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Perimeter & Area Test - 2

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Perimeter & Area Test - 2

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Question 1
1 / -0

A playground is in a shape of a rectangle with a rectangular path all around it. The dimensions of the playground, including the path, is 9 m x 4 m. The path is 0.6 m wide. What is the perimeter of the playground without the path?

A
11.2 m

B
19.2 m

C
20.4 m

D
21.2 m

Solution

Length of the playground = (9 - 0.6 x 2) m
= (9 - 1.2) m = 7.8 m
Breadth of the playground = (4 - (0.6 x 2) m
= (4 - 1.2) m = 2.8 m
So, perimeter of the playground = 2(Length + Breadth)
= 2(7.8 + 2.8)
= 2 x 10.6 m
= 21.2 m
Hence, option (4) is correct.
Question 2
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What is the area of the given figure?

A
23 square units

B
25 square units

C
26 square units

D
28 square units

Solution

As

Then,
Total number of shaded full squares = 24
Area of shaded full squares = 24 square units
Total number of shaded half squares = 4 = 2 full squares
Area of 4 half squares = Area of 2 full squares.
Therefore, total area of shaded squares = 24 + 2 = 26 square units
Hence, option (3) is correct.
Question 3
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In rectangle PQRS, the length of side PQ is 28 cm and of PS is 14 cm. The two circles are equal to each other. What is the area of the shaded portion?

A
60 cm²

B
72 cm²

C
84 cm²

D
96 cm²

Solution

From the above figure it can be seen that the radius is half of side PS.
Radius of circle = 7 cm
Area of circle = πr²
= 154 cm²
Area of shaded portion = Area of rectangle - Area of unshaded portion i.e Area of 2 circles
= (28 x 14) - (2 x 154)
= 392 - 308
= 84 cm²Hence, option (3) is correct.
Question 4
1 / -0

In the given figure, AJ = BC = HI = FG = 3 cm and JI = DC = HG = FE = 2 cm. Find the perimeter of the given figure.

A
30 cm

B
32 cm

C
34 cm

D
36 cm

Solution

Perimeter of the given figure = AJ + JI + IH + HG + GF + FE + ED + DC + CB + AB
= 3 cm + 2 cm + 3 cm + 2 cm + 3 cm + 2 cm + 6 cm + 2 + 3 cm + 10 cm
= 36 cm
Hence, option (4) is correct.
Question 5
1 / -0

MNOP is a square. SN is of MN. Find the perimeter of the square.

A
60 cm

B
62 cm

C
64 cm

D
66 cm

Solution

SN = of MN
8 cm = of MN
MN = 16 cm
Side of square = 16 cm
Perimeter of square = 16 x 4 = 64 cm
Hence, option (3) is correct.
Question 6
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Find area of the shaded portion.

A
24 cm²

B
25 cm²

C
27 cm²

D
29 cm²

Solution

Total number of shaded full squares = 6
Area of shaded full squares = 6 x 3 = 18 cm²

Total number of shaded half squares = 6 = 3 full squares
Area of shaded half squares = area of 3 full squares = 3 x 3 = 9 cm²

Total area of shaded figure = 18 + 9 = 27 cm²
Question 7
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Area of an equilateral triangle is

A
3 a²

B
4 a²

C

D

Solution

An equilateral triangle has all sides equal.

Area =
If side is a, then area =

Hence, option (3) is correct.
Question 8
1 / -0

The perimeter of an isosceles triangle ABC is 180 cm. Given, AB = BC. The length of side AC is 60 cm. Find the length side AB and BC of the triangle.

A
40 cm

B
50 cm

C
60 cm

D
80 cm

Solution

Perimeter of an isosceles triangle = 2a + b
where a and b are the lengths of the sides of isosceles triangle.
From the question we have,
180 cm = 2a + 60 cm
2a = 120 cm
a = cm
a = 60 cm

Hence, option (3) is correct.
Question 9
1 / -0

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

A
100 cm²

B
120 cm²

C
152 cm²

D
160 cm²

Solution

As,

Total number of shaded full squares = 72
Area of shaded full squares = 72 x 2 = 144 cm²

Total number of shaded half squares = 8 = 4 full squares
Area of 8 shaded half squares = area of 4 shaded full squares = 4 x 2 = 8 cm²
Total area of shaded figure = 144 + 8
= 152 cm²
Hence, option (3) is correct.
Question 10
1 / -0

Find the perimeter of the given figure.

A
20 cm

B
30 cm

C
40 cm

D
50 cm

Solution

The perimeter of the given figure
= 7 cm + 2 cm + 2 cm + 5 cm + 2 cm + 2 cm + 7 cm + 2 cm + 2 cm + 5 cm + 2 cm + 2 cm
= 40 cm
Hence, option (3) is correct.
Question 11
1 / -0

Find perimeter of the given shape.

A
71 cm

B
70 cm

C
65.5 cm

D
72.5 cm

Solution

Perimeter is the total distance around the shape.
So, in the above figure start from point B clockwise till you reach at point B through A.
Perimeter of the given shape
= 14 cm + 2 cm + 1 cm + 2 cm + 1.5 cm + 3 cm + 7 cm + 3 cm + 1 cm + 15 cm + 1 cm + 2.5 cm + 7 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1 cm + 2 cm + 1 cm + 2 cm
= 72.5 cm
Hence, option (4) is correct.
Question 12
1 / -0

Find area of the shaded portion.

A
80 cm²

B
110 cm²

C
120 cm²

D
150 cm²

Solution

The area of the outer rectangle = length x width
= 15 x 10
= 150 cm²
Area of unshaded triangle =
= = = 30 cm²
So, area of shaded portion = Area of outer rectangle - Area of unshaded triangle
= 150 - 30
= 120 cm^{2
Hence, option (3) is correct.}
Question 13
1 / -0

Find total area of the given compounded shape.

A
120 cm²

B
135 cm²

C
150 cm²

D
168 cm²

Solution

We have three rectangles in the given compounded shape as shown below.

Area of rectangle ABCD = length x width
= 7 cm x 2 cm
= 14 cm²
Area of rectangle A'B'C'D'
= length x width
= 7 cm x 2 cm
= 14 cm²
Area of rectangle PQRS
= length x width
Now, length of sides PS and QR
= 3 cm + 7 cm + 4 cm
= 14 cm
Length of sides PQ and RS
= 10 cm
Area = 14 x 10
= 140 cm²
Total area = 14 + 14 + 140 = 168 cm²
Hence, option (4) is correct.
Question 14
1 / -0

If ABCD is a square with side 4 cm, find the area of the shaded region.

A
16 cm²

B
8 cm²

C
6 cm²

D
4 cm²

Solution

Area of the square = (Side)²
Side = 4 cm
Area = 4 x 4 = 16 cm²
Area of region AOB =
= = 4 cm²
Question 15
1 / -0

If the area of one square is 8 sq. cm, find the area of the given shape.

A
104 sq. cm

B
100 sq. cm

C
98 sq. cm

D
96 sq. cm

Solution

Area of 1 square = 8 sq. cm
Area of 13 squares = 8 13 = 104 sq. cm
Question 16
1 / -0

What is the area of the shaded portion in the figure given below? (Area of 1 square = 1 sq. unit)

A
10 sq. units

B
8 sq. units

C
7 sq. units

D
4 sq. units

Solution

Number of half-filled squares = 6
Number of fully-filled squares = 4
Total area = 6 + 1 4 = 3 + 4 = 7 sq. units
Hence, (3) is the correct option.
Question 17
1 / -0

If triangle gets removed from the rectangle, then find the remaining area of the given figure.

A
125.41 cm²

B
140.32 cm²

C
160.25 cm²

D
187.54 cm²

Solution

Area of rectangle =

=

= 207.7 cm²
Area of triangle =
=

=

= 20.16 cm²

So, remaining area = 207.7 - 20.16

= 187.54 cm²Hence, option (4) is correct.
Question 18
1 / -0

What is the total distance covered by Ram if he runs 3 laps around a rectangular field of length 24 m and breadth 14 m?

A
248 m

B
228 m

C
76 m

D
114 m

Solution

Perimeter = 2( + b) = 2(24 + 14) m
= 2(38) m
= 76 m
Total distance covered = 3 × 76 m = 228 m
Question 19
1 / -0

An 8 m wide lawn is cultivated all along the outside of a rectangular plot measuring 80 m 50 m. The total area of the lawn is

A
2336 m²

B
6336 m²

C
2663 m²

D
6636 m²

Solution

Length of the plot including the lawn = (80 + 8 + 8) m = 96 m
Width of the plot including the lawn = (50 + 8 + 8) m = 66 m
Area of the lawn = (96 66) m² - (50 80) m²= 6336 m² - 4000 m²= 2336 m²
Question 20
1 / -0

The cost of painting an area of 4 cm²is Rs. 28. Find the cost of painting the shaded portion in the given diagram where a square of side 3 cm is cut out.

A
Rs. 1,000

B
Rs. 1,560

C
Rs. 2,124

D
Rs. 3,087

Solution

Area of rectangle = length x width
= 18 x 25
= 450 cm²
Area of square = side x side
= 3 x 3
= 9 cm²
Area of shaded portion = 450 - 9
= 441 cm²
Cost of painting 1 cm² =
= Rs. 7
Cost of painting the shaded portion = 441 x 7
= Rs. 3,087
Hence, option (4) is correct.
Question 21
1 / -0

Neetu needs to fence the boundary around his garden. His garden is highlighted in the given diagram. How much fence does he need?

A
46 m

B
50 m

C
51 m

D
59 m

Solution

To arrange a fence, he needs to find the perimeter of the given shape.
P = (12 + 4 + 3 + 4 + 15 + 8) m
= 46 m
Hence, option (1) is correct.
Question 22
1 / -0

Find the area of a square park whose perimeter is 512 m.

A
16,834 m²

B
16,384 m²

C
16,884 m²

D
18,634 m²

Solution

Perimeter = 512 m
Side =
Area = (128)²= 16,384 m²
Question 23
1 / -0

Tanu has a garden in regular pentagonal shape with side 7 m. She is trying to arrange a fence around her garden using a wire. For that, she fixed poles at each corner. How much fence will be required altogether, if 1 metre wire is required around every pole?

A
41 m

B
40 m

C
37 m

D
35 m

Solution

For outside boundary, we need perimeter of the shape and for perimeter, we add all the sides together.
P = 7 x 5 (Since all sides are equal)
= 35 m
Fence required = 35 m + 5 m (Extra for poles)
= 40 m
Hence, option (2) is correct.
Question 24
1 / -0

A room is in the form of a square with side 5 m. What is the cost of fitting marble on the floor of the room at the rate of $200 per square metre?

A
$5500

B
$550

C
$5000

D
$500

Solution

Area = 5 5 = 25 m²
Cost of fitting marble = 25 200
= $5000
Question 25
1 / -0

The area of a square field is the same as that of a rectangular field whose length and breadth measure 9 units and 4 units, respectively. What is the measure of each side of the square field?

A
4 units

B
7 units

C
6 units

D
12 units

Solution

Area of square field = (Side)²
Area of rectangular field = b
According to the question,
(Side)² = b = 9 4
(Side)² = 36 unit²
Or
Side = 6 units
Question 26
1 / -0

What is the estimated area of the shaded portion in the figure given below? (1 box = 1 sq. unit)

A
12 sq. units

B
10 sq. units

C
11 sq. units

D
14 sq. units

Solution

Number of fully-filled squares = 8
Number of half-filled squares = 4
Estimated area of the shaded portion = 8 1 + 4 = 8 + 2 = 10 sq. units
Hence, (2) is the correct option.
Question 27
1 / -0

Ritu is going from her home to a shop to buy cookies. Find the distance covered by her from point A to point E, if she passes through point F along the way.

A
29 m

B
25 m

C
21 m

D
20 m

Solution

Distance covered by Ritu from point A to point E, if she passes through point F along the way
= AF + FE
We know that AF = BC + DE
FE = AB + CD
AF + FE = 5 + 9 + 4 + 2 = 20 m
Hence, option (4) is correct.
Question 28
1 / -0

Given shape is made up of identical squares with side 6 cm. Find the perimeter of the shape.

A
72 cm

B
70 cm

C
68 cm

D
62 cm

Solution

Since the upper row is made up of 3 identical squares, of length 6 cm, the length of the side will be = 6 x 3 = 18 cm
Perimeter will be the total of the outside boundary.
= 18 x 4 (as bigger square has 4 sides of 18 cm length) = 72 cm
Hence, option (1) is correct.
Question 29
1 / -0

Which of the following shapes has 16 sq.units area and 20 units perimeter?
(Given:)

A

B

C

D

Solution

Given:

Shape 1:
A = 5 sq.units
P = 12 units

Shape 2:
A = 10 sq.units
P = 16 units

Shape 3:
A = 16 sq.units
P = 20 units

Shape 4:
A = 27 sq.units
P = 28 units
Hence, option (3) is correct.
Question 30
1 / -0

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

A
168 cm²

B
165 cm²

C
171 cm²

D
174 cm²

Solution

Area of the shaded portion = ((14 12) - (3 1)) cm² = (168 - 3) cm² = 165 cm²

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