3}{2}\) minutes \(=200 \pi \,m\)
Therefore, the distance travelled by Manu in \(1\) minute \({\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{ \times 200\pi \,m = 133}}{\rm{.33\pi \,m}}\)
Thus, the speed of Manu \({\rm{ = 133}}{\rm{.33\pi \,m}}\) per minute.
Dhruv covers a distance equal to the circumference of the outer track in \(1\) minute \(32\) seconds, i.e., in \(\left(1+\frac{32}{60}\right)\) minutes, i.e., in \(\frac{23}{15}\) minutes.
Therefore, the distance travelled by Dhruv in \(1\) minute \({\rm{ = }}\frac{{{\rm{15}}}}{{{\rm{23}}}}{\rm{ \times 204\pi \,m = 133}}{\rm{.04\pi\, m}}\)
Thus, the speed of Dhruv \(=133.04 \pi \mathrm{m}\) per minute.
Since the speed of Manu is greater than the speed of Dhruv, therefore, Manu runs faster.

Q.3. A community park is in the form of a square of side 80 cm. A 4 m wide path runs inside it along its sides. Find the cost of repairing the path at the rate of \(₹10\) per \({{\rm{m}}^2}.\)

Ans: Side of the bigger square \(=80 \mathrm{~m}\)
Area of the bigger square \({\rm{ = side \times side}}\)
\({\rm{ = 80 \times 80}}{{\rm{m}}^{\rm{2}}}\)
\(=6400 \mathrm{~m}^{2}\)
Side of smaller square \({\rm{ = (80 – 4 – 4)m}}\)
\(=72 \mathrm{~m}\)
Area of smaller square \(=72 \times 72 \mathrm{~m}^{2}\)
\(=5184 \mathrm{~m}^{2}\)
Area of the path=Area of bigger square-Area of smaller square
\(=6400 \mathrm{~m}^{2}-5184 \mathrm{~m}^{2}=1216 \mathrm{~m}^{2}\)
Cost of repairing \(=₹10×1216=₹12160\)

Q.4. How many square tiles of side \(20\,{\rm{cm}}\) will be needed to pave a footpath that is \(2\,{\rm{m}}\) wide and surrounds a rectangular plot \(40\,{\rm{m}}\) long and \(22\,{\rm{m}}\) wide?

Ans: Since the footpath is \({\rm{2\,m}}\) wide, the length of the outer plot \(=44 \mathrm{~m}\) and its breadth \(=26 \mathrm{~m}\)
Therefore, the area of the outer plot \(=44 \times 26 \mathrm{~m}^{2}=1144 \mathrm{~m}^{2}\)
The area of the inner plot \(=40 \times 22 \mathrm{~m}^{2}=880 \mathrm{~m}^{2}\)
Thus, the area to be covered with tiles \({\rm{ = 1144}}{{\rm{m}}^{\rm{2}}}{\rm{ – 880}}{{\rm{m}}^{\rm{2}}}{\rm{ = 264}}{{\,\rm{m}}^{\rm{2}}}\)
\(=264 \times 100 \times 100 \mathrm{~cm}^{2}\)
Area of one tile \(=20 \times 20 \mathrm{~cm}^{2}=400 \mathrm{~cm}^{2}\)
Hence, the number of tiles required to pave the footpath \(=\frac{264 \times 100 \times 100}{400}=6600\)

Q.5. Find the area of the shaded circular path in the below-given figure. (Take \(\left.\pi=\frac{22}{7}\right)\)

Ans: Two circles are given having the same centre. The width of the shaded region is \(7 \mathrm{~m}\)
The radius of the inner circle \((r)=14 \mathrm{~m}\)
Therefore, area of inner circle \(=\pi r^{2}\)
\(=\frac{22}{7} \times 14 \times 14\)
\(=616 \mathrm{~m}^{2}\)
Now, the radius of the outer circle \((R)=r+7=14+7=21 \mathrm{~m}\)
Therefore, area of outer circle \(=\pi R^{2}\)
\(=\frac{22}{7} \times 21 \times 21\)
\(=1386 \mathrm{~m}^{2}\)
Area of the shaded region \(=\pi R^{2}-\pi r^{2}\)
\(=1386 \mathrm{~m}^{2}-616 \mathrm{~m}^{2}\)
\(=770 \mathrm{~m}^{2}\)

Summary

In this article, we first learnt the basic concept of area and also learnt to find the area, then we learnt to find the area of the borders, crossroads, paths etc. In addition to this, we also learned to find the area and circumference of the circular paths. And, lastly, we learnt to solve some examples based on finding the area of the crossroads, paths, circular tracks to strengthen the grip over the concept.

Learn About Area and Perimeter of Shapes

Frequently Asked Questions (FAQ) – Area of Paths, Roads and Borders

Q.1. How do you find the area?
Ans: The area of a closed figure is the amount of surface enclosed by it.

Q.2. What is the formula for the area of path?
Ans: Formula of the area of the path when it is rectangular in shape, subtract the area of the inner rectangle from the area of the outer rectangle.

Q.3. What are concentric circles?
Ans: Circles having the same centre are called concentric circles.

Q.4. What do you mean by the borders, crossroads and paths?
Ans: We have observed that in parks or gardens, someplace is left around in the form of tracks, paths or in-between cross paths. We also come across some borders or frames, etc., attached to some objects. Thus, this left out the place is referred to as borders, crossroads and paths.

Q.5. What is the perimeter of the path?
Ans: The perimeter of the path is the sum of the lengths of all its sides. The perimeter of a circle is called its circumference.

We hope this detailed article on the area of paths, roads, and borders helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!