Find the Area of a Path by Subtracting Two Rectangles

Subtract the inner area from the outer one. A field 50 m by 30 m with a 2 m path outside it has an outer rectangle of 54 m by 34 m, so the path covers

$$1836-1500=336{\text{ m}}^2$$

The path is simply what is left when the field is taken away. This page covers everything in the ICSE Class 7 Mathematics chapter's second part: areas of rectilinear figures, area of a circle and a ring, combined and shaded figures, and paths and cost problems.

For a circle:

$$A=\pi r^2$$

With $\pi =\frac{22}{7}$ and $r=7$ cm:

$$A=\frac{22}{7}\times 49=154{\text{ cm}}^2$$

For $r=21$ cm: $A=\frac{22}{7}\times 441=1386$ cm².

Working back from an area. If a circle has area 154 cm², then $r^2=\frac{154\times 7}{22}=49$, so $r=7$ cm.

A ring is the region between two concentric circles — circles sharing the same centre. Its area is the difference:

$$A=\pi R^2-\pi r^2=\pi (R^2-r^2)$$

Worked example. Outer radius 14 cm, inner radius 7 cm:

$$A=\frac{22}{7}(196-49)=\frac{22}{7}\times 147=462{\text{ cm}}^2$$

A semicircle is half a circle, so its area is $\frac{1}{2}\pi r^2$ — for $r=7$ cm, that is 77 cm².

A circular flower bed with a paved border round it, or a washer from a hardware shop, is a ring.

Take the common factor out first. Computing $\frac{22}{7}(196-49)$ is far quicker than finding 616 and 154 separately and subtracting — and because $147$ is a multiple of 7, the fraction cancels cleanly.

Break it into simple shapes, then add or subtract their areas.

Adding. A rectangle 20 cm by 14 cm has a semicircle on one shorter side, so the semicircle's diameter is 14 cm and its radius 7 cm.

$$\text{rectangle}=20\times 14=280{\text{ cm}}^2$$
$$\text{semicircle}=\frac{1}{2}\times \frac{22}{7}\times 49=77{\text{ cm}}^2$$
$$\text{total}=280+77=357{\text{ cm}}^2$$

Subtracting. A square of side 14 cm has the largest possible circle cut out of it. That circle's diameter equals the side, so $r=7$ cm.

$$\text{square}=196{\text{ cm}}^2,\text{circle}=154{\text{ cm}}^2$$
$$\text{shaded}=196-154=42{\text{ cm}}^2$$

Both together. An L-shaped floor can be split into two rectangles, say $6\times 4$ and $3\times 2$, giving $24+6=30$ m².

The step that decides these questions is reading the radius off the diagram. The semicircle's radius was 7 cm because its diameter had to match the rectangle's 14 cm side, and the inscribed circle's radius was 7 cm because its diameter had to match the square's side. Neither radius was stated in the question — both had to be deduced, and using 14 as the radius is the error that wrecks the answer.

Find the area in the right unit, then multiply by the rate per unit area.

Path outside a field. A field is 50 m by 30 m with a 2 m wide path all round the outside. The path adds 2 m on each of two opposite sides, so the outer dimensions are $50+4=54$ m and $30+4=34$ m.

$$\text{outer}=54\times 34=1836{\text{ m}}^2,\text{field}=1500{\text{ m}}^2$$
$$\text{path}=336{\text{ m}}^2$$

At ₹15 per m² for levelling:

$$336\times 15=₹5040$$

Path inside a field. A garden 40 m by 25 m has a 2 m path all round inside it. Now the inner lawn is $40-4=36$ m by $25-4=21$ m:

$$1000-756=244{\text{ m}}^2\text{ of path}$$

Carpeting a room. A room is 6 m by 5 m, so its area is 30 m². At ₹120 per m², the cost is $30\times 120=₹3600$.

Note why the width was doubled. A 2 m path on both sides of a length adds $2+2=4$ m, not 2 m — and that single doubling is the most common mistake in the whole chapter. Whether you add or subtract the 4 m depends on the path being outside or inside, so read that word before calculating anything.