Unit Test Questions and Solution: Perimeter and Area | Mathematics for Class 6 PDF Download

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Time: 1 hour M.M. 30 
Attempt all questions. 

• Question numbers 1 to 5 carry 1 mark each. 
• Question numbers 6 to 8 carry 2 marks each. 
• Question numbers  9 to 11 carry 3 marks each. 

Q1. What is the perimeter of a square with a side length of 5 cm?   (1 mark)
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm

Ans: (d) 
Sol: Perimeter of a square = 4 × side length = 4 × 5 = 20 cm.

Q2. The area of a rectangle is 24 cm², and its length is 8 cm. What is the width of the rectangle?   (1 mark)
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm

Ans: (c)
Sol: Area of a rectangle = length × width. 24 = 8 × width, so width = 24 ÷ 8 = 3 cm.

Q3. A triangle has a base of 10 cm and a height of 6 cm. What is its area?   (1 mark)
(a) 30 cm²
(b) 60 cm²
(c) 50 cm²
(d) 15 cm²

Ans:  (a)
Sol: Area of a triangle = (1/2) × base × height = (1/2) × 10 × 6 = 30 cm².

Q4. Find the missing terms:
(a) Perimeter of a rectangle = 20 cm; breadth = 4 cm; length = ? (1 mark)

Ans: We know that the perimeter of a rectangle = 2(l + b)
Here, the perimeter of the rectangle = 20 cm and breadth b = 4 cm, l =?
Thus,
20 = 2(l + 4)
⇒ 20 = 2l + 8
⇒ 2l = 20 – 8 = 12
⇒ l = 12 ÷ 2 = 6 cm

Q5. The area of a rectangular garden that is 60 m long is 1200 sq m. Find the width of the garden.    (1 mark)

Ans: Width of the rectangular garden =
area/length of the garden
= 1200 / 60
= 20 m

Q6: What is the cost of tiling a rectangular plot of land 600 m long and 400 m wide at the rate of ₹10 per hundred sq m?     (2 marks)

Ans: Here, length = 600 m and breadth = 400 m.
Hence, the area of the rectangular plot = length × breadth
= 600 × 400
= 240,000 m²
Now, cost of tiling a rectangular plot = ₹10 per 100 sq m.
Hence, the cost of tiling 240,000 sq m of rectangular plot = ₹10 × 240000 ÷ 100 = ₹24,000.

Q7: Find the length of the third side of a triangle having a perimeter of 60 cm and having two sides of length 22 cm and 18 cm, respectively.    (2 marks)

Ans: Let the length of the third side of the triangle be x cm. Then,
Perimeter of triangle = AB + BC + CA
⇒ 60 = 22 + 18 + x
⇒ 60 = 40 + x
⇒ x = 60 – 40
⇒ x = 20 cm

Q8: What would be the cost of fencing a rectangular park whose length is 180 m and breadth is 130 m if the fence costs ₹50 per meter?          (2 marks)

Ans: The length of the fence is the perimeter of the rectangular park.
Given that the length of the rectangular park = 180 m and breadth = 130 m.
Thus,
Perimeter = 2(l + b)
= 2(180 + 130)
= 2(310)
= 620 m
Now, the cost of fencing per meter = ₹50
Cost of fencing the rectangular park = ₹50 × 620 = ₹31,000

Q9: A farmer has a rectangular field as given. He wants to fence it with 4 rounds of rope. What is the total length of rope needed?​       (3 marks)​​​

Ans: Perimeter of the rectangular field = 2(l + b)
Here, l = 250 m, b = 180 m
Thus, P = 2(250 + 180)
= 2(430)
= 860 m
Distance covered by a farmer in one round = 860 m
Thus, total length of rope needed = 4 × 860 = 3440 m

Q10: A rectangle with side lengths 6 cm and 4 cm is made using a wire. If the wire is straightened and then bent to form a square, what will be the length of the side of the square?       (3 marks)

Ans: Here, the perimeter of the rectangle = 2(6 + 4)
= 2 × 10
= 20 cm
Now the wire is straightened and then bent to form a square.
Thus, perimeter of the square = 20 cm
⇒ 4a = 20 cm
⇒ a = 20 ÷ 4 = 5 cm, the required length of the side of the square.

Q11: The floor of a room is 6 m long and 5 m wide. A square carpet whose sides are 4 m in length is laid on the floor. Find the area that is not carpeted. ​       (3 marks)

Ans: We have a floor with dimensions 6 m in length and 5 m in width.
A square carpet of side 4 m.
Area of the floor = length × breadth
= 6 × 5
= 30 m²
Area of the square carpet = 4 × 4
= 16 m²
Now, we will subtract the square carpet area from the floor’s area to get the area of the floor that is not carpeted.
Hence, the area of the floor that is not carpeted = 30 – 16 = 14 m²
Thus, the area of the floor that is not carpeted is 14 m².