Download CBSE Sample Paper 2024-25 for class 12th to 8th

Sign in

Perimeter & Area Test - 1

Result

Perimeter & Area Test - 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A lawn was fenced leaving a border 1 m wide all around it to protect it from street dogs. The measurement of the fence is 10 m by 6 m. What is the perimeter of the lawn?

A
15 m

B
20 m

C
24 m

D
26 m

Solution

Length of the lawn = [10 - 1 - 1] m = 8 m
Breadth of the lawn = [6 - 1 - 1] m = 4 m
Perimeter of lawn = 2 (Length + Breadth) = 2(8 + 4) = 24 m
Question 2
1 / -0

What is the area of the given figure?

A
24 square units

B
48 square units

C
64 square units

D
36 square units

Solution

There are 23 full squares and 2 halves.
So, total squares = 23 + 1 = 24
Area = 24 x 2 sq. units = 48 sq. units
Question 3
1 / -0

This figure is made up of 4 identical squares as shown below. Find the total area of the shaded region, given that the side of each smaller square is 4 cm.

A
8 cm²

B
64 cm²

C
32 cm²

D
16 cm²

Solution

Area of 1 square = 4 x 4 = 16 cm²But, it is half shaded.
So, area of shaded part of 1 square == 8 cm²Now, there are 4 squares with half portion shaded.
So, area of the shaded region = 4 x 8 = 32 cm²
Question 4
1 / -0

Calculate the perimeter of the figure given below (in cm).

A
100 cm

B
90 cm

C
80 cm

D
60 cm

Solution

Perimeter will be = RS + ST + TU + UV + VW + WP + PQ + QR
Now, TU = PQ + VW = 13 + 9 = 22 cm
UV + WP = RS = 20 cm
QR = ST = 3 cm
Perimeter = 20 + 3 + 22 + 20 + 9 + 13 + 3 = 90 cm
Question 5
1 / -0

ABCD is a rectangle. AE is of AB and BC = 6 cm. Find the perimeter of the given rectangle.

A
30 cm

B
40 cm

C
50 cm

D
60 cm

Solution

Length of the rectangle AB = 3 AE = 3 x 3 = 9 cm
Breadth = 6 cm
Perimeter of the rectangle = 2(Length + Breadth) = 2(9 + 6) = 30 cm
Question 6
1 / -0

What is the area of the shaded portion in the figure given below?

A
20 cm²

B
22 cm²

C
30 cm²

D
44 cm²

Solution

There are 22 squares which are shaded.
Area of 1 square is 1 x 1 = 1 cm²So, total area = 1 x 22 = 22 cm²
Question 7
1 / -0

Area of a square is __________.

A
the sum of its all the sides

B
the product of its sides

C
Both (1) and (2)

D
None of these

Solution

Area of a square is the product of its sides.
Question 8
1 / -0

The perimeter of a rectangular table is 150 cm. If it is 15 cm broad, then find the length of the table.

A
75 cm

B
100 cm

C
60 cm

D
120 cm

Solution

Perimeter = 2(Length + Breadth)
150 = 2(15 + Length)
= 15 + Length
75 - 15 = Length
60 cm = Length
Question 9
1 / -0

Find the total shaded area of the figure made up of identical squares.

A
92 cm²

B
76 cm²

C
66 cm²

D
56 cm²

Solution

There are 23 shaded squares.
Area of 1 square = 4 cm²So, area of 23 squares = 23 × 4 = 92 cm²
Question 10
1 / -0

What is the perimeter of the given figure?

A
36 cm

B
34 cm

C
30 cm

D
20 cm

Solution

Perimeter = AB + BC + CD + DE + EF + FA
Now, AB = FE + CD = 8 + 2 = 10 cm
So, perimeter = 10 + 8 + 2 + 4 + 8 + 4 = 36 cm
Question 11
1 / -0

Find the perimeter of the given figure.

A
39 cm

B
55 cm

C
47 cm

D
40 cm

Solution

Perimeter of the given figure is OA + AB + BC + CD + DE + EF + FG + GH + HI + IO
So, (9 + 3 + 4 + 8 + 2 + 4 + 4 + 5 + 2 + 6 ) cm = 47 cm
Question 12
1 / -0

Find the total area of the shaded parts.

A
15 cm²

B
30 cm²

C
60 cm²

D
40 cm²

Solution

Area of rectangle GIEF = 8 x 5 = 40 cm²
Area of rectangle AHCB = 2 x10 = 20 cm²
Therefore, total area of shaded parts = 20 + 40 = 60 cm²
Question 13
1 / -0

Find the area of the small rectangle if its area is a quarter of the bigger one.

A
7.87 cm²

B
8 cm²

C
6 cm²

D
9.3 cm²

Solution

Area of big rectangle ABDC = 9 x 3.5 =31.5 cm²Now, the small rectangle is exactly one-fourth of the big rectangle ABDC.

So, = 7.87 cm²
Question 14
1 / -0

Find the total area of the shaded portions.

A
110 cm²

B
150 cm²

C
175 cm²

D
200 cm²

Solution

For area of bigger triangle :
Base of triangle = 5 cm + 5 cm + 2.5 cm = 12.5 cm
Height = 4 cm + 4 cm = 8 cm

Area =
=
= 12.5 x 4
= 50 cm²

For area of smaller triangle:
Base of triangle = 5 cm
Height of triangle = 8 cm

Area =
=

= 5 x 4 = 20 cm²

For area of rectangle:
Length = 5 cm + 5 cm = 10 cm
Width = 4 cm
Area = 10 x 4
= 40 cm²

Total area of the shaded portions:
50 + 20 + 40 = 110 cm²
Question 15
1 / -0

Which of the following shaded shapes has the least area?

Area of each square = 4 cm²

A
A

B
B

C
C

D
D

Solution

Area of shape 'A':
Shape 'A' has 5 squares.
So, area of 5 squares = 5 x 4 = 20 cm²

Area of shape 'B':
Shape 'B' has three squares.
So, area of 3 squares = 3 x 4 = 12 cm²

Area of shape 'C'
Shape 'C' has 4 squares.
So, area of 4 squares = 4 x 4 = 16 cm²

Area of shape 'D':
Shape 'D' has 4 squares.
So, area of 4 squares = 4 x 4 = 16 cm²

So, it is clear that shape B has the least area.
Question 16
1 / -0

Find the area of the shaded portion in the given diagram if A, B, C and D are mid-points of the shape and length of each side of the square is 10 cm.

A
50 cm²

B
70 cm²

C
80 cm²

D
100 cm²

Solution

The length of each side of the square is 10 cm and we have four shaded triangles in the given shape.

Area of 1^st triangle =
=
=
= 12.5 cm²

So, area of shaded region = area of 4 such triangles
= 4 x 12.5
= 50 cm²
Question 17
1 / -0

In the given figure PQRS is a square and TVUS is a rectangle. If TVUS is cut away from the square, find the area of the remaining portion.

A
296 m²

B
200 m²

C
324 m²

D
352 m²

Solution

Area of square PQRS = 18 x 18 = 324 m²Area of rectangle TVUS = 7 x 4 = 28 m²
So, total area remaining = 324 - 28 = 296 m²
Question 18
1 / -0

If the area of each small rectangle is 10 cm², then what will be the area of the shaded portion?

A
120 cm²

B
140 cm²

C
150 cm²

D
170 cm²

Solution

The area of the bigger rectangle = length × width
= 20 × 13
= 260 cm²
Area of each smaller rectangle = 10 cm²
There are 9 small rectangles in the bigger rectangle.
So, area of 9 small rectangles = 10 × 9 = 90 cm²
So, the remaining area = 260 - 90 = 170 cm²
Question 19
1 / -0

Find out the cost for painting the unshaded area if the cost for painting 6 cm² is Rs. 24.

A
Rs. 80

B
Rs. 96

C
Rs. 100

D
Rs. 120

Solution

Area of the unshaded portion = Area of inner rectangle - Area of square
Area of inner rectangle = length x width= 6 x 4 = 24 cm²

Area of square = (side)²
= (2)²
= 2 x 2
= 4 cm²
So, area of unshaded portion = 24 - 4 = 20 cm²

Cost for painting of 6 cm²= Rs. 24
Cost for painting area of 1 cm² = 24 ÷ 6 = Rs. 4
So, cost of painting an area of 20 cm² = 20 x 4 = Rs. 80
Question 20
1 / -0

The cost of cementing is Rs. 62 per 2 cm². What will be the cost of cementing the shaded portion?

A
Rs. 2785

B
Rs. 2800

C
Rs. 2852

D
Rs. 2866

Solution

Area of rectangle = 96 cm²
Area of square = 2 x 2 = 4 cm²Total shaded area = 96 - 4 = 92 cm²Cost of cementing 2 cm² = Rs. 62

Cost of cementing 1 cm² = = Rs. 31

Hence, total cost of cementing 92 cm² area = 92 x 31 = Rs. 2852
Question 21
1 / -0

Himanshu decided to buy a carpet for his room. How many square metres will be left in his room after placing the carpet?

A
156

B
210

C
66

D
195

Solution

Total floor area of room = 16 x 15 = 240
Area of the carpet = 5 x 6 =
Area left = 240 - 30 = 210
Question 22
1 / -0

The perimeter of a square hall is 32 m. Calculate the area of that hall.

A
61 m²

B
62 m²

C
64 m²

D
67 m²

Solution

Perimeter is 32 m.
We know it is a square, all the sides are same.
Side of the square is 8 m.
Area = 8 x 8
=64 m²
Question 23
1 / -0

A civil builder is measuring the height of a house with a bob. The height of each floor is 10 m and the thickness of roof top is 0.5 m. If the bob is 0.1 m long, how much thread does he need to measure the height of 4 floors?

A
43 m

B
42 m

C
41.9 m

D
43.9 m

Solution

After adding all of them and reducing the length of the bob, we get
= 4 x 10 + 4 x 0.5 - 0.1
= 40 + 2 - 0.1
= 41.9 m of thread
Question 24
1 / -0

The cost of ploughing a field at the rate of Rs. 8 /sq. m is Rs. 1920. If the width of the field is 15 m, what is the length?

A
29 m

B
25 m

C
20 m

D
16 m

Solution

Total cost = Rs. 1920
Cost of ploughing 1 sq. m = Rs. 8
Area = 1920 8 = 240 m²
A = L x W
240 = 15 x L
L = 240 ÷ 15 = 16 m
Question 25
1 / -0

Nonu is making a rectangular switchboard. If the length and the width of the switchboard are 8 inches and 6 inches respectively, then find its perimeter.

A
19 inches

B
23 inches

C
26 inches

D
28 inches

Solution

P = 8 + 6 + 8 + 6 = 28 inches
Question 26
1 / -0

Find the area of the shaded squares if area of each square is 2 cm² .

A
64 cm²

B
65 cm²

C
69 cm²

D
82 cm²

Solution

Number of shaded squares = 32
We know area of each square is 2 cm².So, area of all shaded squares = 32 x 2 = 64 cm²
Question 27
1 / -0

While playing Yakshab folded a carpet as shown below which is double the area of the floor of a room. Find the area of the floor.

A
72

B
28

C
32

D
36

Solution

Total Length = 5 + 3 = 8 m
Total breadth = 6 + 2 = 8 m
Total area of carpet = 8 x 8 = 64
Area of floor = m²
Question 28
1 / -0

The figure is made up of 9 triangles having total height of 9 m and total length of 12 m. Find the area of a small triangle.

A
5

B
6

C
7

D
8

Solution

There are three triangles which divide the height and base equally.
So, height of a small triangle = m
Breadth of a small triangle = m
Area of a small triangle = = = 6
Question 29
1 / -0

Which of the following figures has an area of 16 m², if the length of small triangle is 1 m and the height is 2 m? (Given that the size of smaller triangle is same in all figures)

A
A

B
C

C
B

D
D

Solution

Area of small triangle = = 1
Total number of triangles in figure B = 16
Area = 16
Question 30
1 / -0

Find the area of the blue shaded area.

A
25 m²

B
56 m²

C
66 m²

D
81 m²

Solution

Area of outer square = 81 m²
Area of inner square = 25 m²
Area of the blue shaded part = 81 - 25 = 56 m²

User

Question Analysis

Correct -
Wrong -
Skipped -

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Self Studies

Get latest Exam Updates
& Study Material Alerts!

No, Thanks

Click on Allow to receive notifications

Allow Notification

Self Studies

Self Studies

To enable notifications follow this 2 steps:

First Click on Secure Icon
Second click on the toggle icon

Allow Notification

Get latest Exam Updates & FREE Study Material Alerts!

Self Studies

×

Open Now

"Enter your details to claim your prize if you win!"

Send OTP

Processing, please wait...

I agree to the Terms of Use & Privacy Policy