Mensuration Test

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Mensuration Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following options represents the area of the pentagonal park as shown in the figure?

A
800 m²

B
700 m²

C
600 m²

D
500 m²

Solution

Given: BC = 20 m
CD = 20 m
DE = 20 m
Height of the pentagon = 40 m
Draw a line from B to E.
We will find the sum of the areas of ABE and square BCDE to find the area of pentagon ABCDE.
Area ABE = = Now, AG = AF + FG
FG = BC = ED = 20 m
AF = 40 - 20 = 20 m
Base = BE = CD = 20 m
Area of ABE = = 200 m²Area of square BCDE = (20 20) = 400 m²
Total area of the pentagon = 200 m² + 400 m²= 600 m²
Question 2
1 / -0

If the area of a rectangle is 60 cm² and its perimeter is 34 cm, then the length of the diagonal is

A
17 cm

B
11 cm

C
15 cm

D
13 cm

Solution

We know that,
Area of rectangle, A = l x w (where l is length and w is width)

Also, diagonal of rectangle, d =
But as we do not know the length and width, we can work with another formula to find diagonal, i.e
(where P is Perimeter of rectangle)

So, d = cm

d = 13 cm

Hence, answer option 4 is correct.
Question 3
1 / -0

Two circular pieces of diameters 2 cm and 3 cm are cut from a rectangular sheet of length 6 cm and width 3 cm. Find the remaining area (approximate) of the rectangular sheet.

A
7.80 cm²

B
4.60 cm²

C
5.90 cm²

D
6.90 cm²

Solution

Area of the rectangular sheet = 6 cm × 3 cm = 18 cm²
Area of first circular piece = = (1)²cm² = 3.14 cm²
Area of second circular piece = (1.5)² cm² = 7.065 cm²
Total area of the circular pieces = (7.065 + 3.14 cm²) = 10.205 cm²
Remaining area = (18 - 10.205) cm² = 7.795 cm² 7.80cm²
Question 4
1 / -0

A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood.

A
53,000 cm³

B
82,125 cm³

C
84,000 cm³

D
85,000 cm³

Solution

Thickness of the wood = 2.5 cm
Now, external dimensions of the wooden box are :
Length = 120 cm, width = 80 cm & height = 40 cm
Volume of wood = (120 80 40) - (115 75 35) = 82125 cm³
Answer: (2)
Question 5
1 / -0

The ratio of the radii of two cylinders is : 1 and their heights are in corresponding ratio of 2 : 3. The ratio of their volumes must be

A
1 : 2

B
2 : 1

C
4 : 9

D
5 : 9

Solution

Let the radius and height of 1^st cylinder be r units and 2h units, respectively.
Thus, radius and height of 2^nd cylinder are r units and 3h units, respectively.
Required ratio is (r)² × 2h : r² × 3h = 6 : 3 = 2 : 1.
Question 6
1 / -0

Find the sum of two opposite sides of a parallelogram whose height is 25 cm and area is 15 m².

A
10000 cm

B
11500 cm

C
11800 cm

D
12000 cm

Solution

Height = 25 cm
Area of the parallelogram = 15 m² = 150000 cm²
Area of the parallelogram = Base × Height
150000 = Base × 25
Base = 6000 cm
Now, sum of parallel sides = 6000 + 6000 = 12000 cm
Question 7
1 / -0

What is the area of the shaded region if the side of the square ABCD is 14 cm?

A
112 cm²

B
196 cm²

C
308 cm²

D
None of these

Solution

In the given figure:

Diameter of the circle = Diagonal of the square

Diagonal of the square = = = Diameter of the circle

Radius of the circle = cm

Area of the shaded region = Area of the circle - Area of the square

= = 14 × 22 - 14 × 14

= 14 × 8 = 112 cm²
Question 8
1 / -0

In the figure below, ABC is a right-angled triangle and CD is perpendicular to AB. If AC = 13 cm, DC = 5 cm and AD = DB, find the area of triangle ABC.

A
90 cm²

B
84.5 cm²

C
82 cm²

D
75 cm²

Solution

According to the question, AC = 13 cm and DC = 5 cm.
So, AC² = AD² + CD²
13² = AD² + 5²
AD = 12 cm
AD = BD
So, BC² = BD² + CD²
Then, BC² = 144 + 25
BC = 13 cm
Area of ABC = × 13 × 13 = 84.5 cm²
Question 9
1 / -0

Four ice cubes of sides 11 cm, 12 cm 13 cm and 14 cm are melted and then frozen together again to form one big ice cube. Find the side of the new ice cube formed.

A
17 cm

B
20 cm

C
40 cm

D
50 cm

Solution

Let the side of the new cube be S.
According to the question,
Total volume of all the cubes = Volume of the new cube
11³ + 12³ + 13³ + 14³= S³
1331 + 1728 + 2197 + 2744 = 8000 cm³ = S³
S = 20 cm
Question 10
1 / -0

The volume of a cylinder is 6280 cm³ and the area of its base is 314 cm². Find the height of the cylinder. (Note: Use π = 3.14)

A
17 cm

B
20 cm

C
22 cm

D
23 cm

Solution

Volume of the cylinder = πr²h = 6280 cm³
Area of the base = πr²= 314 cm²
So, the height of the cylinder will be:
h = = = 20 cm
Question 11
1 / -0

Find the number of rods that can be made from 0.66 m³ of iron, each having length 21 m and radius 2 cm.

A
25

B
40

C
30

D
20

Solution

Volume of 1 rod =

Number of rods = = rods
Question 12
1 / -0

A cylindrical tube, open at both the ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal in the cylinder.

A
700 cu. cm

B
704 cu. cm

C
710 cu. cm

D
720 cu. cm

Solution

Internal radius of cylinder (r) = Diameter/2, i.e. 5.2 cm
External radius (R) = 5.2 cm + 0.8 cm = 6 cm (Since 8 mm = 0.8 cm)
Volume of metal = × (R²-r²) × Length of tube
= × (6 × 6 - 5.2 × 5.2) × 25 cm³= 704 cm³
Question 13
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 14
1 / -0

A cubical box of dimension 26 cm is constructed from a metal of thickness 2 cm. Find out the amount of metal used (in cm³).

A
7589

B
6928

C
6845

D
6752

Solution

Thickness of the cubical box = 2 cm
Internal dimension of the box = 26 cm – [2 + 2] cm = 22 cm
Internal volume of the box = 22 × 22 × 22 = 10,648 cm³
External volume of the box = 26 × 26 × 26 = 17,576 cm³
Volume of the metal used = External volume – Internal volume = 17,576 – 10,648 = 6928 cm³
Question 15
1 / -0

A ditch 8 m long, 4.5 m wide and 2.5 m deep is dug in a field 36 m long and 11 m wide; the earth so dug is spread out evenly on the remaining field. Find the rise in the level of the field.

A
25 cm

B
4 cm

C
0.25 cm

D
20 cm

Solution

Volume of earth dug out = (8 4.5 2.5) m³
Area over which earth is spread = (36 11) - (8 4.5) m²
Rise in level = = ^{m = 0.25 m = 25 cm}
Question 16
1 / -0

A big ice cube's side measures 50 cm. Calculate its weight (in kg), if ice weighs of the weight of the same volume of water and 1 cm³ of water weighs 1 gm.

A
78.6 kg

B
87.5 kg

C
89.5 kg

D
90 kg

Solution

Volume of the ice cube = [50 × 50 × 50] = 1,25,000 cm³
Now, weight of 1 cm³ of water = 1 gm
Weight of 1 cm³ of ice = weight of 1 cm³ of water
So, weight of 1 cm³ of ice = gm
Weight of the ice cube = × 1,25,000 = 87,500 gm = 87.5 kg
Question 17
1 / -0

A cylindrical metal tank whose diameter is 14 dm can hold 2,310 litres of diesel. Taking a litre of diesel to occupy 832 cubic cm, what will be the height of the tank?

A
12.48 dm

B
15.82 dm

C
76.19 dm

D
98.17 dm

Solution

1 dm = 10 cm
r = = 7 dm = 70 cm
litres

h =
h =
Question 18
1 / -0

A rectangular park is 20 m long and 10 m wide. It has a 2 m wide path along the boundary. Find the area of the path.

A
104 m²

B
126 m²

C
150 m²

D
175 m²

Solution

Area of the park = 20 × 10 = 200 m²
Now, length of the park excluding the path = 20 – [2 + 2] = 16
Width = 10 - [2 + 2] = 6
So, area excluding the path = 16 × 6 = 96 m²
So, area of the path = 200 – 96 = 104 m²
Question 19
1 / -0

A basketball court is 15 m long and 8 m wide. What will be the total cost of flooring the court with wood at the rate of Rs. 2000 per square metre?

A
Rs. 2400

B
Rs. 2,40,000

C
Rs. 24,000

D
Rs. 24,00,000

Solution

Area of the court = (15 8) m² = 120 m²
Total cost = 120 2000 = Rs. 2,40,000
Question 20
1 / -0

What is the curved surface area of the solid obtained when two cylinders of height 6 cm each and radii 3 cm and 1 cm are joined base to base?

A

B

C

D

Solution

Curved surface area of the solid = where, r₁ = radius of 1^st cylinder
r₂ = radius of 2^nd cylinder
h = height of each cylinder
Curved surface area of the solid
= = 2 6 cm² = cm²
Question 21
1 / -0

The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height of the room are halved, and length of the room is doubled, then what will be the change in area of the four walls of the room as compared to the original area?

A
It will decrease by 13.64%.

B
It will decrease by 15%.

C
It will decrease by 18.75%.

D
It will decrease by 30%.

Solution

Let the initial dimensions be 3x, 2x and 1x.
After applying the given conditions, final dimensions are 6x, 1x and x.
Initial area of 4 walls = 2(2x² +3x²)
= 10x² sq. units
Final area of 4 walls = 2(3x²) + 2 (0.5x²)
= 7x²
Now, the area decreases and the percentage decrease = = 30%
Question 22
1 / -0

Find the area of the given shape.

A
1150 cm²

B
1256 cm²

C
1560 cm²

D
1250 cm²

Solution

Here, GF + FH = 25 + 15 = 40 cm
Then, GF + FI + IH = 40 cm
KJ = GI = 35 cm
Then, IH = 5 cm
Also, DC + CE = GF = 25 cm
So, DC = 10 cm

Then, Area of the shape = Area of ABCD + Area of BCE + Area of DEFG + Area of EFH + Area of GIJK + Area of HIJ

= [15 × 10] + [ × 15 × 15] + [25 × 10] + [ × 15 × 10] + [15 × 35] + [ × 15 × 5]

= 150 + 112.5 + 250 + 75 + 525 + 37.5 = 1150 cm²

Question 23

1 / -0

Fill in the blanks with the help of the table given and mark the correct option.

(i) The area of a rectangle becomes ___ times its original area if its length and width are doubled.
(ii) The curved surface area of a cylinder becomes _________ if the height is doubled.
(iii) All the faces of a cube are ______ in area.
(iv) If the height (h) of a cylinder is equal to the radius (r) of its base, then the total surface area of the cylinder, in terms of r, is _______.

	(i)	(ii)	(iii)	(iv)
A	2	triple	unequal	4πrh
B	4	double	equal	4πr²
C	8	double	unequal	2πr²
D	4	single	equal	2πrh + 2πr²

A

B

C

D

Solution

(i) The area of a rectangle becomes 4 times its original area if its length and width are doubled.
(ii) The curved surface area of a cylinder becomes double if the height is doubled, using the formula 2πrh
(iii) All the faces of cube are equal in area.
(iv) If height (h) of a cylinder = radius (r) of the base of the cylinder, then
Total surface area of the cylinder = 2πrh + 2πr²
= 2πr × r + 2πr²
= 2πr² + 2πr²
= 4πr²

Question 24
1 / -0

Find the total surface area of the given figures.

A
(A) 1560 cm², (B) 776 cm², (C) 346.5 cm², (D) 957.48 cm²

B
(A) 1176 cm², (B) 736 cm², (C) 326.7 cm², (D) 942.48 cm²

C
(A) 1160 cm², (B) 785 cm², (C) 371.2 cm², (D) 945.8 cm²

D
(A) 1180 cm2^, (B) 740 cm², (C) 330 cm², (D) 940.5 cm²

Solution

(A) Volume = 2744 cm³
So, edge = 14 cm
Total surface area = 6 edge² = 1176 cm²

(B) Total surface area = 2[lb + bh + hl]
= 2 × [16 × 8 + 8 × 10 + 10 × 16]
= 2[128 + 80 + 160] = 736 cm²

(C) Total surface area = = = 326.7 cm²

(D) Total surface area =
= 2 × 3.14 × 5 × 25 + 2 × 3.14 × 5²
= 942.48 cm²
Question 25
1 / -0

The area of a parallelogram is 3380 cm². If the sum of the base and its parallel side is 130 cm, find the height of the parallelogram.

A
52 cm

B
54 cm

C
58 cm

D
62 cm

Solution

Here, sum of DC and AB = 130 cm
We know that the parallel sides of a parallelogram are equal.
So, we can say that DC = AB = = 65 cm
Area of the parallelogram = AH × DC
3380 = AH × 65
So, AH = = 52 cm

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Re-Attempt Weekly Quiz Competition

Mensuration Test - 6

Result

Mensuration Test - 6

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SHARING IS CARING

Question 1

800 m2

700 m2

600 m2

500 m2

Solution

Question 2

17 cm

11 cm

15 cm

13 cm

Solution

Question 3

7.80 cm2

4.60 cm2

5.90 cm2

6.90 cm2

Solution

Question 4

53,000 cm3

82,125 cm3

84,000 cm3

85,000 cm3

Solution

Question 5

1 : 2

2 : 1

4 : 9

5 : 9

Solution

Question 6

10000 cm

11500 cm

11800 cm

12000 cm

Solution

Question 7

112 cm2

196 cm2

308 cm2

None of these

Solution

Question 8

90 cm2

84.5 cm2

82 cm2

75 cm2

Solution

Question 9

17 cm

20 cm

40 cm

50 cm

Solution

Question 10

17 cm

20 cm

22 cm

23 cm

Solution

Question 11

25

40

30

20

Solution

Question 12

700 cu. cm

704 cu. cm

710 cu. cm

720 cu. cm

Solution

800 m²

700 m²

600 m²

500 m²

7.80 cm²

4.60 cm²

5.90 cm²

6.90 cm²

53,000 cm³

82,125 cm³

84,000 cm³

85,000 cm³

112 cm²

196 cm²

308 cm²

90 cm²

84.5 cm²

82 cm²

75 cm²

120 cm²

140 cm²

150 cm²

170 cm²

104 m²

126 m²

150 m²

175 m²

1150 cm²

1256 cm²

1560 cm²

1250 cm²

(A) 1560 cm², (B) 776 cm², (C) 346.5 cm², (D) 957.48 cm²

(A) 1176 cm², (B) 736 cm², (C) 326.7 cm², (D) 942.48 cm²

(A) 1160 cm², (B) 785 cm², (C) 371.2 cm², (D) 945.8 cm²