# Free Mensuration Subjective Test 02 Practice Test - 6th grade

What is Perimeter? [1 MARK]

#### SOLUTION

Solution :

Concept: 1 Mark

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

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 10ft2. How many bags will she need to cover the entire field? [2 MARKS]

#### SOLUTION

Solution :

Steps: 1 Mark
Answer: 1 Mark

Area of the field
= 10×2
= 20ft2

Area covered by one bag of soil
=10ft2

No. of bags required

= Area of the fieldArea covered by one bag of soil=2010=2

Hence, Preeti need 2 bags.

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]

#### SOLUTION

Solution :

Steps: 1 Mark
Answer: 1 Mark

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

50=15+15+y+y

50=2(15+y)

25=(15+y)

y=10

The width of notebook is 10cm

Area covered by the notebook

=15cm×10cm

=150cm2

How many sides a pentagon have? Shakshi was playing in a park which was shaped as a pentagon. Each side of the Pentagon was 30 m. She took one complete round on its boundary. What is the distance covered by her? [3 MARKS]

#### SOLUTION

Solution :

Pentagon: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Pentagon has 5 sides. Given that each side of the pentagon =30m

The perimeter of a regular pentagon

= 5× length of each side

= 5×30

= 150m

Hence, the distance covered by Shakshi is 150m.

Jose wants new a carpet for his living room. His living room is 9m × 9m. If the carpet cost is Rs 12 per square metre, what is the cost of the carpet? [3 MARKS]

#### SOLUTION

Solution :

Steps: 1 Mark
Formula: 1 Mark
Answer: 1 Mark

Area of the living room

= length × breadth = 9m×9m=81m2

The cost per m2  = 12

Cost for 81m2=12×81=972

Find the perimeter of the following figure. [3 MARKS] #### SOLUTION

Solution :

Steps: 2 Marks
Answer: 1 Mark Perimeter of the figure will be:

= 3 cm + 6 cm + 3 cm + 3 cm + 10 cm + 3 cm + 2 cm + 3 cm + 2 cm + 3 cm
(Starting from the topmost edge and going clockwise)
=38cm

If the cost of painting is Rs 250 per km, then what is the cost for painting the boundary of an equilateral triangle shaped garden having its sides as 4 km also find the area of the garden if its height is 3.46 km  [4 MARKS]

#### SOLUTION

Solution :

Steps: 2 Marks
Formula: 1 Mark
Answer: 1 Mark

The perimeter of an equilateral triangle:

= 3 × length of a side.

The total distance to be painted:

= 3 × 4 = 12 km

Hence, the total cost for painting:

= 250 × 12

= Rs 3000

Area of the garden

= 12× base × height.

= 12×42×3.46

= 3.46 km2

Area of a rectangular carpet is 18m2 and its length is 3 m. Find: [4 MARKS]
(a) The breadth of the carpet.
(b) The perimeter of the carpet.

#### SOLUTION

Solution :

Each Option: 2 Marks

(a) Area of a rectangular garden:
= length × breadth = 18
= 3 × breadth

breadth = 183
= 6 m

(b) Perimeter = 2 × (length + breadth)
= 2 × (3+6) = 18m2

Find the area of the figure given below (in sq. inches).  [4 MARKS] #### SOLUTION

Solution :

Steps: 2 Marks
Formula: 1 Mark
Answer: 1 Mark

The area of the figure can be calculated in three steps:

i) Find the area of the small rectangle.

ii) Find the area of the large rectangle.

iii) Find the area of the triangle

i) Area of the small rectangle:
= length × breadth
= 5 × 6
= 30 inches2

ii) Area of the large rectangle:
= 12 × 4
= 48 inches2

iii) Area of the triangle:
= 12× (height × base)
= 12× 4 × 3 = 6 inches2

Area of the shape = Area of the small rectangle + Area of the large rectangle + Area of the triangle
= (30 + 48 + 6) inches2= 84 inches2

The area of a square field is 576m2. Find  [4 MARKS]
(i) The length of its sides
(ii) The length of its perimeter
(iii) Cost of fixing a fence along the boundary of the field at the rate of Rs 3.20 per metre.

#### SOLUTION

Solution : Steps: 2 Marks
Formula: 1 Mark
Answer: 1 Mark

(i) Let l be the length of the square
Area of square = l2
l2=576
l=24 m

(ii) Perimeter of square = 4l

4×(24)=96 m

(iii)  Cost of fixing a fence
=(96×3.20)=307.20