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Name: Area and Perimeter Quiz – 2
Subject: Geometry ( Mathematics / Quantitative aptitude)
Topic: Area and Perimeter
Questions: 12 Objective type
Time Allowed: 20 minutes
Important for: Bank PO, SSC CGL Mains, Railways, Engineering Entrance exam and Class 8, 9 & 10th students.
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 Question 1 of 12
1. Question
1 pointsA wire is looped in the form of a circle of radius 28 cm. It is bent again into a square form. What will be the length of the diagonal of the largest square possible thus?
CorrectThe perimeter would remain the same in both cases.
Circumference of circle = 2 ×π × r = 2 × (22/7) × 28 = 176 cm
Perimeter of square = 176
Greatest side possible = 176/4 = 44 cm
Length of diagonal = √(44^{2}+ 44^{2}) = 62.216
= 88/2 × √2 = 44√2IncorrectThe perimeter would remain the same in both cases.
Circumference of circle = 2 ×π × r = 2 × (22/7) × 28 = 176 cm
Perimeter of square = 176
Greatest side possible = 176/4 = 44 cm
Length of diagonal = √(44^{2}+ 44^{2}) = 62.216
= 88/2 × √2 = 44√2UnattemptedThe perimeter would remain the same in both cases.
Circumference of circle = 2 ×π × r = 2 × (22/7) × 28 = 176 cm
Perimeter of square = 176
Greatest side possible = 176/4 = 44 cm
Length of diagonal = √(44^{2}+ 44^{2}) = 62.216
= 88/2 × √2 = 44√2  Question 2 of 12
2. Question
1 pointsFind the area of the circle circumscribed about a square each side of which is 10 cm.
CorrectDiameter of circle = diagonal of square
√(10^{2}+ 10^{2}) = √200 = 10√2
∴ Radius = (10√2)/2 = 5√2
Area of circle = π × r^{2}
∴ 50 × π = 50 × 3.14 = 157.14 cm^{3}IncorrectDiameter of circle = diagonal of square
√(10^{2}+ 10^{2}) = √200 = 10√2
∴ Radius = (10√2)/2 = 5√2
Area of circle = π × r^{2}
∴ 50 × π = 50 × 3.14 = 157.14 cm^{3}UnattemptedDiameter of circle = diagonal of square
√(10^{2}+ 10^{2}) = √200 = 10√2
∴ Radius = (10√2)/2 = 5√2
Area of circle = π × r^{2}
∴ 50 × π = 50 × 3.14 = 157.14 cm^{3}  Question 3 of 12
3. Question
1 pointsA road that is 7 m wide surrounds a circular path whose circumference is 352 m. What will be the area of the road?
CorrectLet us consider the inner radius of the road = r
Then Circumference of road = 2 × π × r = 352 m.
∴ r = 56
Then outer radius = r + 7 = 63 = R
Now Area of the Road = Area of the road with outer radius – Area of the road with inner radius
= π × R^{2} – π × r^{2} = π × (R^{2} – r^{2}) =2618 m^{2}IncorrectLet us consider the inner radius of the road = r
Then Circumference of road = 2 × π × r = 352 m.
∴ r = 56
Then outer radius = r + 7 = 63 = R
Now Area of the Road = Area of the road with outer radius – Area of the road with inner radius
= π × R^{2} – π × r^{2} = π × (R^{2} – r^{2}) =2618 m^{2}UnattemptedLet us consider the inner radius of the road = r
Then Circumference of road = 2 × π × r = 352 m.
∴ r = 56
Then outer radius = r + 7 = 63 = R
Now Area of the Road = Area of the road with outer radius – Area of the road with inner radius
= π × R^{2} – π × r^{2} = π × (R^{2} – r^{2}) =2618 m^{2}  Question 4 of 12
4. Question
1 pointsA bicycle wheel makes 5000 revolutions in moving 11 km. What is the radius of the wheel?
CorrectLet the radius of the wheel be = r
Then 5000 × 2 × π × r = 11×1000×100 cm
∴ r = 35 cmIncorrectLet the radius of the wheel be = r
Then 5000 × 2 × π × r = 11×1000×100 cm
∴ r = 35 cmUnattemptedLet the radius of the wheel be = r
Then 5000 × 2 × π × r = 11×1000×100 cm
∴ r = 35 cm  Question 5 of 12
5. Question
1 pointsThe circumference of a circle exceeds its diameter by 16.8 cm. Find the circumference of the circle.
CorrectLet the radius of the circle be = r
Then 2 × π × r – 2 × r = 16.8
∴ r = 3.92 cm
Then 2 × π × r = 24.6 cmIncorrectLet the radius of the circle be = r
Then 2 × π × r – 2 × r = 16.8
∴ r = 3.92 cm
Then 2 × π × r = 24.6 cmUnattemptedLet the radius of the circle be = r
Then 2 × π × r – 2 × r = 16.8
∴ r = 3.92 cm
Then 2 × π × r = 24.6 cm  Question 6 of 12
6. Question
1 pointsThe length, breadth and height of a room are in the ratio of 3 : 2 : 1. If its volume be 1296 m^{3}, find its breadth
CorrectLet the common ratio be = x
Then; length of the room = 3x,
The breadth of the room = 2x and
The height of the room = x
Then; as per question 3x × 2x × x = 1296
6x^{3} = 1296
x = 6 m
∴ Breadth = 2x = 12 mIncorrectLet the common ratio be = x
Then; length of the room = 3x,
The breadth of the room = 2x and
The height of the room = x
Then; as per question 3x × 2x × x = 1296
6x^{3} = 1296
x = 6 m
∴ Breadth = 2x = 12 mUnattemptedLet the common ratio be = x
Then; length of the room = 3x,
The breadth of the room = 2x and
The height of the room = x
Then; as per question 3x × 2x × x = 1296
6x^{3} = 1296
x = 6 m
∴ Breadth = 2x = 12 m  Question 7 of 12
7. Question
1 pointsA circular racing track is in the form of a ring whose inner and outer circumference are 352 meter and 396 meter respectively. Find the width of the track.
CorrectLet us consider the inner radius of the track= r and
the outer radius of the track = R
∴ Width of the track = R – r = 396/2π 352/2π = 44/2π = 7 meterIncorrectLet us consider the inner radius of the track= r and
the outer radius of the track = R
∴ Width of the track = R – r = 396/2π 352/2π = 44/2π = 7 meterUnattemptedLet us consider the inner radius of the track= r and
the outer radius of the track = R
∴ Width of the track = R – r = 396/2π 352/2π = 44/2π = 7 meter  Question 8 of 12
8. Question
1 pointsA rectangular piece of paper 11cm ×4cm is folded without overlapping to make a cylinder of height 4cm. What will be the volume of the cylinder?
CorrectLength of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let the radius of the cylinder is ‘r’ and the height is ‘h’.
Perimeter of the base of the cylinder = 2 × π × r = 11
2 × (22/7) × r = 11
∴ r = 7/4 cm
∴ Volume of the cylinder = V = π × r^{2} × h
= 22/7 × 7/4 × 7/4 × 4 cm^{3} = 38.5 cm^{3}
Hence the volume of the cylinder is 38.5 cm^{3}.IncorrectLength of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let the radius of the cylinder is ‘r’ and the height is ‘h’.
Perimeter of the base of the cylinder = 2 × π × r = 11
2 × (22/7) × r = 11
∴ r = 7/4 cm
∴ Volume of the cylinder = V = π × r^{2} × h
= 22/7 × 7/4 × 7/4 × 4 cm^{3} = 38.5 cm^{3}
Hence the volume of the cylinder is 38.5 cm^{3}.UnattemptedLength of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let the radius of the cylinder is ‘r’ and the height is ‘h’.
Perimeter of the base of the cylinder = 2 × π × r = 11
2 × (22/7) × r = 11
∴ r = 7/4 cm
∴ Volume of the cylinder = V = π × r^{2} × h
= 22/7 × 7/4 × 7/4 × 4 cm^{3} = 38.5 cm^{3}
Hence the volume of the cylinder is 38.5 cm^{3}.  Question 9 of 12
9. Question
1 pointsA doorframe of dimensions 3 m × 2 m is fixed on the wall of dimension 10 m × 10 m. Find the total labor charges for painting the wall if the labor charges for painting 1 m^{2} of the wall is Rs 2.50.
CorrectPainting of the wall has to be done excluding the area of the door.
Area of the door = l × b = 3 × 2 m^{2} = 6 m^{2}
Area of wall including door = side × side = 10 m × 10 m = 100 m^{2}
Area of wall excluding door = (100 – 6) m^{2} = 94 m^{2}
Total labor charges for painting the wall = Rs 2.50 × 94 = Rs 235IncorrectPainting of the wall has to be done excluding the area of the door.
Area of the door = l × b = 3 × 2 m^{2} = 6 m^{2}
Area of wall including door = side × side = 10 m × 10 m = 100 m^{2}
Area of wall excluding door = (100 – 6) m^{2} = 94 m^{2}
Total labor charges for painting the wall = Rs 2.50 × 94 = Rs 235UnattemptedPainting of the wall has to be done excluding the area of the door.
Area of the door = l × b = 3 × 2 m^{2} = 6 m^{2}
Area of wall including door = side × side = 10 m × 10 m = 100 m^{2}
Area of wall excluding door = (100 – 6) m^{2} = 94 m^{2}
Total labor charges for painting the wall = Rs 2.50 × 94 = Rs 235  Question 10 of 12
10. Question
1 pointsThe length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. What will be its area?
CorrectArea of the rectangle = l × b = 500 × 300 = 150000 m^{2}.
IncorrectArea of the rectangle = l × b = 500 × 300 = 150000 m^{2}.
UnattemptedArea of the rectangle = l × b = 500 × 300 = 150000 m^{2}.
 Question 11 of 12
11. Question
1 pointsShazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area.
CorrectGiven that the length of the wire = circumference of circular shape = 44 cm
∵ We know that the circumference of circle = 2 × π × r = 44 cm
∴ r = 7 cm
Area of the circular shape = π × r^{2} = 154 cm^{2}IncorrectGiven that the length of the wire = circumference of circular shape = 44 cm
∵ We know that the circumference of circle = 2 × π × r = 44 cm
∴ r = 7 cm
Area of the circular shape = π × r^{2} = 154 cm^{2}UnattemptedGiven that the length of the wire = circumference of circular shape = 44 cm
∵ We know that the circumference of circle = 2 × π × r = 44 cm
∴ r = 7 cm
Area of the circular shape = π × r^{2} = 154 cm^{2}  Question 12 of 12
12. Question
1 pointsThe cost of fencing a circular field at the rate of Rs 24 per meter is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m^{2}. Find the cost of ploughing the field.
CorrectLength of the fence (in meters) = (Total cost)/Rate = 5280/24 = 220
So, circumference of the field = 220 m
Therefore, if r meters is the radius of the field, then
2 × π × r = 220
or, 2 × (22/7) ×r = 220
or, r= (220 × 7)/(2 × 22)=35
i.e., radius of the field is 35 m.
Therefore, area of the field = π r^{2} = 22/7 × 35 × 35 m^{2} = 22 × 5 × 35 m^{2}
Now, cost of ploughing 1 m^{2} of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925IncorrectLength of the fence (in meters) = (Total cost)/Rate = 5280/24 = 220
So, circumference of the field = 220 m
Therefore, if r meters is the radius of the field, then
2 × π × r = 220
or, 2 × (22/7) ×r = 220
or, r= (220 × 7)/(2 × 22)=35
i.e., radius of the field is 35 m.
Therefore, area of the field = π r^{2} = 22/7 × 35 × 35 m^{2} = 22 × 5 × 35 m^{2}
Now, cost of ploughing 1 m^{2} of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925UnattemptedLength of the fence (in meters) = (Total cost)/Rate = 5280/24 = 220
So, circumference of the field = 220 m
Therefore, if r meters is the radius of the field, then
2 × π × r = 220
or, 2 × (22/7) ×r = 220
or, r= (220 × 7)/(2 × 22)=35
i.e., radius of the field is 35 m.
Therefore, area of the field = π r^{2} = 22/7 × 35 × 35 m^{2} = 22 × 5 × 35 m^{2}
Now, cost of ploughing 1 m^{2} of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925