Mensuration Test

Result

Mensuration Test - 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

What will be the breadth of a rectangle whose length and perimeter are 60 cm and 200 cm, respectively?

A
40 m

B
50 cm

C
40 cm

D
90 cm

E
50 m

Solution

Let the length of the rectangle be denoted by l and the breadth be denoted by b. (both being in cm)
Now, l = 60.
Also, 2(l + b) = 200
Substituting for the value of l in the above equation, we get b = 40.
Thus, the breadth of the rectangle = 40 cm.
Question 2
1 / -0

The dimensions of a rectangular park are 90 m × 70 m. How much time will a man take in taking 5 rounds of the park if he walks at the rate of 4 km/h?

A
24 minutes

B
22 minutes

C
19.2 minutes

D
15.36 minutes

E
48 minutes

Solution

Length of park = 90 m
Breadth of park = 70 m
Distance covered in one round = Perimeter of park
= 2 × (90 + 70) m
= 2 ×160
= 320 m
Distance covered in five rounds = 320 5
= 1600 m
= 1.6 km
Speed of walking = 4 km/h

Time taken =
= hour

= = 24 minutes
Question 3
1 / -0

What will be the cost of fencing an orchard of length 250 m and breadth 200 m by 4 rounds of barbed wire at a rate of Rs. 30 per metre?

A
Rs. 1,35,000

B
Rs. 1,80,000

C
Rs. 1,08,000

D
Rs. 27,000

E
Rs. 1,00,800

Solution

Length of orchard = 250 m, and Breadth = 200 m
Barbed wire required for 1 round = Perimeter of the orchard
= 2 (250 + 200) = 2 450 = 900 m
Number of rounds = 4
Length of the wire required for fencing = 900 4 = 3600 m
Total cost = 3600 30 = Rs. 1,08,000
Question 4
1 / -0

What is the area of a square whose perimeter is equal to that of a rectangle having dimensions 14 cm 6 cm?

A
225 sq. cm

B
25 sq. cm

C
400 sq. cm

D
100 sq. cm

E
40 sq. cm

Solution

Length of rectangle = 14 cm, Breadth = 6 cm

Perimeter of rectangle = 2 (Length + Breadth) = 2 (14 + 6) = 2 20 = 40 cm

Perimeter of square = 40 cm

Side of square = = = 10 cm
Area of square = Side²= 10²= 10 10 = 100 sq. cm
Question 5
1 / -0

I am a rectangular cloth. I have whole number dimensions. I occupy 24 square metre area. What should be the magnitudes of my length and breadth so that I can cover a rectangular wooden cardboard with largest possible perimeter?

A
Length = 6 cm, breadth = 4 cm

B
Length = 8 cm, breadth = 3 cm

C
Length = 12 cm, breadth = 2 cm

D
Length = 24 cm, breadth = 1 cm

E
Length = 15 cm, breadth = 9 cm

Solution

Here the area of the table cloth = 24 square cm
Then we have to find the length and breadth with largest possible perimeter.
The possible factors of 24
Case I : 24 = 124
If length is 24 cm and breadth = 1 cm
The perimeter = 2 (24 + 1)
= 2 25
= 50 cm
Case II: 24 = 212
If length is 12 cm and breadth = 2 cm
then perimeter = 2 (12 + 2)
= 2 14
= 28 cm
Case III:- 24 = 38
If length is 8 cm and breadth = 3 cm
the perimeter = 2 (8 + 3)
= 2 11
= 22 cm
Case IV: 24 = 4 6
If length is 6 cm and breadth = 4 cm
then perimeter = 2 (6 + 4)
= 2 10
= 20 cm
Hence, the length = 24 and breadth = 1 cm
Question 6
1 / -0

Match the following format and choose the correct option.
Column A Column B
(a) Area of rectangle i. ( side)²
(b) Area of isosceles right triangle ii. Side²(c) Area of square iii. base height
(d) Area of equilateral triangle iv Length breadth

A
(a) - iv, (b) - iii, (c) - ii, (d) - i

B
(a) - i, (b) - ii, (c) - iii, (d) - iv

C
(a) - iii, (b) - iv, (c) - ii, (d) - i

D
(a) - iv, (b) - ii, (c) - iii, (d) - i

E
(a) - iv, (b) - iii, (c) - i, (d) - ii

Solution

The area of rectangle is determined by the formula:
Area = length breadth
Area of isosceles right triangle = base height
Area of square = (side)²Area of equilateral triangle = ( side)²
Question 7
1 / -0

A wire in the form of square of side 30 cm is bent into a regular hexagon. What will be the length of the side of the new hexagon?

A
10 cm

B
16 cm

C
20 cm

D
25 cm

E
30 cm

Solution

The side of square = 30 cm
Its perimeter = 4 Side
= 4 30 = 120 cm
Now, the side of hexagon =
= (120/6) = 20 cm
Question 8
1 / -0

A wire of length 100 cm is cut into three pieces. One piece is used to make a square of side 4 cm, the second piece is used for making a regular pentagon of side 6 cm and a regular hexagon is framed with the remaining piece of wire. What will be the length of this hexagon?

A
4 cm

B
9 cm

C
7.66 cm

D
6 cm

E
14 cm

Solution

The length of the rope = 100 cm
The length of first piece = the perimeter of square
= 4 side
= 4 4 = 16 cm
The length of second piece = the perimeter of pentagon
= 5 side
= 5 6 = 30 cm
Total length of first piece and second piece = 16 + 30
= 46 cm
The remaining length of third piece = 100 46
= 54 cm
The side of regular hexagon =
=
= 9 cm
Question 9
1 / -0

If each side of a square having perimeter 48 cm is decreased by 25% and the resulting square is reshaped into a rectangle of length 10 cm, what will be the magnitude of its breadth?

A
8 cm

B
20 cm

C
36 cm

D
2 cm

E
9 cm

Solution

The perimeter of square = 48 cm
The side of square = = 12 cm
Now 25% of 12 = = 3 cm
The side of new square = 12 3 = 9 cm
The perimeter of the resulting square = 9 4
= 36 cm
The perimeter of rectangle = 36 cm
Length = 10 cm
Breadth = (perimeter ) length
= 36 10
= 18 10
= 8 cm
Question 10
1 / -0

The dimensions of the floor of a dining room are 15 m 12 m. A square carpet of side 10 m is spread in the middle of this room. How much area of this room is not carpeted?

A
80 m²

B
50 m²

C
36 m²

D
17 m²

E
140 m²

Solution

Length of the room = 15 m
Breadth of the room = 12 m
The area of the room = Length Breadth
= 15 12
= 180 m²
The side of square carpet = 10 m
Area of the carpet = (Side)² = 10 10
= 100 m²
Uncarpeted area = 180 100
= 80 m²
Question 11
1 / -0

How many postcards of dimensions 14 cm × 10 cm will be made out of a sheet of paper of dimensions 120 cm × 84 cm?

A
60

B
66

C
72

D
78

E
84

Solution

Length of sheet = 120 cm
Breadth of sheet = 84 cm
Area of sheet = Length Breadth = 120 84 sq. cm
Length of postcard = 14 cm
Breadth of postcard = 10 cm
Area of a postcard = 14 10 sq. cm
The number of postcards = = 72
Question 12
1 / -0

If the length and the area of a rectangular plot are 30 m and 600 m², respectively. What is the simplest ratio of its breadth and perimeter?

A
20 : 100

B
1 : 5

C
3 : 10

D
6 : 1

E
3 : 2

Solution

The length of plot = 30 m
The area of the plot = 600 m²
The breadth of plot = m = 20 m
The perimeter of the plot = 2 (length + breadth) = 2(30 + 20) = 2 50 = 100 m
The required ratio =
Simplest ratio = = 1 : 5
Question 13
1 / -0

If the length of a rectangle is tripled and the breadth is doubled, then by what percent will its area increase?

A
200%

B
300%

C
500%

D
800%

E
700%

Solution

Let the length of the rectangle be x metres and the breadth be y metres.
Its area = x × y square metres
According to the question:
The length of the new rectangle = (3x) metres
And the breadth of the new rectangle = (2y) metres
Therefore, area of the new rectangle = 3x 2y square metres = 6xy square metres
Increase in area = 6xy - xy = 5xy

Percentage increase in area = % = 500%
Question 14
1 / -0

Four equal glass panes of size 40 cm by 30 cm are fitted on a wooden door of size 2 m 40 cm by 1 m 20 cm. What is the area of wooden part on the door?

A
2.16 m²

B
2.24 m²

C
2.32 m²

D
2.40 m²

E
2.64 m²

Solution

The length of the door = 2 m 40 cm = 240 cm
The breadth of the door = 1 m 20 cm = 120 cm
The area of the door = 240 120 square cm
= 28800 square cm
The length of a glass pane = 40 cm
The breadth of glass pane = 30 cm
The area of a glass pane = 40 30
= 1200 square cm
The area of 4 glass panes = 4 1200
= 4800 cm² The area of wooden part = 28800 – 4800
= 24000 square cm

= ( because 1 square m = 10000 square cm)
= 2.40 m²
Question 15
1 / -0

What is the length of the altitude of a triangle with base 30 cm and area 240 square cm?

A
8 cm

B
24 cm

C
16 cm

D
14 cm

E
26 cm

Solution

area = 1/2 x altitude x base
240 cm² = 1/2 x alt. x 30 xcm
therefore, altitude = 16 cm