NCERT Solutions for Class 7 Maths Chapter 9: Perimeter and Area — Free PDF

Perimeter is the total length of the boundary of a closed figure. It is measured in linear units (cm, m, km).

Area is the amount of surface enclosed by a closed figure. It is measured in square units ($\text{cm}^2$, $\text{m}^2$, $\text{km}^2$).

Parallelogram: A quadrilateral whose opposite sides are parallel and equal. The area formula uses the perpendicular height, not the slant side.

Triangle: The area of any triangle is exactly half the area of a parallelogram with the same base and height.

A circle is the set of all points equidistant from a fixed centre point.
- Circumference $= 2\pi r = \pi d$
- Area $= \pi r^2$
- $\pi \approx \dfrac{22}{7}$ or $3.14$

Annulus (Ring): The region between two concentric circles.

• $1$ m $= 100$ cm, so $1$ m$^2 = 10{,}000$ cm$^2$
  - $1$ km $= 1000$ m, so $1$ km$^2 = 1{,}000{,}000$ m$^2$
  - $1$ hectare $= 10{,}000$ m$^2$ (a square of side $100$ m)

Remember: when converting area units, you must square the conversion factor. This is a very common mistake.

Find the area of a parallelogram with base $8$ cm and height $5$ cm.

The area of a parallelogram is $54$ cm$^2$ and its base is $9$ cm. Find the height.

Find the area of a triangle with base $12$ cm and height $7$ cm.

The area of a triangle is $36$ cm$^2$ and its height is $8$ cm. Find the base.

A parallelogram and a rectangle have the same base $10$ cm. The height of each is $6$ cm. Compare their areas.

Area of rectangle $= 10 \times 6 = 60$ cm$^2$.
Area of parallelogram $= 10 \times 6 = 60$ cm$^2$.

Both areas are equal. A rectangle is a special parallelogram where the height equals the side perpendicular to the base.

ABCD is a parallelogram with base AB $= 14$ cm and height $= 6.5$ cm. DE is the height on side BC $= 10$ cm. Find DE.

Area using AB as base: $14 \times 6.5 = 91$ cm$^2$.

Using BC as base: $91 = 10 \times DE$.

This shows that a parallelogram has the same area regardless of which side you choose as the base, as long as you use the corresponding height.

Find the circumference of a circle with radius $14$ cm. (Use $\pi = \dfrac{22}{7}$)

Find the area of a circle with radius $7$ cm.

The circumference of a circle is $44$ cm. Find its radius and area.

Find the area of a circular garden with diameter $28$ m.

Radius $= 14$ m.

A circular track has inner radius $28$ m and outer radius $35$ m. Find the area of the track.

Q6. The minute hand of a clock is $14$ cm long. Find the area swept by it in one hour.

The minute hand sweeps a full circle in one hour.

Q7. A wire is in the shape of a rectangle $40$ cm $\times$ $22$ cm. If it is re-bent into a circle, find the radius.

Perimeter of rectangle $= 2(40 + 22) = 124$ cm.

This perimeter becomes the circumference:

Q1. Convert $5$ m$^2$ to cm$^2$.

Q2. A rectangular field is $80$ m long and $60$ m wide. Find its area in hectares.

Find the area of a path $2$ m wide surrounding a rectangular garden $20$ m by $15$ m.

Outer dimensions: $(20 + 4)$ m $\times$ $(15 + 4)$ m $= 24 \times 19$ m.
(Adding $2$ m on each side, so $+4$ total.)

A rectangular park $60$ m $\times$ $40$ m has two crossroads running through the middle, each $3$ m wide. Find the area of the crossroads.

We subtract $3 \times 3$ because the intersection is counted twice.

A garden is $90$ m long and $75$ m wide. A path $5$ m wide is to be built outside all around it. Find the area of the path and the cost at Rs $50$ per m$^2$.

Outer dimensions: $(90 + 10) \times (75 + 10) = 100 \times 85$ m.

Area of path $= (100 \times 85) - (90 \times 75) = 8500 - 6750 = 1750$ m$^2$.

Cost $= 1750 \times 50 =$ Rs $87{,}500$.

Two crossroads, each $3$ m wide, run at right angles through the centre of a rectangular park $70$ m $\times$ $45$ m. Find the area of the roads and the cost of tiling at Rs $80$ per m$^2$.

Area of roads $= (70 \times 3) + (45 \times 3) - (3 \times 3) = 210 + 135 - 9 = 336$ m$^2$.

Cost $= 336 \times 80 =$ Rs $26{,}880$.

A figure is made by attaching a semicircle of diameter $14$ cm to one side of a rectangle of dimensions $14$ cm $\times$ $10$ cm. Find the area and perimeter.

Solution:
Area of rectangle $= 14 \times 10 = 140$ cm$^2$.
Radius of semicircle $= 7$ cm.
Area of semicircle $= \dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \times \dfrac{22}{7} \times 49 = 77$ cm$^2$.

Total area $= 140 + 77 = 217$ cm$^2$.

Perimeter $= 10 + 14 + 10 + \dfrac{1}{2} \times 2\pi r = 34 + \dfrac{22}{7} \times 7 = 34 + 22 = 56$ cm.

A parallelogram has area $252$ cm$^2$. If the base is $18$ cm, find the height. If a triangle has the same base and height, find its area.

Solution:
Height of parallelogram $= \dfrac{252}{18} = 14$ cm.

Area of triangle $= \dfrac{1}{2} \times 18 \times 14 = 126$ cm$^2$.

Notice: the triangle's area is exactly half the parallelogram's area when they share the same base and height.

A circular field has radius $21$ m. Find the cost of fencing it at Rs $15$ per metre.

Circumference $= 2 \times \dfrac{22}{7} \times 21 = 132$ m.

Cost $= 132 \times 15 =$ Rs $1{,}980$.

A square of side $14$ cm has four quarter-circles drawn at its corners, each with radius $7$ cm. Find the area of the shaded region (the region inside the square but outside the quarter-circles).

Solution:
Area of square $= 14^2 = 196$ cm$^2$.
Total area of $4$ quarter-circles $= 4 \times \dfrac{1}{4} \pi r^2 = \pi \times 7^2 = \dfrac{22}{7} \times 49 = 154$ cm$^2$.

Shaded area $= 196 - 154 = 42$ cm$^2$.

The height of a parallelogram is the perpendicular distance between the base and the opposite side. If a parallelogram has base $10$ cm, slant side $8$ cm, and height $6$ cm, the area is $10 \times 6 = 60$ cm$^2$, NOT $10 \times 8 = 80$ cm$^2$.

When a problem says "diameter is $28$ cm", the radius is $14$ cm. Always halve the diameter before using formulas with $r$. Using $28$ instead of $14$ in $\pi r^2$ gives an answer $4$ times too large.

When two roads cross at right angles, the intersection area gets counted twice. Always subtract one intersection rectangle: Area $= (l \times w_1) + (b \times w_2) - (w_1 \times w_2)$.

Students often write $1$ m$^2 = 100$ cm$^2$. This is wrong! Since $1$ m $= 100$ cm, we have $1$ m$^2 = 100 \times 100 = 10{,}000$ cm$^2$. Square units require squaring the conversion factor.

Area must always be in square units ($\text{cm}^2$, $\text{m}^2$) and perimeter in linear units (cm, m). Omitting units loses marks in exams. Always write the unit alongside your numerical answer.

Find the area of a parallelogram with base $15$ cm and height $9$ cm.

Answer: Area $= 15 \times 9 = 135$ cm$^2$.

The area of a triangle is $60$ cm$^2$ and its base is $12$ cm. Find the height.

Answer: $60 = \dfrac{1}{2} \times 12 \times h \implies h = \dfrac{60}{6} = 10$ cm.

Find the circumference and area of a circle with radius $21$ cm. (Use $\pi = \dfrac{22}{7}$)

Answer: Circumference $= 2 \times \dfrac{22}{7} \times 21 = 132$ cm. Area $= \dfrac{22}{7} \times 21^2 = \dfrac{22}{7} \times 441 = 1386$ cm$^2$.

A rectangular park is $50$ m $\times$ $30$ m. A path $2$ m wide runs along the outside. Find the area of the path.

Answer: Outer dimensions $= 54 \times 34$. Area of path $= (54 \times 34) - (50 \times 30) = 1836 - 1500 = 336$ m$^2$.

Convert $3.5$ hectares to m$^2$.

Answer: $3.5 \times 10{,}000 = 35{,}000$ m$^2$.