Mensuration Test

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Mensuration Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

If area of a triangle is equal to area of a square and side of square is equal to height of triangle, then what will be the base of the triangle?

A
It will be double of the length

B
It will be half of the length of side of square.

C
It will be equal to the length of side of square.

D
It will be smaller than the length of side of square.

Solution

Area of triangle = or

Area of square = where is length of side of square.

Now, according to question, we get

=

= 2

b =

b = 2

So, base will be double of the length of side of square.
Question 2
1 / -0

Find out the area of the given shape, if the area of each square is 3 cm².

A
84 cm²

B
75 cm²

C
60 cm²

D
50 cm²

Solution

Area of one square = 3 cm²
There are 28 squares = 28 x 3
= 84 cm²
Question 3
1 / -0

If area of a parallelogram is 24 cm², then which of the following pairs of base and height would not give area of the parallelogram = 24 cm²?

A
Base = 6 cm, Height = 4 cm

B
Base = 8 cm, Height = 3 cm

C
Base = 12 cm, Height = 2 cm

D
Base = 48 cm, Height = 2 cm

Solution

Area of parallelogram = base x height
So, 6 cm x 4 cm = 24 cm²
8 cm x 3 cm = 24 cm²
12 cm x 2 cm =24 cm²
48 cm × 2 cm = 96 cm² which is not correct (Because answer for all is 24), but 48 × 2 = 96
So, statement (4) is not correct and it does not represent correct area of parallelogram i.e 24 cm².
Question 4
1 / -0

Find the area of the given shape.

(Note: It is symmetrical along the line AB)

A
150.2 cm²

B
165.4 cm²

C
186.5 cm²

D
200.3 cm²

Solution

So, we have 5 rectangles in the given figure.
The opposite sides of a rectangle are equal.
Area of 1^st rectangle =
= 15 x 3
= 45 cm²
Area of 2^nd rectangle = 17 x 5
= 85 cm²
Area of 3^rd rectangle
= 2 x 15
= 30 cm²
Area of 4^th rectangle = 19 x 1
= 19 cm²
Area of 5^th rectangle
= 0.5 x 15
= 7.5 cm²
Total area = 45 + 85 + 30 + 19 + 7.5
= 186.5 cm²
Question 5
1 / -0

A table top measures 6 m 14 cm and 86 cm. Find the area of the table top.

A
52704 cm²

B
52804 cm²

C
528334 cm²

D
52844 cm²

Solution

Length = 6 m 14 cm = 614 cm
Breadth = 86 cm
Therefore, area = 614 x 86
= 52804 cm²
Question 6
1 / -0

Find the perimeter of the shaded region of the given shape.(side of each square is 1 unit)

A
16 units

B
17 units

C
18 units

D
19 units

Solution

Perimeter is sum of length of boundary of a figure.
There are 18 faces on boundary of this figure and length of each face is 1 unit.
Perimeter of the shaded region = 18 x 1 = 18 units
Question 7
1 / -0

Find the perimeter of the given shape.

A
50 cm

B
54 cm

C
58 cm

D
70 cm

Solution

Side GF + Side AB = Side DE + Side CI
4 + 10 = ? + 4
? = 10 (Side DE = 10 cm)
Perimeter of figure is sum of all sides at boundary of a figure.
= 10 cm + 12 cm + 4 cm + 3 cm + 10 cm + 6 cm + 4 cm + 9 cm = 58 cm
Question 8
1 / -0

If breadth of a rectangle is of its length and its area is 2904 cm², then find its perimeter.

A
200 cm

B
205 cm

C
220 cm

D
235 cm

Solution

Let the length be L cm.
Area =
2904 = ⁽⁾
2904 x =
L = 66 cm
Therefore, breadth = = 44 cm
Perimeter = 2(L + B)
= 2(66 + 44)
= 220 cm
Question 9
1 / -0

If area of a triangle is 16 cm² and its height is 8 cm, then find the base of the triangle.

A
2 cm

B
3 cm

C
4 cm

D
5 cm

Solution

Area of triangle = x base x height or x b x h
16 =
32 =

b = 4
So, base is 4 cm.
Question 10
1 / -0

Which of the following figures has the largest perimeter? (Given: Area of each small square is same)

A

B

C

D

Solution

No of faces on boundary of Figure 1 = 27
So perimeter = 27 units
No of faces on boundary of Figure 2 = 25
So perimeter = 25 units

No of faces on boundary of Figure 3 = 26
So perimeter = 26 units

No of faces on boundary of Figure 4 = 26
So perimeter = 26 units

Figure in option (1) has the most number of faces on boundary, so its perimeter is the largest (27 units).
Question 11
1 / -0

If length of a rectangle is doubled and its width is halved, then what will happen to its area?

A
It will get doubled.

B
It will get halved.

C
It will decrease by 75%.

D
It will remain the same.

Solution

Let the length of rectangle be L cm and its width be W cm.
Therefore, area of the original rectangle = length x width
= L x W
If length is doubled and width is halved, then area of the new rectangle = 2L x
= L x W
Therefore, there will be no change in the area of the rectangle.
Question 12
1 / -0

The perimeter of a rectangular swimming pool is 56 m. If the width of the swimming pool is 12.5 m, then what is the double of the area of swimming pool?

A
387.5 m²

B
387.25 m²

C
387 m²

D
386.5 m²

Solution

Perimeter of swimming pool = 56 m
Width = 12.5 m
Perimeter =
56 = 2 × length (L) + 2 × 12.5
56 = 2 × L + 25
2 × L = 56 - 25
L = 31 ÷ 2
Length = 15.5 m
Area of the swimming pool = (15.5 × 12.5) m²= 193.75 m²
Double the area of the swimming pool = 193.75 + 193.75
= 387.5 m²
Question 13
1 / -0

Find the missing side of the given parallelogram, if its perimeter is 38 cm.

A
13 cm

B
12 cm

C
11 cm

D
14 cm

Solution

The given shape is a parallelogram. The opposite sides of a parallelogram are the same.
Perimeter = 2 x length + 2 x width
38 = 2 x 8 + 2 x width
38 = 16 + 2 x width
Width =
Width =
Width = 11 cm
Question 14
1 / -0

Perimeter of a square is double of its area. What is the measure of side of the square(in cm)?

A
8

B
6

C
4

D
2

Solution

Perimeter of square = 2(Area of square)
Now, let the side be a.

On solving this, we will get,

a = 2
side of square = 2 cm
Question 15
1 / -0

Directions: Read the given information carefully and choose the correct answer.

The following figure is made using two similar rectangles. Find the perimeter of the given shape formed by two rectangles, if the length and breadth of one rectangle is 16 cm and 8 cm.

A
64 cm

B
59 cm

C
49 cm

D
32 cm

Solution

Length of one rectangle = 16 cm
Length of the whole shape = 16 cm
Breadth of one rectangle = 8 cm
Breadth of the whole shape = 8 + 8 = 16 cm
Perimeter of rectangle = 2 (16 + 16) = 64 cm
Question 16
1 / -0

There are 96 square tiles which make up the floor of the hall in a wedding palace. If the perimeter of each tile is 8 m, what is the area of the floor of the hall?

A
356 m²

B
372 m²

C
384 m²

D
392 m²

Solution

Perimeter of each square tile = side × number of sides
8 = side × 4
8 ÷ 4 = side
side = 2 m
Area of each tile = side²
= 2²
= 4 m²
Area of 96 tiles on the floor = 4 × 96
= 384 m²
Question 17
1 / -0

There is a rectangular park with width 45.5 m and length 65.25 m. A fencing is to be put around it. What length of fencing is required to surround the whole park, if there is an opening of 2.5 m for the gate?

A
212 m

B
216 m

C
219 m

D
221.5 m

Solution

Perimeter of park = 2(length + width)
= 2(65.25 + 45.5)
= 2(110.75)
= 221.5 m

But, there is an opening of 2.5 m for the gate.
Therefore, fencing required to surround the park = Perimeter of the park - Length of the gate
= 221.5 - 2.5
= 219 m
Question 18
1 / -0

The area of a square shaped swimming pool is 225 m². What is the perimeter of the swimming pool?

A
60 m

B
56 m

C
48 m

D
36 m

Solution

Area of swimming pool = 225 m²
We know that area of square = Side²
Area of swimming pool = Side²
225 = Side²
Side =
Side = 15 m
Perimeter of the swimming pool = 4 x length of side
= 4 x 15
= 60 m
Question 19
1 / -0

A sheet of tin with dimensions 165 cm by 396 cm is to be cut, such that smaller sheets of dimensions 36 cm by 24 cm are obtained to make small cans. How many smaller sheets of tin can be obtained?

A
74

B
75

C
76

D
77

Solution

Area of the big sheet of tin =
= 65340 cm²Area of the smaller sheets of tin =
= 864 cm²
Number of smaller sheets of tin that can be obtained =
= 75, remainder 540

Therefore, 75 smaller sheets of tin can be obtained from the bigger sheet.
Question 20
1 / -0

How many sheets of paper of dimensions 9 cm by 8 cm are required to fully cover the surface of a table measuring 72 cm by 36 cm?

A
24

B
36

C
48

D
60

Solution

Area of the table =
= 2592 cm²
Area of the sheet = 9 x 8
= 72 cm²
Number of sheets required to fully cover the table =
= 36
Question 21
1 / -0

There is a square garden with a pathway around it. The length of a side of the square garden is 40 m and the path is 3 m in width. Find the area of the pathway.

A
230 m²

B
249 m²

C
250 m²

D
263 m²

Solution

Area of square = Length x Length

Side of square with pathway = 40 + 3 = 43 m

Side of square = 40 m

Area of square with sideway = 43 x 43 = 1849 m²

Area of square without pathway = 40 x 40 = 1600 m²

Area of pathway = 1849 - 1600 = 249 m²
Question 22
1 / -0

A person wants to paint a rectangular wall in his house. Length of the wall is 15 m and width is 8 m. Find the cost of painting the wall at the rate of Rs. 50 per sq.m.

A
Rs. 6,000

B
Rs. 6,012

C
Rs. 6,015

D
Rs. 6,020

Solution

Length = 15 m
Width = 8 m
Area of wall = 15 × 8 = 120 m²
Cost = 120 × 50 = Rs. 6,000
Question 23
1 / -0

There is a rectangular floor made from square tiles. Side of one tile is 2 m. Length of the floor is 12 m and width is 10 m. There were 10 tiles set up on the floor. How much area (in sq.m) is left in which tiles have yet to be set up?

A
20

B
40

C
80

D
100

Solution

Area of one tile = Side x Side
= 2 x 2 = 4 sq.m
Area covered by 10 tiles = 40 sq.m
Area of floor = Length x Breadth
= 12 x 10 = 120 sq.m
Area of floor left to set up = 120 - 40
= 80 sq.m
Question 24
1 / -0

There is a track in shape of a circle with radius of 20 m. How much distance is covered by Hari, if he runs 5 laps around it?

A
628 m

B
620 m

C
615 m

D
610 m

Solution

Radius = 20 m
Circumference = 2πr
= 2 x 3.14 x 20
= 125.6 m
Distance covered = 125.6 x 5
= 628 m
Question 25
1 / -0

Jaggi goes for a walk every Thursday. He walks 12 times around a rectangular park of length 40 m and breadth 15 m. How much distance does he cover?

A
1320 m

B
1040 m

C
950 m

D
860 m

Solution

Length = 40 m
Width = 15 m
Perimeter of park = 2(Length + Breadth)
= 2(40 + 15) = 110 m
When he walks around it 12 times, distance covered = 12 x 110
= 1320 m
Question 26
1 / -0

Which of following figures has the minimum shaded area?

A
A

B
C

C
B

D
All have the same area

Solution

Figure (A)
Area = (10 x 2) + (8 x 2)
= 20 + 16
= 36 sq. m

Figure (B)
Area = (7 x 2) + (7 x 2) + (4 x 2)
= 14 + 14 + 8
= 36 sq. m

Figure (C)
Area = (8 x 7) - (6 x 5)
= 56 - 30
= 26 sq. m

So we can say that Figure (C) has minimum shaded area.
Question 27
1 / -0

Directions: State 'T' for true and 'F' for false.

(i) Perimeter of a square and area of a square with side 2 cm is equal.

(ii) Circumference of circle is the measure of its boundary.

(iii) Area of a rectangle with length 8 cm and width 4 cm is twice the area of a square with side 4 cm.

A

(i) (ii) (iii)

T F T

B

(i) (ii) (iii)

T F F

C

(i) (ii) (iii)

F T T

D

(i) (ii) (iii)

T T T

Solution

(i) Perimeter of square with side 2 cm = 4 x length of side = 4 x 2 = 8 cm
Area of square = side² = 2 x 2 = 4 sq. cm (False)

(ii) Circumference is the measure of outer boundary. (True)

(iii) Area of rectangle with length 8 cm and width 4 cm = 8 x 4 = 32 sq. cm
Area of square with side 4 cm = 4 x 4 = 16 sq. cm
So, area of rectangle is twice the area of square. (True)
Question 28
1 / -0

Find the area of given figure if all the circles are equal to each other having radius 4 cm.

A
200.96 sq. cm

B
214.23 sq. cm

C
220.12 sq. cm

D
225.25 sq. cm

Solution

Radius of circles = 4 cm
Area of 1 circle = π x r²
= 3.14 x 4 x 4
= 50.24 sq. cm
Area of whole figure= 50.24 x 4 = 200.96 sq. cm

Question 29

1 / -0

Directions: Find the values of P, Q, R and S.

Radius of circle (cm)	Area (cm²)	Circumference (cm)
12	P	Q
15	R	S

P Q R S

452.16 75.36 706.5 94.2

P Q R S

200.22 75.36 706.5 94.2

P Q R S

452.16 80.14 706.5 94.2

P Q R S

452.16 75.36 706.5 100.5

Solution

Circumference of circle = 2πr
Area of circle = π x r²
P = 3.14 x 12 x 12 = 452.16 cm²Q = 2 x 3.14 x 12 = 75.36 cm
R = 3.14 x 15 x 15 = 706.5 cm²
S = 2 x 3.14 x 15 = 94.2 cm

Question 30
1 / -0

Figure (a) is made up of ten identical squares. Four squares were added to figure (a) to form figure (b). The area of figure (a) is 810 cm². What is the area of figure (b)?

A
1134 cm²

B
1144 cm²

C
2134 cm²

D
2234 cm²

Solution

Let the side of each square be x.
So, area of Figure (a) = 10 x x²
x² x 10 = 810
x² = 81
x = 9 cm
Therefore, area of Figure b = 14 x x²B = 14 x 81 = 1134 cm²

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(i)	(ii)	(iii)
T	F	T

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T	F	F

(i)	(ii)	(iii)
F	T	T

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T	T	T

Mensuration Test - 12

Result

Mensuration Test - 12

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Rank

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SHARING IS CARING

Question 1

It will be double of the length

It will be half of the length of side of square.

It will be equal to the length of side of square.

It will be smaller than the length of side of square.

Solution

Question 2

84 cm2

75 cm2

60 cm2

50 cm2

Solution

Question 3

Base = 6 cm, Height = 4 cm

Base = 8 cm, Height = 3 cm

Base = 12 cm, Height = 2 cm

Base = 48 cm, Height = 2 cm

Solution

Question 4

150.2 cm2

165.4 cm2

186.5 cm2

200.3 cm2

Solution

Question 5

52704 cm2

52804 cm2

528334 cm2

52844 cm2

Solution

Question 6

16 units

17 units

18 units

19 units

Solution

Question 7

50 cm

54 cm

58 cm

70 cm

Solution

Question 8

200 cm

205 cm

220 cm

235 cm

Solution

Question 9

2 cm

3 cm

4 cm

5 cm

Solution

Question 10

Solution

Question 11

It will get doubled.

It will get halved.

It will decrease by 75%.

It will remain the same.

Solution

Question 12

387.5 m2

387.25 m2

387 m2

386.5 m2

Solution

Question 13

13 cm

12 cm

11 cm

84 cm²

75 cm²

60 cm²

50 cm²

150.2 cm²

165.4 cm²

186.5 cm²

200.3 cm²

52704 cm²

52804 cm²

528334 cm²

52844 cm²

387.5 m²

387.25 m²

387 m²

386.5 m²

356 m²

372 m²

384 m²

392 m²

230 m²

249 m²

250 m²

263 m²