Mensuration(6th Grade > Mathematics ) Questions and answers for exam Preparation

6th Grade > Mathematics

MENSURATION MCQs

Total Questions : 95 | Page 4 of 10 pages

Question 31. A large pizza has a circumference of 88 inches. What will be the smallest area of the box holding the pizza? [4 MARKS]

Discuss Question

:
Steps: 2 Marks
Formula: 1 Mark
Answer: 1 Mark
Circumference of a circle
=

2 \times π \times r = d \times π = 88 i n c h e s

Where, d = diameter and

π = \frac{22}{7}

\Rightarrow 88 = \frac{22}{7} \times d

\Rightarrow d = \frac{88 \times 7}{22}

\Rightarrow d = 28 i n c h e s

The diameter of the pizza will be equal to the length of the side of the box.
Side of box = 28 inches
Area of the box = (side)

^{2}

= (28 inches)

^{2}

= 784 inches

^{2}

Question 32. The blackboard of Akshit's classroom has length 6m and its width is 5m. If it has to be framed by a steel plate at the borders, then what will be the length of steel plate required? [3 MARKS]

Discuss Question

:
Steps: 1 Mark
Formula: 1 Mark
Answer: 1 Mark
The perimeter of the blackboard = Length of the steel plate required.
Perimeter of a rectangle

= 2 \times (L + B)

.
Length of steel plate
=

2 \times (6 + 5) = 2 \times 11 = 22 m

Question 33. What is Perimeter? [1 MARK]

Discuss Question

:
Concept: 1 Mark
The perimeter is the distance covered along the boundary forming a closed figure when you go round the figure once.

Question 34. Jose wants new a carpet for his living room. His living room is 9m

\times

9m. If the carpet cost is Rs 12 per square metre, what is the cost of the carpet? [3 MARKS]

Discuss Question

:
Steps: 1 Mark
Formula: 1 Mark
Answer: 1 Mark
Area of the living room
= length

\times

breadth =

9 m \times 9 m = 81 m^{2}

The cost per

m^{2}

= ₹12
Cost for

81 m^{2} = 12 \times 81 = ₹ 972

Question 35. The length of a rectangular notebook is 15 cm and the perimeter is 50 cm. Find the width and area of the notebook. [2 MARKS]

Discuss Question

:
Steps: 1 Mark
Answer: 1 Mark
Let the breadth of the notebook be

y c m

Length = 15 cm (Given)
Perimeter = 50 cm (Given)
Perimeter of the rectangular notebook = Sum of the lengths of its four sides.

\Rightarrow 50 = 15 + 15 + y + y

\Rightarrow 50 = 2 (15 + y)

\Rightarrow 25 = (15 + y)

\Rightarrow y = 10

∴

The width of notebook is

10 c m

Area covered by the notebook

= 15 c m \times 10 c m

= 150 c m^{2}

Question 36. Preeti has a rectangular field that is 10 ft long and 2 ft wide and she wants to cover it with soil. One bag of soil can cover

10 {f t}^{2}

. How many bags will she need to cover the entire field? [2 MARKS]

Discuss Question

:
Steps: 1 Mark
Answer: 1 Mark
Area of the field
=

10 \times 2

20 f t^{2}

Area covered by one bag of soil
=

10 f t^{2}

∴

No. of bags required
=

\frac{Area of the field}{Area covered by one bag of soil} = \frac{20}{10} = 2

Hence, Preeti need2 bags.

Question 37. Area of a rectangular carpet is

18 m^{2}

and its length is 3 m. Find: [4 MARKS]
(a) The breadth of the carpet.
(b) The perimeter of the carpet.

Discuss Question

:
Each Option: 2 Marks
(a) Area of a rectangular garden:
= length

\times

breadth = 18
= 3

\times

breadth

\Rightarrow

breadth =

\frac{18}{3}

= 6 m
(b) Perimeter = 2

\times

(length + breadth)
= 2

\times

(3+6) = 18m

^{2}

Question 38. The amount of surface enclosed by a closed figure is called its _____.

perimeter
area
volume
height

Discuss Question

Answer: Option B. -> area
:
B
The amount of surface enclosed by aclosed figure is called its area. An area expresses the extent of a two-dimensionalfigure.

Question 39. A piece of square land has a perimeter equal to 20 m. A cow is standing in the middle of the square land as shown in the figure. What is the minimum distance the cow has to travel in order to go out of the field and drink water?