Perimeter and Area Questions

Perimeter and area questions are provided here, along with their solutions and explanations. Students must practice these questions, which will help them excel in their school examinations as well as any type of competitive exams.

The perimeter of any two-dimensional shape is the length of its boundary whereas the area is the measure of the region enclosed by that shape. The perimeter is basically a length hence measured in units of lengths. Area is length squared hence measured in square units.

Below listed are some important formulae for solving perimeter and area questions:

Learn more about perimeter and area.

Perimeter and Area Questions with Solutions

Practice these questions to easily attempt questions related to perimeter and area in various examinations.

Question 1: A wire is bent in the form of a circle. The radius of the circle is 28 cm. The wire is then rebent to form a square. Find the side of the square formed?

Solution:

Radius of the circle = 28 cm

Perimeter of the circle = 2𝜋r = 2 × 22/7 × 28

= 2 × 22 × 4 = 176 cm

Now, Perimeter of square = Perimeter of the circle

⇒ 4a = 176

⇒ a = 44 cm.

Question 2: The area of a right triangle is 28 cm². One of its perpendicular sides exceeds the other by 10 cm. Find the length of the longest perpendicular side.

Solution:

Let the length of shorter perpendicular be x cm then the length of longer pendicular will be x + 10 cm.

Area of right triangle = ½ (Product of perpendicular side) = 28 cm²

⇒ ½ [x(x + 10)] = 28

⇒ x² + 10x – 56 = 0

⇒ (x + 14)(x – 4) = 0

Therefore, x = 4 (neglecting the negative value as length cannot be negative)

Thus, length of the longer perpendicular = 4 + 10 = 14 cm.

Question 3: Find the area of an isosceles triangle of sides 10 cm, 10 cm and 12 cm.

Solution:

Here, a = 10 cm and b = 12 cm

Area of isosceles triangle = b/4[4a² – b²]^1/2 = 12/4 [4(10)² – 12²]^1/2

= 3 [400 – 144]^1/2 = 3(256)^1/2 = 3 × 16 = 48 cm².

Question 4: The diagonal of a square field is 30 m. Find the area of the square field?

Solution:

Area of a square = (Diagonal)²/2 = 30²/2 = 900/2 = 450 m².

Question 5: Find the maximum length of any object that can be kept inside a rectangular box of dimensions 8 cm × 6 cm × 2 cm.

Solution:

Length of the box = 8 cm

Breadth of the box = 6 cm

Heigth of the box = 2 cm

Maximum length of any object = Diagonal of the rectangular box = (l² + b² + h²)^1/2

= (8² + 6² + 2²)^1/2

= √104

= 2√26 cm

Question 6: A circular track runs around a circular park. If the difference between the circumference of the track and the park is 66 m, then find the width of the track.

Solution:

Let R be the radius of the track and r be the radius of the park.

Given,

Circumference of track – circumference of park = 66 cm

Or, 2𝜋R – 2𝜋r = 66

⇒ 2𝜋(R – r) = 66

⇒ (R – r) = 66/2𝜋 = 10.5 m

∴ Width of the park is 10.5 m.

Also Read:

• Sector of a Circle
• Segment of a Circle
• Surface Area and Volume
• Tangent of a Circle

Question 7: In a circle of radius 21 cm, an arc subtends an angle of 30^o at the centre. Find

(i) area of the sector formed.

(ii) area of the minor segment.

(iii) area of the major segment.

Solution:

Radius of the circle = 21 cm

Angle subtended by the arc at the centre = 30^o

(i) Area of the sector = 𝜃/360^o ( 𝜋r²)

= 30^o/360^o (22/7 × 21²)

= 1/12 × 22/7 × 21²

= ½ × 231

= 115.5 cm²

(ii) Area of the minor segment = 𝜃/360^o ( 𝜋r²) – ½ r² sin 𝜃

= 30^o/360^o (22/7 × 21²) – ½ × 21² sin 30^o

= 115.5 – ½ × 441 × ½ (since, sin 30^o =½)

= 115.5 – 110.25

= 5.25 cm²

(iii) Area of the major segment = Area of the circle – area of the minor segment

= 22/7 × 21² – 5.25

= 1386 – 5.25

= 1380.75 cm²

Question 8: The perimeter of a sector of a circle of radius 14 cm is 68 cm. Find the area of the sector.

Solution:

Radius of the circle = 14 cm

Perimeter of the sector = 2r + l where l = length of the arc

⇒ 2r + l = 68 cm

⇒ 28 + l = 68

⇒ l = 68 – 28 = 40 cm

∴ Area of sector = ½ rl = ½ × 14 × 40 = 280 cm².

Question 9: Find the area of a rhombus of perimeter 60 cm and one of its diagonals is 24 cm.

Solution:

Let ABCD be the rhombus

Length of one diagonal (let AC) = 24 cm.

Let the length of sides of rhombus be ‘a’.

Perimeter of rhombus = 4a = 60

⇒ a = 60/4 = 15 cm

By the properties of rhombus, diagonals of a rhombus bisect at right angles.

Then, triangle AOD is a right-angled triangle, with AD = 15 cm and AO = 12 cm

By Pythagoras theorem,

AD² = AO² + OD² ⇒ OD = (AD² – AO²)^½ = (15² – 12²)^½

⇒ OD = 9 cm

∴ length of other diagonal (BD) = 18 cm

Area of rhombus = ½ × product of diagonals = ½ × 24 × 18 = 216 cm².

Question 10: The area of a square field is 36 m². How long would it take for a bird to cross it diagonally, if the bird is flying at the rate of 30√2 m/min?

Solution:

Let ‘a’ be the length of side of square field.

Area of square field = a² = 36 m²

⇒ a = 6 m

Diagonal of square = a√2 = 36√2 m

Now, we know that speed = Distance/Time

∴ 30√2 = 36√2/Time

⇒ Time taken to fly = 36√2/30√2 = 6/5 = 1.2 min = 72 seconds.

Question 11: What is the cost of levelling the field in the form of parallelogram at the rate of ₹ 75 per 10 m², whose base and perpendicular distance from the other side being 54 m and 24 m respectively?

Solution:

Base of parallelogram = 54 m

Height of parallelogram = 24 m

Area of field = 54 × 24 = 1296 m²

Cost of levelling per 10 m² = ₹75

Cost of levelling per 1 m² = ₹ 75/10 = ₹ 7.5

Cost of levelling per 1296 m² = ₹ (1296 × 7.5) = ₹ 9720.

Question 12: The diagonals of two squares are in the ratio 2:5. Find the ratio of their areas.

Solution:

Let the diagonals of the squares be 2x and 5x, respectively.

Area of one square = (2x)²/2 = 2x²

Area of another square = (5x)²/2 = 25x²/2

Ratio of the area of squares = (2x²)/(25x²/2) = 4/25 = 4:25.

Question 13: If it costs ₹ 420 to fence a square park at the rate of ₹ 6 per metre. Find the length and area of the square park.

Solution:

Let ‘a’ be the length of the square park.

Cost of fencing the park = 4a × 6 = ₹ 420

⇒ a = 420/24 = 17.5 m

Area of square park = 17.5 × 17.5 =306.25 m²

∴ Length of the square park is 17.5 m and its area is 306.25 m².

Question 14: Sammy has 16 square stamps of sides 4 cm each. She glues them onto an envelope to form a bigger square. What area of the envelope does the bigger square cover?

Solution:

Length of each stamps = 4 cm

Number of stamps = 16

Area of each stamp = 4 × 4 = 16 cm²

Area of 16 stamps = 16 × 16 = 256 cm².

Thus, an area of 256 cm² will be covered on the envelope.

Question 15: Find the area of a kite, if the length of its diagonals is 20 cm and 15 cm.

Solution:

Area of kite = ½ × product of its diagonals

= ½ × 20 × 15

= 150 cm²

Practice Questions

1. If the diagonal length of a square is 7 cm. Find the area and perimeter of the square?

2. The perimeter of a semi-circular region is 144 cm. What is its area?

3. A verandah 15 m long and 12 m broad is to be paved with tiles each measuring 500 cm × 300 cm. Find the number of tiles needed to cover the whole verandah?

4. What is the height of an equilateral triangle, if the length of its sides is 144 cm?

5. The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

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