Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > Objective Type Questions: Mensuration
Class 6 Maths Chapter 9 Question Answers - Mensuration
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Page 1
Mark the correct alternative in each of the following:
1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Solution:
The option (b) is the correct answer.
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 × 9x
So we get
72 = 18x
By division
x = 72/18 = 4
Hence, the length of the rectangle = 5x = 5 × 4 = 20 cm
2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Solution:
The option (b) is the correct answer.
We must find the perimeter of the rectangle for fencing the field.
The dimensions of the rectangle are
Length = 34m
Breadth = 18m
We know that Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter of the rectangle = 2 (34 + 18) = 2 × 52 = 104 m
So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 × 2.25 = Rs. 234
3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24
m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Solution:
The option (d) is the correct answer.
Page 2
Mark the correct alternative in each of the following:
1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Solution:
The option (b) is the correct answer.
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 × 9x
So we get
72 = 18x
By division
x = 72/18 = 4
Hence, the length of the rectangle = 5x = 5 × 4 = 20 cm
2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Solution:
The option (b) is the correct answer.
We must find the perimeter of the rectangle for fencing the field.
The dimensions of the rectangle are
Length = 34m
Breadth = 18m
We know that Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter of the rectangle = 2 (34 + 18) = 2 × 52 = 104 m
So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 × 2.25 = Rs. 234
3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24
m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Solution:
The option (d) is the correct answer.
It is given that cost of fencing the rectangular field = Rs. 600
So the rate of fencing the field = Rs. 7.50 per m
We know that perimeter of the field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the field = 600/7.50 = 80 m
Length of the field = 24 m
So we get breadth of the field = Perimeter/2- Length = 80/2- 24 = 16 m
4. The cost of putting a fence around a square field at Rs 2.50 per meter is Rs 200. The length of each side
of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these
Solution:
The option (c) is the correct answer.
It is given that cost of fencing the square field = Rs. 200
So the rate of fencing the field = Rs. 2.50
We know that, perimeter of the square field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the square field = 200/2.50 = 80 m
Perimeter of square = 4 × Side of the square
It can be written as
Side of the square = Perimeter/4 = 80/4 = 20 m
5. The length of a rectangle is three times of its width. If the length of the diagonal is 8 v10 m, then the
perimeter of the rectangle is
(a) 15 v10 m
(b) 16 v10 m
(c) 24 v10 m
(d) 64 m
Solution:
The option (d) is the correct answer.
Consider ABCD as a rectangle.
Assume that the width of the rectangle BC = x m
We know that the length is three times width of the rectangle.
So, length of the rectangle AB = 3x m
Page 3
Mark the correct alternative in each of the following:
1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Solution:
The option (b) is the correct answer.
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 × 9x
So we get
72 = 18x
By division
x = 72/18 = 4
Hence, the length of the rectangle = 5x = 5 × 4 = 20 cm
2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Solution:
The option (b) is the correct answer.
We must find the perimeter of the rectangle for fencing the field.
The dimensions of the rectangle are
Length = 34m
Breadth = 18m
We know that Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter of the rectangle = 2 (34 + 18) = 2 × 52 = 104 m
So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 × 2.25 = Rs. 234
3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24
m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Solution:
The option (d) is the correct answer.
It is given that cost of fencing the rectangular field = Rs. 600
So the rate of fencing the field = Rs. 7.50 per m
We know that perimeter of the field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the field = 600/7.50 = 80 m
Length of the field = 24 m
So we get breadth of the field = Perimeter/2- Length = 80/2- 24 = 16 m
4. The cost of putting a fence around a square field at Rs 2.50 per meter is Rs 200. The length of each side
of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these
Solution:
The option (c) is the correct answer.
It is given that cost of fencing the square field = Rs. 200
So the rate of fencing the field = Rs. 2.50
We know that, perimeter of the square field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the square field = 200/2.50 = 80 m
Perimeter of square = 4 × Side of the square
It can be written as
Side of the square = Perimeter/4 = 80/4 = 20 m
5. The length of a rectangle is three times of its width. If the length of the diagonal is 8 v10 m, then the
perimeter of the rectangle is
(a) 15 v10 m
(b) 16 v10 m
(c) 24 v10 m
(d) 64 m
Solution:
The option (d) is the correct answer.
Consider ABCD as a rectangle.
Assume that the width of the rectangle BC = x m
We know that the length is three times width of the rectangle.
So, length of the rectangle AB = 3x m
AC is the diagonal of rectangle
Consider ABC as a right angled triangle.
AC
2
= AB
2
+ BC
2
By substituting the values
640 = 9x
2
+ x
2
We get
640 = 10x
2
On further calculation
x
2
= 640/10 = 64
x = v64 = 8 m
So the breadth of the rectangle x = 8 m
Length of the rectangle 3x = 3 × 8 = 24 m
Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter = 2 (24 + 8) = 2 × 32 = 64 m
6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio
(a) 3: 1
(b) v3: 1
(c) v2 : 1
(d) 2 v2: 1
Solution:
The option (d) is the correct answer.
Assume that the length of the smaller side of the rectangle BC = x
Length of the larger side AB = y
It is given that the length of the diagonal is three times that of the smaller side.
Diagonal of the rectangle 3x = AC
By using Pythagoras theorem
(AC)
2
= (AB)
2
+ (BC)
2
By substituting the values
(3x)
2
= (x)
2
+ (y)
2
On further calculation
9x
2
= x
2
+ y
2
We get
8x
2
= y
2
By taking square roots of both sides,
2 v2 x = y
Hence, the ratio of the larger side to the smaller side is 2v2: 1.
Page 4
Mark the correct alternative in each of the following:
1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Solution:
The option (b) is the correct answer.
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 × 9x
So we get
72 = 18x
By division
x = 72/18 = 4
Hence, the length of the rectangle = 5x = 5 × 4 = 20 cm
2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Solution:
The option (b) is the correct answer.
We must find the perimeter of the rectangle for fencing the field.
The dimensions of the rectangle are
Length = 34m
Breadth = 18m
We know that Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter of the rectangle = 2 (34 + 18) = 2 × 52 = 104 m
So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 × 2.25 = Rs. 234
3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24
m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Solution:
The option (d) is the correct answer.
It is given that cost of fencing the rectangular field = Rs. 600
So the rate of fencing the field = Rs. 7.50 per m
We know that perimeter of the field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the field = 600/7.50 = 80 m
Length of the field = 24 m
So we get breadth of the field = Perimeter/2- Length = 80/2- 24 = 16 m
4. The cost of putting a fence around a square field at Rs 2.50 per meter is Rs 200. The length of each side
of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these
Solution:
The option (c) is the correct answer.
It is given that cost of fencing the square field = Rs. 200
So the rate of fencing the field = Rs. 2.50
We know that, perimeter of the square field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the square field = 200/2.50 = 80 m
Perimeter of square = 4 × Side of the square
It can be written as
Side of the square = Perimeter/4 = 80/4 = 20 m
5. The length of a rectangle is three times of its width. If the length of the diagonal is 8 v10 m, then the
perimeter of the rectangle is
(a) 15 v10 m
(b) 16 v10 m
(c) 24 v10 m
(d) 64 m
Solution:
The option (d) is the correct answer.
Consider ABCD as a rectangle.
Assume that the width of the rectangle BC = x m
We know that the length is three times width of the rectangle.
So, length of the rectangle AB = 3x m
AC is the diagonal of rectangle
Consider ABC as a right angled triangle.
AC
2
= AB
2
+ BC
2
By substituting the values
640 = 9x
2
+ x
2
We get
640 = 10x
2
On further calculation
x
2
= 640/10 = 64
x = v64 = 8 m
So the breadth of the rectangle x = 8 m
Length of the rectangle 3x = 3 × 8 = 24 m
Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter = 2 (24 + 8) = 2 × 32 = 64 m
6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio
(a) 3: 1
(b) v3: 1
(c) v2 : 1
(d) 2 v2: 1
Solution:
The option (d) is the correct answer.
Assume that the length of the smaller side of the rectangle BC = x
Length of the larger side AB = y
It is given that the length of the diagonal is three times that of the smaller side.
Diagonal of the rectangle 3x = AC
By using Pythagoras theorem
(AC)
2
= (AB)
2
+ (BC)
2
By substituting the values
(3x)
2
= (x)
2
+ (y)
2
On further calculation
9x
2
= x
2
+ y
2
We get
8x
2
= y
2
By taking square roots of both sides,
2 v2 x = y
Hence, the ratio of the larger side to the smaller side is 2v2: 1.
7. The ratio of the areas of two squares, one having its diagonal double than the other, is
(a) 1: 2
(b) 2: 3
(c) 3: 1
(d) 4: 1
Solution:
The option (d) is the correct answer.
Consider ABCD and PQRS as the two squares. We know that, the diagonal of square PQRS is twice the diagonal
of square ABCD.
PR = 2 AC
Area of the square = Diagonal
2
/2
Area of PQRS = PR
2
/2
In the same way, area of ABCD = AC
2
/2
From the question, we know that:
If AC = x units, we get PR = 2x units
Area of PQRS/Area of ABCD = (PR
2
×2)/ (2×AC
2
)
By substituting the value
Area of PQRS/Area of ABCD = [(2x)
2
×2]/ [2×x
2
] = 4/1
Hence, the ratio of the areas of squares PQRS and ABCD is 4: 1.
8. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is
(a) 225 : 256
(b) 256 : 225
(c) 15 : 16
(d) 16 : 15
Solution:
The option (c) is the correct answer.
Consider ABCD and PQRS as the two squares.
Let the lengths of each side of ABCD and PQRS be x and y.
We know that
Area of sq. ABCD/Area of sq. PQRS = x
2
/y
2
So we get x
2
/y
2
= 225/256
By taking square roots on both sides,
x/y = 15/16
Page 5
Mark the correct alternative in each of the following:
1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm
Solution:
The option (b) is the correct answer.
Consider the sides of the rectangle as 5x and 4x.
We know that, perimeter of rectangle = 2 (Length + Breadth)
By substituting the values
72 = 2 (5x + 4x)
On further calculation
72 = 2 × 9x
So we get
72 = 18x
By division
x = 72/18 = 4
Hence, the length of the rectangle = 5x = 5 × 4 = 20 cm
2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334
Solution:
The option (b) is the correct answer.
We must find the perimeter of the rectangle for fencing the field.
The dimensions of the rectangle are
Length = 34m
Breadth = 18m
We know that Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter of the rectangle = 2 (34 + 18) = 2 × 52 = 104 m
So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 × 2.25 = Rs. 234
3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24
m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m
Solution:
The option (d) is the correct answer.
It is given that cost of fencing the rectangular field = Rs. 600
So the rate of fencing the field = Rs. 7.50 per m
We know that perimeter of the field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the field = 600/7.50 = 80 m
Length of the field = 24 m
So we get breadth of the field = Perimeter/2- Length = 80/2- 24 = 16 m
4. The cost of putting a fence around a square field at Rs 2.50 per meter is Rs 200. The length of each side
of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these
Solution:
The option (c) is the correct answer.
It is given that cost of fencing the square field = Rs. 200
So the rate of fencing the field = Rs. 2.50
We know that, perimeter of the square field = Cost of fencing/Rate of fencing
By substituting the values
Perimeter of the square field = 200/2.50 = 80 m
Perimeter of square = 4 × Side of the square
It can be written as
Side of the square = Perimeter/4 = 80/4 = 20 m
5. The length of a rectangle is three times of its width. If the length of the diagonal is 8 v10 m, then the
perimeter of the rectangle is
(a) 15 v10 m
(b) 16 v10 m
(c) 24 v10 m
(d) 64 m
Solution:
The option (d) is the correct answer.
Consider ABCD as a rectangle.
Assume that the width of the rectangle BC = x m
We know that the length is three times width of the rectangle.
So, length of the rectangle AB = 3x m
AC is the diagonal of rectangle
Consider ABC as a right angled triangle.
AC
2
= AB
2
+ BC
2
By substituting the values
640 = 9x
2
+ x
2
We get
640 = 10x
2
On further calculation
x
2
= 640/10 = 64
x = v64 = 8 m
So the breadth of the rectangle x = 8 m
Length of the rectangle 3x = 3 × 8 = 24 m
Perimeter = 2 (Length + Breadth)
By substituting the values
Perimeter = 2 (24 + 8) = 2 × 32 = 64 m
6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio
(a) 3: 1
(b) v3: 1
(c) v2 : 1
(d) 2 v2: 1
Solution:
The option (d) is the correct answer.
Assume that the length of the smaller side of the rectangle BC = x
Length of the larger side AB = y
It is given that the length of the diagonal is three times that of the smaller side.
Diagonal of the rectangle 3x = AC
By using Pythagoras theorem
(AC)
2
= (AB)
2
+ (BC)
2
By substituting the values
(3x)
2
= (x)
2
+ (y)
2
On further calculation
9x
2
= x
2
+ y
2
We get
8x
2
= y
2
By taking square roots of both sides,
2 v2 x = y
Hence, the ratio of the larger side to the smaller side is 2v2: 1.
7. The ratio of the areas of two squares, one having its diagonal double than the other, is
(a) 1: 2
(b) 2: 3
(c) 3: 1
(d) 4: 1
Solution:
The option (d) is the correct answer.
Consider ABCD and PQRS as the two squares. We know that, the diagonal of square PQRS is twice the diagonal
of square ABCD.
PR = 2 AC
Area of the square = Diagonal
2
/2
Area of PQRS = PR
2
/2
In the same way, area of ABCD = AC
2
/2
From the question, we know that:
If AC = x units, we get PR = 2x units
Area of PQRS/Area of ABCD = (PR
2
×2)/ (2×AC
2
)
By substituting the value
Area of PQRS/Area of ABCD = [(2x)
2
×2]/ [2×x
2
] = 4/1
Hence, the ratio of the areas of squares PQRS and ABCD is 4: 1.
8. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is
(a) 225 : 256
(b) 256 : 225
(c) 15 : 16
(d) 16 : 15
Solution:
The option (c) is the correct answer.
Consider ABCD and PQRS as the two squares.
Let the lengths of each side of ABCD and PQRS be x and y.
We know that
Area of sq. ABCD/Area of sq. PQRS = x
2
/y
2
So we get x
2
/y
2
= 225/256
By taking square roots on both sides,
x/y = 15/16
By taking the ratio of their perimeters, we get
Perimeter of sq. ABCD/Primeter of sq. PQRS = (4 × side of sq. ABCD)/ (4 × side of sq. PQRS) = 4x/4y
By removing common terms in numerator and denominator
Perimeter of sq. ABCD/Perimeter of sq. PQRS = x/ y
So perimeter of sq. ABCD/Perimeter of sq. PQRS = 15/16
Hence, the ratio of their perimeters = 15: 16
9. If the sides of a square are halved, then its area
(a) remains same
(b) becomes half
(c) becomes one fourth
(d) becomes double
Solution:
The option (c) is the correct answer.
Consider x as the side of the square.
We know that area of a square = Side × Side = x × x = x
2
If the sides are halved, we get new side = x/2
So the new area = (x/2)
2
= x
2
/4
From this we know that the area has become one fourth of its previous value.
10. A rectangular carpet has area 120 m
2
and perimeter 46 meters. The length of its diagonal is
(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m
Solution:
The option (c) is the correct answer.
It is given that area of the rectangle = 120 m
2
Perimeter of the rectangle = 46 m
Consider l and b as the length and breadth.
Area of the rectangle = l × b = 120 m
2
Perimeter of the rectangle = 2 (l + b) = 46
So we get
(l + b) = 46/2 = 23 m
Length of the diagonal of the rectangle = l
2
+ b
2
It can be written as
(l
2
+ b
2
) = (l + b)
2
- 2 (l × b)
By substituting the values
(l
2
+ b
2
) = (23)
2
- 2 (120) = 529 - 240 = 289
By adding square roots on both sides
Length of the diagonal of the rectangle = vl
2
+ b
2
= v289 = 17 m
Hence, the length of the diagonal of the rectangle is 17m.
11. If the ratio between the length and the perimeter of a rectangular plot is 1: 3, then the ratio between the
length and breadth of the plot is
(a) 1: 2
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Objective Type Questions: Mensuration Class 6
The "Objective Type Questions: Mensuration Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Objective Type Questions: Mensuration on the App
Students of Class 6 can study Objective Type Questions: Mensuration alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Objective Type Questions: Mensuration,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Objective Type Questions: Mensuration is prepared as per the latest Class 6 syllabus.
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