Rectangle Questions

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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 = √(l 2 + b 2 ) [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

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

Length of rectangle = 17 cm

breadth of rectangle = 13 cm

Perimeter of rectangle = 2 (length + width)

= 2 (17 + 13)

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.

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 cm 2

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

Length of a rectangle = 43 m

Width of a rectangle = 13 m

Area of a rectangle = l × b

= 43 × 13 m 2

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?

Length of rectangle = 6cm

Width of rectangle = 4cm

Area of the first rectangle = L × W = 6 × 4 = 24 cm 2

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

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?

Length of sheet = 100 cm

Width of sheet = 75 cm

Area of the sheet = length x width = 100 x 75 = 7500 cm 2

Length of envelope = 20 cm

Width of envelope = 5 cm

Area of envelope = 20 x 5 = 100 cm 2

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

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,

BD 2 = DC 2 + BC 2

⇒ 17 2 = DC 2 + 8

⇒ 289 – 64 = DC

So, length of rectangle = 15 cm

Thus, area of rectangle = length × breadth

= 15 × 8 cm 2

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

= 2 × 23 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.

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?

Length of bricks = 26 cm

Breadth of bricks = 10 cm

Therefore, the area of a bricks = length × breadth

= 26 cm × 10 cm

Length of footpath = 260 m

Breadth of footpath = 15 m

So, the area of the footpath = 260 × 15 × 10000 cm 2

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

= (260 × 15 × 10000)/260

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.

Perimeter of rectangle = 30 cm

30 = 2(10 + width of rectangle)

30 = 2 (10) + 2w

Width = 5 cm

Diagonal of rectangle = √(l 2 + b 2 ) = √(10 2 + 5 2 )

= √(5 x 5 x 5) cm

Video Lesson on Properties of rectangles

problem solving involving area of rectangle

Related Articles

  • Area of Rectangle
  • Perimeter of Rectangle
  • Properties of Rectangle

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