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

### Question 1

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.

### Question 2

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

= 20ft2Area covered by one bag of soil

=10ft2

∴ No. of bags required

= Area of the fieldArea covered by one bag of soil=2010=2Hence, Preeti need 2 bags.

### Question 3

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

### Question 4

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.

### Question 5

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

### Question 6

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

### Question 7

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

### Question 8

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

### Question 9

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 inches2Area of the shape = Area of the small rectangle + Area of the large rectangle + Area of the triangle

= (30 + 48 + 6) inches2= 84 inches2

### Question 10

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