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CBSE Class 5  >  Class 5 Test  >  Mathematics  >  Area & Perimeter - 2 - Class 5 MCQ

Area & Perimeter - 2 - Free MCQ Test with solutions for Class 5 Mathematics


MCQ Practice Test & Solutions: Area & Perimeter - 2 (10 Questions)

You can prepare effectively for Class 5 Mathematics for Class 5 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Area & Perimeter - 2". These 10 questions have been designed by the experts with the latest curriculum of Class 5 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

Sign up on EduRev for free to attempt this test and track your preparation progress.

Area & Perimeter - 2 - Question 1
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Calculate the area of rectangle if one square represents an area of 1 cm2.

Detailed Solution: Question 1

There are 6×4=24 squares and each square has an area of = 1cm2. So, the area of rectangle =6×4×1=24cm2.

Area & Perimeter - 2 - Question 2
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Each side of square is of 10m. What will be the area of the square?

Detailed Solution: Question 2

- A square has four equal sides.
- Each side measures 10 meters.
- The area of a square is found by multiplying one side by itself.
- Therefore, Area = 10 m × 10 m = 100 m².

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Area & Perimeter - 2 - Question 3
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Find the missing side in the figure given below if the perimeter of the figure is 28m?

Detailed Solution: Question 3

Given figure is a rectangle with length (l) = 8 m.
Let the breadth be b.
Perimeter of a rectangle = 2 × (length + breadth)
Given perimeter = 28 m
So, 2 × (8 + b) = 28
=> 8 + b = 28 ÷ 2 = 14
=> b = 14 - 8 = 6 m
Hence, the missing side (breadth) is 6 m.

Area & Perimeter - 2 - Question 4
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The area of a triangle whose base is 12 cm and height is twice the base is:

Detailed Solution: Question 4

Solution:

  • Base of the triangle = 12 cm
  • Height of the triangle = 2 × 12 = 24 cm
  • Area of the triangle = (1/2) × 12 × 24 = 144 sq. cm

Area & Perimeter - 2 - Question 5
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The breadth of a rectangle is increased by 2 units. Its perimeter is now increased by?

Detailed Solution: Question 5

The perimeter of a rectangle is given by the formula: P=2×(l+b) Where:

  • l = length of the rectangle
  • b = breadth of the rectangle

For the new rectangle, the new length is increased by 2 units, i.e., l′=l+2, while the breadth remains the same.

The perimeter of the new rectangle is calculated as: P′=2×(l′+b)
Substitute l′=l+2l' into the equation: P′=2×((l+2)+b)
Simplifying: P′=2l+2b+4
Thus, the new perimeter becomes: P′=P+4
This shows that the perimeter of the rectangle increases by 4 units when the breadth is increased by 2 units, while the length  remains the same.

Area & Perimeter - 2 - Question 6
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What is the formula for the area of a square?

Detailed Solution: Question 6

The area of a square is calculated by multiplying one side by itself. This is often expressed as side × side.

Area & Perimeter - 2 - Question 7
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Which of these figures represents the area of 24m2 if given that each block=2m2

Detailed Solution: Question 7

Each block has area 2m2.
Number of blocks required = 24 ÷ 2 = 12 blocks.
Option C contains 12 blocks (2 × 6 = 12), so it represents an area of 24m2.

Answer: C

Area & Perimeter - 2 - Question 8
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Find area of unshaded region if each box = 3m2.

Detailed Solution: Question 8

Total Boxes = 88
No. of shaded boxes = 28,
No. of unshaded boxes = 88 -28 = 60
Area of 1 unshaded box =3m2
So, area of unshaded region = 60×3=180m2.

Area & Perimeter - 2 - Question 9
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Raj walks around a rectangular park with a length of 150 meters and a width of 100 meters. If he takes 3 rounds of the park, how much distance does he cover?

Detailed Solution: Question 9

Distance covered in one round = Perimeter of the park
= 2 × (Length + Width)
= 2 × (150 m + 100 m)
= 2 × 250 m
= 500 m
Distance covered in 3 rounds = 3 × 500 m
= 1500 m

Area & Perimeter - 2 - Question 10
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Find the cost of tiling a hall 40 m long and 20 m broad at the rate of ₹ 10 per sq. m.

Detailed Solution: Question 10

Tiling the hall means tiling the area of the hall.
Area (A) = Length × Width
= 40 m × 20 m
= 800 sq. m
Cost of tiling 1 sq. m = ₹ 10
Cost of tiling 800 sq. m = 800 × ₹ 10 = ₹ 8000
Thus, the cost of tiling the hall is ₹ 8000.

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About this Class 5 Practice Test

This test contains 10 MCQs on Area & Perimeter - 2 with detailed solutions for each question. It is part of the Mathematics for Class 5 course on EduRev, designed to align with the current Class 5 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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