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
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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