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Perimeter & Area Test - 3

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Perimeter & Area Test - 3

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Question 1
1 / -0

Some tiles are laid in the centre of a rectangular garden that is 9 m by 6 m, leaving a border of 1.5 m around them. What is the perimeter of rectangle in which the tiles are laid?

A
18 m

B
16 m

C
14 m

D
12 m

Solution

Length of rectangle formed by joining tiles = (9 - 3) m = 6 m
Breadth of rectangle formed by joining tiles = (6 - 3) m = 3 m
So, perimeter of rectangle formed by tiles = 2(Length + Breadth)
= 2(6 + 3) = 18 m
Question 2
1 / -0

What is the area of the given figure (in sq.units)?

A
51 sq.units

B
52 sq.units

C
54 sq.units

D
56 sq.units

Solution

Number of squares = 28
Now, 1 square = 2 sq.units
Therefore, total area = 28 x 2 = 56 sq.units
Question 3
1 / -0

The figure is made up of 8 identical rectangles. Find the total area of the shaded figure (in m²).

A
240 m²

B
300 m²

C
350 m²

D
400 m²

Solution

Number of shaded triangles in the figure = 8
Height of shaded triangle = 20 m
Base of shaded triangle = 3 m

Area of 1 shaded triangle =× B × H
=× 3 × 20 = 30 m²
Total shaded area = 8 × 30 = 240 m²
Question 4
1 / -0

In the given figure, PQ = 8 m, PR = 1 m, RS = 3 m, UV = 2 m and QX = 15 m. Find the perimeter of the given figure (in m).

A
45 m

B
44 m

C
43 m

D
40 m

Solution

Perimeter of the given figure is the sum of all the sides.

= 8 + 1 + 3 + 3 + 3 + 2 + 8 + 15 = 43 m
Question 5
1 / -0

ABDC is a square and AB = 2AE. What is the perimeter of the square?

A
32 m

B
30 m

C
28 m

D
26 m

Solution

Length of AE = 4 m
According to the question,
AB = 2AE
So, AB = 8 m
Length of AB = 8 m
Perimeter of square
= 8 + 8 + 8 + 8 = 32 m
Question 6
1 / -0

What is the area of the shaded figure?
(Each square is of dimensions 1 m × 1 m)

A
62 m²

B
66 m²

C
72 m²

D
76 m²

Solution

Area of 1 small square = 1 × 1 = 1 m²
Number of squares which are fully shaded = 65
Area of squares which are fully shaded = 65 x 1 = 65 m²Number of half shaded squares = 2 = 1 fully shaded square
Total area of shaded figure = 65 + 1 = 66 m²
Question 7
1 / -0

Perimeter of any irregular shape can be determined by

A. adding the lengths of the sides of the shape
B. multiplying the lengths of the sides of the shape
C. dividing the figure into different types of shapes
D. Both A and C

A
A

B
B

C
C

D
D

Solution

Perimeter can be determined by both of the following ways:

1. Splitting the figure into many small figures
2. Adding the total length of outer boundary of the shape
Question 8
1 / -0

The perimeter of a rectangular garden is 180 m. It is 40 m wide. How long is it?

A
60 m

B
50 m

C
30 m

D
20 m

Solution

Perimeter = 180 m
Breadth = 40 m
Therefore, 2(Length + Breadth) = 180 m
Length + 40 = 90 m
Length = 90 - 40 = 50 m
Question 9
1 / -0

Find the total shaded area of the figure made up of identical squares, if dimensions of 1 square are 2 m × 2 m.

A
172 m²

B
184 m²

C
190 m²

D
195 m²

Solution

The above figure is divided into 2 parts namely 1 and 2.
Area of shaded figure = Area of figure 1 + Area of figure 2

Area of 1 small square = 2 × 2 = 4 m²Number of fully shaded squares in figure 2 = 38
Area of fully shaded squares in figure 2 = 38 × 4 = 152 m²

Area of figure 1 (triangle) = × base × height
= × 16 × 4 ( base = side of 8 squares = 2 x 8 = 16 m)
= 32 m²Area of shaded figure = 152 + 32 = 184 m²
Question 10
1 / -0

Find the perimeter of the given figure.

A
182 cm

B
172 cm

C
162 cm

D
150 cm

Solution

Perimeter of the shape
= 35 × 2 + 4 × 4 + 20 × 4 + 8 × 2
= 70 + 16 + 80 + 16
= 182 cm
Question 11
1 / -0

Find the perimeter of the shape. (All the dimensions in the figure are in cm)

A
22 cm

B
24 cm

C
30 cm

D
48 cm

Solution

Perimeter of the given figure: (A perimeter is a path that surrounds a two-dimensional shape)

= 2 + 6 + 3 + 5 + 3 + 3 = 22 cm
Question 12
1 / -0

Find the total area of the shaded parts of the rectangle BADC.

A
400 cm²

B
500 cm²

C
800 cm²

D
900 cm²

Solution

Area of triangle XCY =× 40 × 30 = 600 cm²
Area of rectangle EADY = 30 × 10 = 300 cm²

Total area of shaded parts of the figure = 600 + 300 = 900 cm²
Question 13
1 / -0

The given figure is made up of two rectangles. Find the total area.

A
12.5 m²

B
13.5 m²

C
14.5 m²

D
10.5 m²

Solution

Area of rectangle ABDC = 2 × 4 = 8 m²
Length of GF = 7 - 4 = 3 m
Area of rectangle GDEF = 3 × 1.5 = 4.5 m²
Therefore, total area = 8 + 4.5 = 12.5 m²
Question 14
1 / -0

Find the area of the shaded figure.
(Area of 1 square = 3 cm × 3 cm)

A
34 cm²

B
44 cm²

C
54 cm²

D
60 cm²

Solution

The 4 triangles in the above figure are named as 1, 2, 3 and 4.
Area of 1 small square = 3 × 3 = 9 cm²
Area of figure 1 = × base × height
= × 3 × 6 = 9 cm²

Area of figure 2 = × base × height
= × 6 × 6 = 18 cm²

Area of figure 3 = × base × height
= × 6 × 6 = 18 cm²

Area of figure 4 = × base × height
= × 6 × 3 = 9 cm²

Total area of shaded figure = 9 + 18 + 18 + 9 = 54 cm²
Question 15
1 / -0

Which of the following shapes in the grid has the smallest area?

(Area of 1 square = 4 sq.units)

A
B

B
A

C
C

D
D

Solution

Area of 1 square = 4 sq.units
Area of shape A = 12 x 4 = 48 sq.units
Area of shape B = 18 x 4 = 72 sq.units
Area of shape C = 16 x 4 = 64 sq.units
Area of shape D = 25 x 4 = 100 sq.units
So, shape A has the smallest area.
Question 16
1 / -0

From a rectangle, a part is cut out of the given dimensions. Find the area of the shaded figure.

A
100 m²

B
95 m²

C
80 m²

D
70 m²

Solution

Area of rectangle PQTV = 9 × 15 = 135 m²
Area of rectangle RSUV = 5 × 8 = 40 m²
Required area = 135 - 40 = 95 m²
Question 17
1 / -0

In the given figure, ABCD is a square and PQSR is a rectangle. Find the remaining area, if PQSR is removed from the figure.

A
375 cm²

B
445 cm²

C
464 cm²

D
475 cm²

Solution

Area of square = 25 × 25 = 625 cm²
Length of rectangle PQSR = 15 cm
Breadth of rectangle PQSR = 10 cm
So, area of rectangle PQSR = 15 × 10 = 150 cm²
Remaining area = 625 - 150 = 475 cm²
Question 18
1 / -0

The given figure is made up of 14 similar squares and 2 triangles. Hypotenuse of triangles is 4 cm. The perimeter of each square is 16 cm. What is the perimeter of the figure?

A
70 cm

B
72 cm

C
80 cm

D
88 cm

Solution

Perimeter of each square = 16 cm

4 × Side of square = 16 cm

Side of square = 16 4 = 4 cm

Now we know that a perimeter is a path that surrounds a two-dimensional shape, So the following figure can be drawn:

So, total perimeter of a figure = 20 × 4 = 80 cm
Question 19
1 / -0

Find the area of the shaded part in the figure.

A
454 m²

B
354 m²

C
300 m²

D
250 m²

Solution

Area of rectangle = 30 × 15 = 450 m²
Area of unshaded square = 4 × 4 = 16 m²
Area of unshaded rectangle = 20 × 4 = 80 m²
Total unshaded = 16 + 80 = 96 m²
Area of shaded part = 450 - 96 = 354 m²
Question 20
1 / -0

The shaded part in the given figure is covered with grass in a house. If the cost is Rs. 80 to put grass on 2 m², then find the total cost of putting grass.

A
Rs. 8,000

B
Rs. 6,000

C
Rs. 5,000

D
Rs. 4,000

Solution

Area of larger rectangle = 30 x 15 = 450 m²
Also, area of smaller rectangle = 250 m²
Area of shaded part = 450 - 250 = 200 m²
Now, cost of putting grass in 2 m²= Rs. 80
Cost of putting grass in 1 m²= Rs. 40
Cost in 200 m²= 200 x 40 = Rs. 8,000
Question 21
1 / -0

Mr. Sharma is going to make a car parking along the side of his house and decides to put tiles in the rest of the area. How much area will be left for the tiles?

A
325 m²

B
225 m²

C
200 m²

D
125 m²

Solution

Area of rectangular car parking = 20 × 10 = 200 m²
Total area = 35 × 15 = 525 m²
Area left for putting tiles = 525 - 200 = 325 m²
Question 22
1 / -0

Area of the floor of a rectangular classroom is 160 sq.m. What is its perimeter, if its length is 20 m?

A
45 m

B
50 m

C
55 m

D
56 m

Solution

Area of rectangular classroom = 160 sq.m
Length × Breadth = 160 sq.m
20 × Breadth = 160 sq.m
Breadth = 8 m
Therefore, perimeter = 2(Length + Breadth)
= 2(20 + 8)
= 2 × 28
= 56 m
Question 23
1 / -0

Four poles are stuck into a square ground of side 40 m at the four corners. A rope is to be put around the poles. What length of rope will be required if 2 m is required for tying the each knot?

A
164 m

B
168 m

C
169 m

D
178 m

Solution

Length of rope required = 40 + 40 + 40 + 40 + 2 + 2 + 2 + 2 = 168 m
Question 24
1 / -0

The cost of painting a rectangular wall is Rs. 11 per square metre and total cost to paint the complete rectangular wall is Rs. 1,177. If the length of wall is 15 m, then its breadth is

A
5.19 m

B
6.35 m

C
7.13 m

D
8.42 m

Solution

Total cost = Rs. 1,177
Cost of painting 1 sq.m = 11
Area of wall = 1177 11 = 107 sq.m
Length × Breadth = 107 sq.m
Breadth × 15 m = 107
Breadth = 107 15 = 7.13 m
Question 25
1 / -0

The design of a toy building is rectangular. A boy marked the figure to help him out with the perimeter of the figure. What is the perimeter of the rectangular building?

A
60 m

B
70 m

C
75 m

D
76 m

Solution

is one part

Length of the building = 10 parts

Breadth of the building = 9 parts

Perimeter of the building = 2(10 + 9) × 2

= 2 × 19 × 2 = 76 m
Question 26
1 / -0

A big square is made up of 121 small squares. After removing some small squares from the big square, the following figure is obtained. The length of a small square is 4 cm. Find the total area of all the squares removed (in cm²).

A
757 cm²

B
656 cm²

C
565 cm²

D
425 cm²

Solution

Area of 1 small square removed.
Area = 4 x 4 = 16 cm²
Number of squares removed = 41
Therefore, total area of squares removed = 41 x 16 = 656 cm²
Question 27
1 / -0

Two corners of a paper are folded as shown in figure. Find the area when it is unfolded.

A
4500 cm²

B
5000 cm²

C
5200 cm²

D
5600 cm²

Solution

Unfolded form of given paper is given below.

Area of paper = 80 × 70 = 5600 cm²
Question 28
1 / -0

The figure is made up of 16 squares of the same size. The area of the figure is 144 cm². Find the perimeter of the figure.

A
44 cm

B
54 cm

C
66 cm

D
78 cm

Solution

Area of figure = 144 cm²
Number of squares = 16
Area of 1 square = 144 / 16 = 9 cm²
Side of 1 small square = 3 cm
Perimeter of figure = 18 × 3 = 54 cm
Question 29
1 / -0

Which of the following figures has an area of 35 sq.m and a perimeter of 24 m?

A
(A)

B
(D)

C
(B)

D
(C)

Solution

Side of 1 square = 1 m
Area of 1 square = 1 m²

(A) Area of 21 squares = 21 m²
Perimeter = 2(7 + 3) = 20 m

(B) Area of 15 squares = 15 m²
Perimeter of 15 squares = 2(5 + 3) = 16 m

(C) Area of 14 squares = 14 m²
Perimeter of 14 squares = 2(7 + 2) = 18 m

(D) Area of 35 squares = 35 m²
Perimeter = 2(7 + 5) = 24 m
Question 30
1 / -0

The given figure shows the top of a building that has a swimming pool in centre with an area of 144 m². Find the area of the building not occupied by the swimming pool.

A
1356 sq.m

B
1456 sq.m

C
1556 sq.m

D
1600 sq.m

Solution

Area of rectangle AFKB = 30 x 40 = 1200 sq.m
Area of rectangle KEDC = (60 - 40) x 15
= 20 x 15 = 300 sq.m
Total area = 1200 + 300 = 1500 sq.m
Given, area of pool = 144 sq.m
Area not covered by pool = 1500 - 144 = 1356 sq.m

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