Mensuration Test

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Mensuration Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

How many times does the area of a square become if its side is doubled?

A
1/2 times the old one

B
4 times the old one

C
1/5 times the old one

D
8 times the old one

Solution

Let side = x, then area of the square = x²
If side is doubled;
New side of the square = 2x,
Then, new area of the square = 4x²Hence, the area becomes 4 times after the side of the square is doubled.
Question 2
1 / -0

If the area of one square is 5 cm², find the area of the given figure.

A
58 cm²

B
55 cm²

C
60 cm²

D
50 cm²

Solution

The area of one square is 5 cm².
There are 10 squares.
So, the total area will be 5 × 10 = 50 cm².
Question 3
1 / -0

If the area of a rectangle is 64 cm², then which of the following can be the length and width of that rectangle?

A
Length = 7 cm and width = 8 cm

B
Length = 16 cm and width = 4 cm

C
Length = 31 cm and width = 2 cm

D
Length = 8 cm and width = 10 cm

Solution

Putting the values as follows:
Length = 16 cm
Width = 4 cm
Area of the rectangle = 16 × 4 = 64 cm²
Question 4
1 / -0

Find the area of the given figure.

A
78.4 cm²

B
71.7 cm²

C
67.7 cm²

D
60.5 cm²

Solution

Area of the figure = Area of rectangle ABCD + Area of rectangle IJKL + Area of rectangle GEFH
(6 × 8.5) + (2 × (8.5 - 3 - 1.5)) + ((6 - 3.5 - 1) × 1)
= 51 + 8 + 1.5
= 60.5 cm²
Question 5
1 / -0

A glass sheet for a building measures 7 m 600 cm by 1 m 300 cm. Find the perimeter of the glass sheet.

A
17.8 m

B
21.2 m

C
28.9 m

D
30.9 m

Solution

Length of the glass sheet = 7.6 m
Width of the glass sheet = 1.3 m
Perimeter of the glass sheet = 2(l + w)
= 2(7.6 + 1.3)
= 2 × 8.9 = 17.8 m
Question 6
1 / -0

Find the perimeter of the shaded portion in the given figure.

A
22 units

B
24 units

C
28 units

D
30 units

Solution

Length of each side of a small square = 1 unit
Number of sides which are included in the perimeter of shaded portion = 22
Perimeter of the shaded portion = 22 x 1 = 22 units
Question 7
1 / -0

Find the perimeter of the given shape.

A
42 cm

B
24 cm

C
16 cm

D
34 cm

Solution

Perimeter of the shape = 8.5 + 8.5 + 6 + 6 + 1.5 + 1.5 + 1 + 1 = 34 cm
Question 8
1 / -0

The breadth of a rectangle is 68 cm and its length isof the breadth. Find the area of the rectangle.

A
1780 cm²

B
5780 cm²

C
5870 cm²

D
4590 cm

Solution

Breadth of the rectangle = 68 cm

Length of the rectangle =

Area of the rectangle = 68 × 85

= 5780 cm²
Question 9
1 / -0

If the perimeter of an isosceles triangle with the unequal side measuring 5 cm is 37 cm, find the length of the other two sides.

A
17.5 cm

B
17 cm

C
16 cm

D
6 cm

Solution

An isosceles triangle is a triangle that has two sides of equal length.
37 - 5 = 32 cm
Length of the other two sides = 32 ÷ 2 = 16 cm
Question 10
1 / -0

Which of the following figures has the largest perimeter?

(Note: Side of all the squares in the question is 1 unit)

A

B

C

D

Solution

Length of each side of a square = 1 unit
So, perimeter of the figure in option 1 = 10 units
Perimeter of the figure in option 2 = 12 units
Perimeter of the figure in option 3 = 12 units
Perimeter of the figure in option 4 = 14 units
Thus, the figure in option (4) has the largest perimeter.
Question 11
1 / -0

If the length of each side of an equilateral triangle is multiplied by 7, then the new perimeter is __ times the original perimeter.

A
7

B
3.5

C
5

D
14

Solution

Let the perimeter of the equilateral triangle be X.
The new perimeter will be 7X.
So, the new perimeter is 7 times the original perimeter.
Question 12
1 / -0

The area of a rectangular sheet is 425 cm² and its breadth is 17 cm. Which of the following options is 4 times the length of the rectangle?

A
50 cm

B
100 cm

C
150 cm

D
200 cm

Solution

Area = L × B (L = length and B = breadth)
B = 17
425 = L × 17
L =
L = 25 cm
4 times the length = 4 × 25 = 100 cm
Question 13
1 / -0

Find the missing length in the given figure.

A
14 cm

B
15 cm

C
16 cm

D
17 cm

Solution

Perimeter = 10 + 29 + 23 + ?
77 = 62 + ?
? = 15 cm
Question 14
1 / -0

If the perimeter of a square is numerically equal to the area of a rectangle and the breadth of the rectangle is equal to the side of the square, then what is the length of the rectangle?

A
1 unit

B
2 units

C
3 units

D
4 units

Solution

Perimeter of the square = Area of the rectangle
4 x Side = Length × Breadth
But, Breadth of the rectangle = Side of the square
So,
4 x Side = Length × Side
Length = 4 units
Question 15
1 / -0

A shape is formed by putting 9 squares together as shown in the figure below. If the length of each side is 11 cm, find the perimeter of the figure.

A
220 cm

B
215 cm

C
127 cm

D
137 cm

Solution

Perimeter of the figure = Sum of all sides
Total number of sides = 20
So, perimeter = 20 × 11 = 220 cm
Question 16
1 / -0

Find the ratio of the perimeter to the area of a square with side 12 cm.

A
1 : 3

B
3 : 1

C
2 : 3

D
4 : 5

Solution

Ratio of the perimeter to the area of the square = 4 × side/(side)² = (4 × 12)/(12 × 12) = 4/12 = 1 : 3
Question 17
1 / -0

There is a garden with length 80.5 m and width 70 m. Find the perimeter of the fence required for fencing the garden.

A
345 m

B
198 m

C
405 m

D
301 m

Solution

Perimeter = 2(Length + Breadth)
= 2(80.5 + 70)
= 2 × 150.5
= 301 m
So, required length = 301 m
Question 18
1 / -0

The lid of a square box of side 40 cm is sealed all round with the tape. What is the length of the tape required if the lid is sealed 4 times?

A
540 cm

B
640 cm

C
160 cm

D
120 cm

Solution

Length of tape required = 4(perimeter of the lid) = 4(4 × 40) = 640 cm
Question 19
1 / -0

A certain number of tiles, each of area 46 cm², are to be placed in a warehouse of area 69,000 cm². Find the number of tiles that can be placed in the warehouse.

A
1300

B
1450

C
1500

D
1880

Solution

Area of one tile = 46 cm²
Number of tiles that can placed in the warehouse with area 69,000 cm²
Question 20
1 / -0

Find the number of tiles, each measuring 25 cm by 6 cm, which can cover a wall of dimensions 12 m by 8 m.

A
6400

B
9800

C
8000

D
7200

Solution

Area of a tile = 25 × 6 = 150 cm²
Area of the wall (in cm) = 1200 × 800 = 9,60,000 cm²
Number of tiles required = 9,60,000 ÷ 150
= 6400
Question 21
1 / -0

Piyush runs twice around a rectangular field of length 18 m and breadth 16 m. Mohit runs thrice around a square field of side 24 m. Who covers more distance and by how much?

A
Piyush, 152 m

B
Mohit, 152 m

C
Piyush, 162 m

D
Mohit, 162 m

Solution

Side of the square field = 24 m

Perimeter of the square field = 4 × 24 = 96 m

Length of the rectangular field = 18 m

Breadth of the rectangular field = 16 m

Perimeter of the rectangular field = 2(18 + 16)

= 2(34) = 68 m

Distance covered by Piyush = 2 × 68 m = 136 m

Distance covered by Mohit = 3 × 96 m = 288 m

So, Mohit covers more distance than Piyush, the difference being 152 m.
Question 22
1 / -0

Manish wants to plough a rectangular field with the help of a tractor. The length of the rectangular field is 100 m and the breadth is 50.5 m. Find the total cost of ploughing the field if the cost of ploughing is Rs. 15 per metre.

A
Rs. 4,315

B
Rs. 4,415

C
Rs. 4,515

D
Rs. 4,615

Solution

Length of the rectangular field = 100 m
Breadth of the rectangular field = 50.5 m
Perimeter of the field = 2(100 m + 50.5 m) = 2(150.5) m = 301 m
Cost of ploughing = Rs. 15 per metre
Cost of ploughing 301 m = 301 15 = Rs. 4,515
Question 23
1 / -0

Seven square-shaped flower beds, each of side 1.5 m, are dug on a piece of land, 5.8 m long and 5.2 m wide. What is the area of the remaining part of the land?

A
13.31 m²

B
14.41 m²

C
15.51 m²

D
16.61 m²

Solution

Area of the land = 5.8 m 5.2 m = 30.16 m²

Area of one flower bed = 1.5 m 1.5 m = 2.25 m²

Area of 7 flower beds = 7 2.25 m² = 15.75 m^{2

So, area of the remaining part of the land = 30.16 m² - 15.75 m²}= 14.41 m²
Question 24
1 / -0

Find the distance covered by Mohan if he takes 5 rounds of an equilateral triangular park of side 110 m.

A
1550 m

B
1600 m

C
1650 m

D
1700 m

Solution

Length of each side of the park = 110 m
Perimeter of the equilateral triangular park = 3 × 110 m = 330 m
Distance covered by Mohan in 5 rounds = 5 × 330 m = 1650 m
Question 25
1 / -0

An athlete takes 18 rounds of a rectangular park, 35 m long and 25 m wide. The total distance covered by him is

A
2060 m

B
2110 m

C
2160 m

D
2210 m

Solution

Length of the rectangular park = 35 m

Breadth of the rectangular park = 25 m

Perimeter of the park = 2(35 + 25) = 2(60) = 120 m

So, distance travelled by the athlete in 18 rounds = 18 × 120 = 2160 m
Question 26
1 / -0

Which of the following figures has/have the maximum shaded area?

A
(a)

B
(b)

C
(c)

D
(d)

Solution

(a)
Area of A = 10 x 2 = 20 m²
Area of B = 6 x 1 = 6 m²
Area of the shaded region = 20 + 6 = 26 m²(b)

Area of A = 12 x 2 = 24 m²
Area of B = 7 x 1.5 = 10.5 m²
Area of C = 9 x 1 = 9 m²
Area of the shaded region = 24 + 9 + 10.5 = 43.5 m²(c)

Area of A = 9 x 2 = 18 m²
Area of B = 10 x 1.5 = 15 m²
Area of shaded region = 18 + 15 = 33 m²Figure (b) has the maximum shaded area.

Question 27

1 / -0

Directions: State T for true and F for false.

(i) The perimeter of a rectangle is four times its sides.
(ii) In a blackboard, the region available for writing is its area.
(iii) The distance covered along the boundary of a closed figure is called its perimeter.

	(i)	(ii)	(iii)
(A)	T	F	F
(B)	F	T	T
(C)	F	F	T
(D)	F	T	F

(A)

(B)

(C)

(D)

Solution

(i) The perimeter of a rectangle is four times its sides. FALSE; perimeter of a rectangle = 2(l + b)
(ii) In a blackboard, the region available for writing is its area. TRUE
(iii) The distance covered along the boundary of a closed figure is called its perimeter. TRUE; the perimeter of a closed figure is the distance covered in one round along the boundary of the figure.

Question 28
1 / -0

Rahul bent a rope to form the given figure (not drawn to scale). The figure is made up of 5 identical squares and 7 equilateral triangles. Find the length of the rope.

A
760 cm

B
1240 cm

C
940 cm

D
1040 cm

Solution

Side of a square = Side of an equilateral triangle = 80 4 = 20 cm
Perimeter of 1 square = 4 × 20 = 80 cm
Perimeter of 1 equilateral triangle = 3 × 20 = 60 cm

Length of the rope = Perimeter of the figure
Perimeter of the figure = 14 × (Length of one side of equilateral triangle) + 5 × (Length of one side of square) = 14 × 60 + 5 × 80 = 1240 cm

Question 29

1 / -0

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

Length of rectangle (cm)	Breadth of rectangle (cm)	Area (cm²)	Perimeter (cm)
25	P	400	Q
18	R	S	132

	P	Q	R	S
(A)	16	80	44	854
(B)	16	82	48	864
(C)	18	84	52	874
(D)	18	86	56	884

(A)

(B)

(C)

(D)

Solution

Area = 400 cm²
25 × P = 400
P = 16 cm
Perimeter = 2(16 + 25) = 82 cm
So, P = 16 cm, Q = 82 cm

Also, perimeter = 132
2(18 + R) = 132
18 + R = 66
R = 48 cm
Area = S = 48 × 18 = 864 cm²

Question 30
1 / -0

Figure P is made up of ten identical squares. Six squares are removed from figure P to form figure Q. If the perimeter of figure P is 304 cm, what will be the perimeter of figure Q?

A
170 cm

B
180 cm

C
190 cm

D
200 cm

Solution

Let the length of each side of a square be x.
Number of sides of squares which are included in the perimeter of figure P = 16
Perimeter of figure P = 16x = 304

Length of each side of a square will be:
x = 304/16 = 19 cm

Number of sides of squares which are included in the perimeter of figure Q = 10
Perimeter of figure Q = 10x = 10 × 19 = 190 cm

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Mensuration Test - 11

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Mensuration Test - 11

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SHARING IS CARING

Question 1

1/2 times the old one

4 times the old one

1/5 times the old one

8 times the old one

Solution

Question 2

58 cm2

55 cm2

60 cm2

50 cm2

Solution

Question 3

Length = 7 cm and width = 8 cm

Length = 16 cm and width = 4 cm

Length = 31 cm and width = 2 cm

Length = 8 cm and width = 10 cm

Solution

Question 4

78.4 cm2

71.7 cm2

67.7 cm2

60.5 cm2

Solution

Question 5

17.8 m

21.2 m

28.9 m

30.9 m

Solution

Question 6

22 units

24 units

28 units

30 units

Solution

Question 7

42 cm

24 cm

16 cm

34 cm

Solution

Question 8

1780 cm2

5780 cm2

5870 cm2

4590 cm

Solution

Question 9

17.5 cm

17 cm

16 cm

6 cm

Solution

Question 10

Solution

Question 11

7

3.5

5

14

Solution

Question 12

50 cm

100 cm

150 cm

200 cm

Solution

Question 13

14 cm

15 cm

16 cm

58 cm²

55 cm²

60 cm²

50 cm²

78.4 cm²

71.7 cm²

67.7 cm²

60.5 cm²

1780 cm²

5780 cm²

5870 cm²

13.31 m²

14.41 m²

15.51 m²

16.61 m²