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

Chapter 9 Overview: Perimeter and Area

This chapter extends your Class 6 knowledge of perimeter and area to new shapes including parallelograms, triangles, and circles. Perimeter and area are two of the most practically useful concepts in mathematics — whether you are tiling a floor, painting a wall, fencing a garden, or wrapping a gift, you need to compute either the boundary length (perimeter) or the surface measure (area).

In Class 6, you learned the formulas for rectangles and squares. Now in Class 7, the NCERT textbook introduces curved boundaries (circles) and slanted shapes (parallelograms), both of which appear frequently in real life and in competitive exams.

The chapter has three exercises. Exercise 9.1 covers the area of parallelograms and triangles. Exercise 9.2 covers circumference and area of circles. Exercise 9.3 covers conversion of units and mixed problems including paths and crossroads.

Key Concepts and Definitions

Before diving into the exercises, make sure you are comfortable with the following definitions and formulas.

Perimeter and Area Basics

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 and Triangle

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.

Circle Formulas

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.

where $R$ is the outer radius and $r$ is the inner radius.

Unit Conversions

- $1$ 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.

Exercise 9.1 — Area of Parallelograms and Triangles

The key formula: Area of a parallelogram $= \text{base} \times \text{height}$. The height must be the perpendicular distance from the base to the opposite side.

Q1. Area of a parallelogram

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

Q2. Finding the height

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

Q3. Area of a triangle

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

Q4. Finding the base of a triangle

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

$$36 = 4b$$

$$b = 9 \text{ cm}$$

Q5. Comparing parallelogram and rectangle areas

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.

Q6. Two different bases and heights of a parallelogram

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.

Exercise 9.2 — Circumference and Area of Circles

Circumference $= 2\pi r = \pi d$ and Area $= \pi r^2$. Use $\pi = \dfrac{22}{7}$ or $3.14$ as specified.

Q1. Circumference of a circle

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

Q2. Area of a circle

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

Q3. Finding radius from circumference

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

$$2 \times \frac{22}{7} \times r = 44$$

$$r = \frac{44 \times 7}{2 \times 22} = 7 \text{ cm}$$

Q4. Area of a circular garden

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

Radius $= 14$ m.

Q5. Area of a circular track (ring)

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

$$= \frac{22}{7}(1225 - 784) = \frac{22}{7} \times 441 = 22 \times 63 = 1386 \text{ m}^2$$

Q6–Q7. Clock hand and wire reshaping

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:

$$r = \frac{124 \times 7}{2 \times 22} = \frac{868}{44} = 19.73 \text{ cm (approx.)}$$

Exercise 9.3 — Conversion of Units and Mixed Problems

This exercise covers unit conversions and practical problems involving paths and crossroads.

Q1–Q2. Basic unit conversions

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.

$$4800 \text{ m}^2 = \frac{4800}{10{,}000} = 0.48 \text{ hectares}$$

Q3. Path around a rectangle

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.)

Q4. Crossroads problem

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.

$$= 180 + 120 - 9 = 291 \text{ m}^2$$

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

Q5. Path with cost calculation

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 $87{,}500$.

Q6. Crossroads with tiling cost

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 $26{,}880$.

Worked Examples — Additional Practice

These extra examples cover combined figures and reverse calculations frequently asked in exams.

Example 1: Rectangle plus semicircle

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.

Example 2: Parallelogram and triangle relationship

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.

Example 3: Fencing a circular field

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 $1{,}980$.

Example 4: Area of a shaded region

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$^2$.

Common Mistakes to Avoid

These are the most frequent errors students make in this chapter.

Mistake 1: Using slant side as height

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$^2$.

Mistake 2: Confusing radius and diameter

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.

Mistake 3: Forgetting to subtract overlap in crossroads

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

Mistake 4: Wrong unit conversion for area

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.

Mistake 5: Forgetting units in the answer

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.

Practice Questions with Answers

Try these on your own, then check the answers below.

Q1. Area of a parallelogram

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

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

Q2. Height of a triangle

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.

Q3. Circle calculations

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$^2$.

Q4. Path around a park

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$^2$.

Q5. Hectare conversion

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

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

Key Formulas to Remember

Tips for Scoring Full Marks

Ten Common Mistakes Students Make in Perimeter and Area

After years of marking Class 7 papers, teachers keep seeing the same handful of errors on perimeter and area questions. Here they are with the exact wrong vs correct examples you need to memorise.

Mistake 1: Using the slant side as the height of a parallelogram
❌ Wrong: Base $= 10$ cm, slant side $= 8$ cm, area $= 10 \times 8 = 80$ cm$^2$.
✅ Correct: Height is the perpendicular distance between base and opposite side. If height $= 6$ cm, area $= 60$ cm$^2$.

Mistake 2: Confusing radius with diameter
❌ Wrong: Diameter $= 14$ cm, area $= \pi \times 14^2 = 616$ cm$^2$.
✅ Correct: Radius $= 7$ cm, area $= \pi \times 7^2 = 154$ cm$^2$.
Always halve the diameter before plugging into $\pi r^2$ or $2\pi r$.

Mistake 3: Forgetting the $\tfrac{1}{2}$ in the triangle area formula
❌ Wrong: Triangle with base $8$ and height $6$ has area $48$.
✅ Correct: Area $= \tfrac{1}{2} \times 8 \times 6 = 24$ cm$^2$.

Mistake 4: Wrong unit conversion for area
❌ Wrong: $1$ m$^2 = 100$ cm$^2$.
✅ Correct: $1$ m$^2 = 10{,}000$ cm$^2$ because you square the conversion factor.

Mistake 5: Forgetting to subtract the overlap in crossroad problems
❌ Wrong: Area of two crossing roads $=$ area of road $1 +$ area of road $2$.
✅ Correct: Subtract the intersection rectangle once, because it was counted twice.

Mistake 6: Omitting units in the final answer
❌ Wrong: Writing $48$ as the final answer.
✅ Correct: Writing $48$ cm$^2$ for area or $48$ cm for perimeter.

Mistake 7: Using $\pi \approx 3$ without noting it
❌ Wrong: Writing $\pi r^2 = 3 \times 7^2 = 147$ without any note.
✅ Correct: Use $\pi = \tfrac{22}{7}$ or the value given in the question. If asked to approximate, state it.

Mistake 8: Confusing perimeter with area
❌ Wrong: For a square of side $5$ cm, writing area $= 20$ cm$^2$.
✅ Correct: Perimeter $= 20$ cm. Area $= 25$ cm$^2$.

Mistake 9: Forgetting that the path around a rectangle needs outer minus inner
❌ Wrong: Area of path $=$ path width times rectangle perimeter.
✅ Correct: Area of path $=$ outer area $-$ inner area.
The path width formula only works in a simple approximation, not for exact answers.

Mistake 10: Adding the areas of a combined figure before separating them
❌ Wrong: For a rectangle with a semicircle on top, multiplying the rectangle width by total height.
✅ Correct: Compute the rectangle area and the semicircle area separately, then add them.

Previous Year Questions (PYQs) with Solutions

These are CBSE Class 7 style perimeter and area questions from the past few years.

Q1. Find the area of a square whose perimeter is $48$ cm. `(CBSE 2021, 2 marks)`
Solution: Side $= \tfrac{48}{4} = 12$ cm. Area $= 12^2 = 144$ cm$^2$.
Answer: $144$ cm$^2$

Q2. Find the circumference of a circle with radius $14$ cm. (Use $\pi = \tfrac{22}{7}$) `(CBSE 2022, 2 marks)`
Solution: $C = 2\pi r = 2 \times \tfrac{22}{7} \times 14 = 88$ cm.
Answer: $88$ cm

Q3. The area of a triangle is $36$ cm$^2$ and its base is $9$ cm. Find its height. `(CBSE 2023, 2 marks)`
Solution: $36 = \tfrac{1}{2} \times 9 \times h$, so $h = 8$ cm.
Answer: $8$ cm

Q4. A rectangular park is $40$ m long and $25$ m wide. A path $3$ m wide runs along the outside. Find the area of the path. `(CBSE 2022, 3 marks)`
Solution: Outer dimensions $= 46 \times 31$.
Outer area $= 1426$ m$^2$.
Inner area $= 40 \times 25 = 1000$ m$^2$.
Path area $= 1426 - 1000 = 426$ m$^2$.
Answer: $426$ m$^2$

Q5. The radius of a circular field is $21$ m. Find the cost of fencing it at Rs. $20$ per metre. `(CBSE 2023, 3 marks)`
Solution: $C = 2 \times \tfrac{22}{7} \times 21 = 132$ m.
Cost $= 132 \times 20 =$ Rs. $2640$.
Answer: Rs. $2640$

Q6. Find the area of a parallelogram with base $15$ cm and corresponding height $8$ cm. `(CBSE 2021, 2 marks)`
Solution: Area $= 15 \times 8 = 120$ cm$^2$.
Answer: $120$ cm$^2$

Q7. Two cross roads, each of width $2$ m, run at right angles through the centre of a rectangular park $70$ m long and $45$ m wide. Find the area of the roads. `(CBSE 2024, 4 marks)`
Solution:
Area of road $1$ $= 70 \times 2 = 140$ m$^2$.
Area of road $2$ $= 45 \times 2 = 90$ m$^2$.
Overlap $= 2 \times 2 = 4$ m$^2$.
Total area of roads $= 140 + 90 - 4 = 226$ m$^2$.
Answer: $226$ m$^2$

Q8 (HOTS). A circular pool of radius $7$ m is surrounded by a concrete path $1.5$ m wide. Find the area of the path. `(CBSE 2024, 4 marks)`
Solution:
Outer radius $= 7 + 1.5 = 8.5$ m.
Area of outer circle $= \tfrac{22}{7} \times 8.5^2 = \tfrac{22}{7} \times 72.25 \approx 227.07$ m$^2$.
Area of pool $= \tfrac{22}{7} \times 49 = 154$ m$^2$.
Area of path $\approx 227.07 - 154 = 73.07$ m$^2$.
Answer: approximately $73.07$ m$^2$

Practice This Chapter on SparkEd

Perimeter and Area is a chapter where every mark depends on remembering the right formula and applying it cleanly. The only way to get there is through deliberate practice with enough variety that your brain learns to match the question type to the formula instantly.

The SparkEd Perimeter and Area module has more than 60 questions across three difficulty levels. Level 1 gives you clean single shape problems: area of a rectangle, perimeter of a square, circumference of a circle. Level 2 moves into combined figures, path problems, and reverse calculations where you are given the area and have to find a missing dimension. Level 3 is the CBSE exam zone with crossroad problems, fencing costs, and tricky HOTS questions that involve inscribed or surrounding shapes.

The instant AI feedback catches the most common mistakes like using slant side instead of height, or forgetting the overlap subtraction in crossroad problems, and tells you exactly where your working went off. The gamification keeps you coming back, and the adaptive engine feeds you the question types you struggle with most.

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