Download free pdf of Chapter 11 **Perimeter and Area Worksheets for Class 7 Maths with answers**. Students can easily get all the Class 7 Maths Worksheets and Exercises in Free Printable ( PDF Format) as per the latest syllabus and examination pattern in CBSE and NCERT pattern. Below you will find all the Perimeter and Area related sums that are asked in 7th Grade Mathematics Exams Across Indian Schools.

**Perimeter and Area** Worksheets for Class 7 in PDF Format

- CBSE Class 7 Perimeter and Area Worksheet with answers – 6
- CBSE Class 7 Perimeter and Area Worksheet with answers – 5
- CBSE Class 7 Perimeter and Area Worksheet with answers – 4
- CBSE Class 7 Perimeter and Area Worksheet with answers – 3
- CBSE Class 7 Perimeter and Area Worksheet with answers – 1
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- Perimeter and Area CBSE Class 7 Maths Worksheet
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- Perimeter and Area Class 7 Maths Worksheet in PDF
- Perimeter and Area Grade 7 Worksheet for Maths
- Perimeter and Area CBSE Grade 7 Maths Worksheet
- Maths MCQ Questions for Class 7 Perimeter and Area – 4
- Maths MCQ Questions for Class 7 Perimeter and Area – 3
- MCQ Questions for Class 7 Maths Chapter 11 Perimeter and Area
- MCQ Perimeter and Area Chapter 11 Class 7 Maths

### CBSE Class 7 **Perimeter and Area** Worksheet with answers

**Perimeter and Area**

**Q.1)** A door-frame of dimensions 3 m × 2 m is fixed on the wall of dimension 10 m × 10 m. Find the total labour charges for painting the wall if the labour charges for painting 1 m2 of the wall is Rs 2.50.

**Solution:**

Area of the wall = Length × Width = 10 m × 10 m = 100 m^2 Area of the door-frame = (Length + Width) × 2 = (3 m + 2 m) × 2 = 10 m × 2 = 20 m^2

Area to be painted = Area of the wall – Area of the door-frame = 100 m^2 – 20 m^2 = 80 m^2

Labor charges for painting 1 m^2 = Rs 2.50

Total labor charges = Area to be painted × Labor charges per m^2 Total labor charges = 80 m^2 × Rs 2.50/m^2 = Rs 200

**Answer: The total labor charges for painting the wall are Rs 200.**

**Q.2)** The area of a rectangular sheet is 500 cm2. If the length of the sheet is 25 cm, what is its width? Also find the perimeter of the rectangular sheet.

**Solution** :

Given: Area of the rectangular sheet = 500 cm^2 Length of the sheet = 25 cm

We can find the width using the formula for the area of a rectangle: Area = Length × Width

So, Width = Area / Length = 500 cm^2 / 25 cm = 20 cm

To find the perimeter: Perimeter = 2 × (Length + Width) = 2 × (25 cm + 20 cm) = 2 × 45 cm = 90 cm

**Answer: The width of the rectangular sheet is 20 cm, and the perimeter is 90 cm.**

**Q.3)** Anu wants to fence the garden in front of her house, on three sides with lengths 20 m, 12 m and 12 m. Find the cost of fencing at the rate of Rs 150 per metre.

**Solution**:

Total length of fencing needed = 20 m + 12 m + 12 m = 44 m

Cost of fencing per meter = Rs 150

Total cost of fencing = Total length × Cost per meter = 44 m × Rs 150/m = Rs 6,600

**Answer: The cost of fencing is Rs 6,600.**

**Q.4) **The diameter of the wheels of a bus is 140 cm. How many revolutions per minute must a wheel make in order to move a t a speed of 66km/hr? (a) 240 (b) 250 (c) 260 (d) 270

**Solution**:

Given: Diameter of the wheel = 140 cm = 1.4 m (converted to meters) Speed of the bus = 66 km/hr = 66000 m/hr (converted to meters per hour)

The distance covered by one revolution of the wheel is equal to the circumference of the wheel.

Circumference = π × Diameter = π × 1.4 m ≈ 4.4 m (approximately)

To find the number of revolutions per minute, we need to convert the speed from meters per hour to meters per minute.

Speed in meters per minute = Speed in meters per hour / 60 minutes = 66000 m / 60 min = 1100 m/min

Now, to find the number of revolutions per minute: Number of revolutions per minute = Speed in meters per minute / Circumference = 1100 m/min / 4.4 m/rev ≈ 250 rev/min

**Answer: The wheel must make approximately 250 revolutions per minute.**

**Q.5) **The perimeter of regular polygon is(a) no. of sides x lengths of one side (b) no. of sides – lengths of one side

(c) no. of sides + lengths of one side (d) no. of sides ÷ lengths of one side

**Answer: (a) no. of sides × length of one side**

**Q.6)** A wire is in the shape of a square of side 10 cm. If the wire is re-bent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle?

Given: Side of the square = 10 cm

The wire is re-bent into a rectangle of length 12 cm. Let’s find its breadth (B).

Perimeter of the square = 4 × Side = 4 × 10 cm = 40 cm Perimeter of the rectangle = 2 × (Length + Breadth) = 2 × (12 cm + B)

Since the wire remains the same, the perimeters of both shapes are equal.

So, 40 cm = 2 × (12 cm + B)

Now, solve for B: 2 × (12 cm + B) = 40 cm 12 cm + B = 20 cm B = 20 cm – 12 cm B = 8 cm

So, the breadth of the rectangle is 8 cm.

To determine which shape encloses more area, we can compare their areas:

Area of the square = Side × Side = 10 cm × 10 cm = 100 cm^2 Area of the rectangle = Length × Breadth = 12 cm × 8 cm = 96 cm^2

The square encloses more area compared to the rectangle.

**Answer: The breadth of the rectangle is 8 cm, and the square encloses more area than the rectangle.**

**Q.7)** The area of a square and a rectangle are equal. If the side of the square is 40 cm and the breadth of the rectangle is 25 cm, find the length of the rectangle. Also, find the perimeter of the rectangle.

Given: Side of the square = 40 cm Breadth of the rectangle = 25 cm

We know that the area of a square is given by: Area of square = Side × Side

And the area of a rectangle is given by: Area of rectangle = Length × Breadth

Since the areas are equal, we can set up the equation: Side × Side = Length × Breadth

Plugging in the given values: 40 cm × 40 cm = Length × 25 cm

Now, solve for the length (L): Length = (40 cm × 40 cm) / 25 cm Length = 1600 cm^2 / 25 cm Length = 64 cm

So, the length of the rectangle is 64 cm.

To find the perimeter of the rectangle: Perimeter = 2 × (Length + Breadth) = 2 × (64 cm + 25 cm) = 2 × 89 cm = 178 cm

**Answer: The length of the rectangle is 64 cm, and the perimeter of the rectangle is 178 cm.**

**Q.8)** In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm.

**Solution:**

The area of a triangle is given by the formula: Area = (1/2) × Base × Height

Given: Area of triangle ABC = 36 cm² Height AD = 3 cm

We need to find the base, which is BC.

36 cm² = (1/2) × BC × 3 cm

Now, solve for BC: BC = (2 × 36 cm²) / 3 cm BC = 72 cm² / 3 cm BC = 24 cm

**Answer: BC = 24 cm.**

**Q.9)** Find BC, if the area of the triangle ABC is 36 cm^{2} and the height AD is 3 cm.

**Solution:**

In a parallelogram, opposite sides are equal in length.

So, BC = AD = 4.5 cm.

**Q.10)** What is the circumference of a circle of diameter 10 cm (Take **π** = 3.14)?

**Solution:**

Circumference of a circle = π × Diameter

Given: Diameter = 10 cm π = 3.14 (approximated)

Circumference = 3.14 × 10 cm = 31.4 cm

**Answer: The circumference is 31.4 cm.**

**Q.11)** What is the circumference of a circular disc of radius 14 cm?

**Solution:**

Circumference of a circle = 2 × π × Radius

Given: Radius = 14 cm π = 3.14 (approximated)

Circumference = 2 × 3.14 × 14 cm = 87.92 cm (rounded to two decimal places)

**Answer: The circumference is approximately 87.92 cm.**

**Q.12)** The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m^{2}. Find the cost of ploughing the field.

**Solution:**

Given: Cost of fencing = Rs 5280 Rate of fencing = Rs 24 per meter

To find the perimeter of the circular field, we need to know the length of the fencing required. The perimeter of a circle = 2 × π × Radius.

Let R be the radius of the circular field, then 2 × π × R = Length of fencing

So, 2 × 3.14 × R = Length of fencing

Given the cost of fencing is Rs 5280, we can write the equation: Length of fencing × Rate of fencing = Cost of fencing

2 × 3.14 × R × 24 = 5280

Now, solve for R: R = 5280 / (2 × 3.14 × 24)

R ≈ 5.5 meters

Now that we know the radius, we can calculate the area of the circular field.

Area of the circular field = π × (Radius)^2

Area ≈ 3.14 × (5.5 meters)^2 ≈ 94.985 square meters

Now, find the cost of plowing: Cost of plowing = Area × Rate of plowing = 94.985 m² × Rs 0.50/m² ≈ Rs 47.49

**Answer: The cost of plowing the field is approximately Rs 47.49.**

**Q.13)** The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

**Solution:**

Area of a circle = π × (Radius)^2

Given: Radius of the first circle (R1) = 8 cm Radius of the second circle (R2) = 6 cm

Area of the first circle = π × (8 cm)^2 = 64π cm² Area of the second circle = π × (6 cm)^2 = 36π cm²

We want to find the radius (R3) of the circle with the sum of the areas of the two circles: Area of the third circle = Area of first circle + Area of second circle

R3 = √[(64π cm² + 36π cm²) / π] R3 = √[100 cm²] = 10 cm

**Answer: The radius of the circle is 10 cm.**

**Q.14)** The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

**Solution:**

Circumference of a circle = 2 × π × Radius

Given: Radius of the first circle (R1) = 19 cm Radius of the second circle (R2) = 9 cm

Circumference of the first circle = 2 × π × 19 cm ≈ 119.38 cm Circumference of the second circle = 2 × π × 9 cm ≈ 56.52 cm

We want to find the radius (R3) of the circle with the sum of the circumferences of the two circles: Circumference of the third circle = Circumference of first circle + Circumference of second circle

R3 = (119.38 cm + 56.52 cm) / (2 × π) R3 ≈ 175.9 cm / (2 × 3.14) R3 ≈ 27.97 cm (rounded to two decimal places)

****Answer**: The radius of the circle

**Perimeter and Area: Understanding the Basics**

Perimeter and area are fundamental concepts in geometry that help us understand and quantify the size and shape of two-dimensional figures. They play a crucial role in various real-life scenarios, from building a fence around a garden to calculating the amount of paint needed to cover a wall. Let’s delve into the basics of perimeter and area to get a better understanding.

**Perimeter: The Boundary Length**

Imagine you have a piece of string, and you use it to outline the shape of a figure, such as a rectangle, square, or circle. The length of that string, which traces the boundary of the figure, is what we call the “perimeter.” In simpler terms, the perimeter is the total distance around the edge of a shape.

**Calculating Perimeter:**

**Rectangle or Square:**To find the perimeter of a rectangle or a square, add up all four sides’ lengths. If the rectangle has sides of length “a” and “b,” then the perimeter (P) is given by P = 2a + 2b.**Circle:**In the case of a circle, we use a special constant called π (pi), approximately equal to 3.14. The perimeter of a circle, known as the “circumference,” is calculated as C = 2πr, where “r” is the radius of the circle.

**Area: The Enclosed Space**

Now, let’s shift our focus to area. Area measures the amount of space enclosed by a two-dimensional figure. Think of it as the surface area you would need to paint or cover with a carpet. Area is expressed in square units, such as square centimeters (cm²) or square meters (m²).

**Calculating Area:**

**Rectangle or Square:**The area of a rectangle or square is found by multiplying the length (l) and the width (w). So, Area (A) = l × w.**Circle:**For a circle, the area can be calculated using the formula A = πr², where “r” represents the radius of the circle.**Triangle:**In the case of a triangle, you can find the area using the formula A = (1/2) × base × height, where the base and height are perpendicular to each other.**Other Shapes:**Various other shapes have their own specific formulas for calculating area, such as the formula for the area of a trapezoid or a parallelogram.

**Connecting Perimeter and Area:**

There’s an interesting relationship between perimeter and area. If you think about it, perimeter deals with the outer boundary, while area focuses on the interior space. Sometimes, you might need to consider both when solving real-world problems.

For instance, imagine you want to build a fence around a rectangular garden. You’ll need to calculate the perimeter to determine how much fencing material is required. On the other hand, to find out how much grass seed or fertilizer you need to cover the garden, you’ll need to calculate the area.

In conclusion, perimeter and area are essential tools in geometry that help us measure and understand the size and shape of two-dimensional figures. Whether you’re solving math problems or tackling practical projects, a solid grasp of these concepts will serve you well in various aspects of life. So, the next time you encounter a shape, think about both its perimeter and area to get a complete picture of its dimensions.