Rectangle Questions

Rectangle Questions with solutions are available here. These questions are majorly based on the area and perimeter of rectangles. The rectangle questions are very helpful for students of Class 7 and 8 to practise for exams. The rectangle problems are prepared as per the NCERT guidelines and the latest CBSE syllabus (2022-2023). Learn about rectangles at BYJU’S.

Definition: A rectangle is a four-sided polygon that has its opposite sides parallel and equal. Also, all the four angles of a rectangle are right angles.

  • Area of rectangle = Length x Width [square units]
  • Perimeter of rectangle = 2(Length + Width) [units]
  • Diagonal of rectangle = √(l2 + b2) [units]

Rectangle Questions and Answers

Q.1: Find the perimeter of the rectangle with a length equal to 10 cm and width equal to 5 cm.

Solution: Given,

Length of rectangle = 10 cm

Width of rectangle = 5cm

Perimeter of rectangle = 2(length + width)

= 2(10 + 5) cm

= 2 x 15 cm

= 30 cms

Q.2: Find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm.

Solution: Given,

Length of rectangle = 17 cm

breadth of rectangle = 13 cm

Perimeter of rectangle = 2 (length + width)

= 2 (17 + 13)

= 2 (30)

= 60 cms

Area of rectangle = Length x Breadth

= 17 x 13 sq.cm.

= 221 sq.cm.

Q.3: If the perimeter of a rectangle is 48 cm and its breadth is 6 cm, then find the area of a rectangle.

Solution: Given,

perimeter of rectangle = 48 cm

breadth of rectangle = 6 cm

Using perimeter of formula;

48 = 2(length + 6)

24 = length + 6

length of rectangle = 24 – 6 = 18 cm

Area of rectangle = length x breadth = 18 x 6 = 108 cm2

Q.4: Find the area of a rectangle of length 43 m and width 13 m.

Solution: Given,

Length of a rectangle = 43 m

Width of a rectangle = 13 m

Area of a rectangle = l × b

= 43 × 13 m2

= 559 m2

Q.5: The length of a rectangle is 6 cm and the width is 4 cm. If the length is increased by 2 cm, what should be the width of the new rectangle that has the same area as the first one?

Solution: Given,

Length of rectangle = 6cm

Width of rectangle = 4cm

Area of the first rectangle = L × W = 6 × 4 = 24 cm2

Now, the new length is = 6 + 2 = 8 cm

As per the given question, the area of the new rectangle is equal to the area of an old rectangle. Thus,

8 × Width = 24

Then,

Width = 24 ÷ 8 = 3 cm

Thus, the required width is 3 cm.

Q.6: If 1 envelope requires a 20 cm by 5 cm piece of paper, how many envelopes can be made out of a sheet of paper 100 cm by 75 cm?

Solution: Given,

Length of sheet = 100 cm

Width of sheet = 75 cm

Area of the sheet = length x width = 100 x 75 = 7500 cm2

Length of envelope = 20 cm

Width of envelope = 5 cm

Area of envelope = 20 x 5 = 100 cm2

Number of envelopes = (Area of the sheet)/(Area of envelope)

= 7500/100

= 75

Q.7: The width of the rectangle is 8 cm and its diagonal is 17 cm. Find the area and perimeter of the rectangle.

Solution:Given,

Width of rectangle = 8cm

Diagonal = 17 cm

Rectangle questions

In the above figure, using Pythagoras theorem,

BD2 = DC2 + BC2

⇒ 172 = DC2 + 8

⇒ 289 – 64 = DC

⇒ 225 = DC

⇒ 15 = DC

So, length of rectangle = 15 cm

Thus, area of rectangle = length × breadth

= 15 × 8 cm2

= 120 cm2

And, perimeter of rectangle = 2 (15 + 8) cm

= 2 × 23 cm

= 46 cm

Q.8: The area of a rectangular fence is 500 square feet. If the width of the fence is 20 feet, then find its length.

Solution: Given,

Area of rectangular fence = 500 sq.ft.

Width = 20 ft.

As per the formula of area, we have;

Area = length x width

500 = length x 20

Length = 500/20 = 25 ft.

Q.9: How many bricks each 26 cm long and 10 cm broad will be required to lay a footpath 260 cm long and 15 cm wide?

Solution:

Given,

Length of bricks = 26 cm

Breadth of bricks = 10 cm

Therefore, the area of a bricks = length × breadth

= 26 cm × 10 cm

= 260 cm2

Length of footpath = 260 m

Breadth of footpath = 15 m

So, the area of the footpath = 260 × 15 × 10000 cm2

Hence, required number of slabs = Area of path/Are of each slab

= (260 × 15 × 10000)/260

= 150000

Therefore, 150000 bricks will be required to build a footpath of given dimensions.

Q.10: The perimeter of the rectangle is 30 cm and its length is 10 cm. Calculate the length of the diagonal of the rectangle.

Solution: Given,

Perimeter of rectangle = 30 cm

Length of rectangle = 10 cm

Thus,

30 = 2(10 + width of rectangle)

30 = 2 (10) + 2w

10 = 2w

Width = 5 cm

Diagonal of rectangle = √(l2 + b2) = √(102 + 52)

= √125 cm

= √(5 x 5 x 5) cm

= 5√5 cm

Video Lesson on Properties of rectangles

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