# Mensuration Questions and Answers updated daily – Aptitude

Mensuration Questions: Solved 782 Mensuration Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Mensuration Questions

41. Area of 1st circle and circumference of 2nd circle is 1386 cm

^{2}and 176 cm respectively. There is a square whose side is 35(5/7)% of twice of sum of the radius of both the circles. Find the perimeter of the square (in cm)?SHOW ANSWER

Correct Ans:140

Explanation:

Given, Area of 1st circle = 1386 cm

---> π r

---> r

---> r

---> r = √441

---> r = 21

Hence,

Given, Circumference of 2nd circle = 176 cm

---> 2πr = 176

---> r = 176 * (7/22) * (1/2)

---> r = 4 * 7

---> r = 28

Hence,

Now, Side of the square = 35(5/7)% of 2(radius of 1st circle + radius of 2nd circle)

= (250/700) * 2(21 + 28)

= (25/70) * 2(49)

= (5/14) * 98

= 5 * 7

= 35

Then,

^{2}---> π r

^{2}= 1386---> r

^{2}= 1386 * (7/22)---> r

^{2}= 63 * 7 = 441---> r = √441

---> r = 21

Hence,

**radius of 1st circle = 21**Given, Circumference of 2nd circle = 176 cm

---> 2πr = 176

---> r = 176 * (7/22) * (1/2)

---> r = 4 * 7

---> r = 28

Hence,

**radius of 2nd circle = 28**Now, Side of the square = 35(5/7)% of 2(radius of 1st circle + radius of 2nd circle)

= (250/700) * 2(21 + 28)

= (25/70) * 2(49)

= (5/14) * 98

= 5 * 7

= 35

**Side of the square = 35 cm**Then,

**Perimeter of the square**= 4 * side = 4 * 35 =**140 cm**
Workspace

42. The angles of a quadrilateral are in the ratios 2 : 4 : 7 : 5. The smallest angle of the quadrilateral equal to the smallest angle of a triangle. One of the angles of the triangle is twice the smallest angle of triangle. What is the second largest angle of triangle?

SHOW ANSWER

Correct Ans:60°

Explanation:

W.K.T:- Sum of angles of quadrilateral =360°

Given, ratio of angles of quadrilateral = 2 : 4 : 7 : 5

Then, 2x + 4x + 7x + 5x = 360°

---> 18x = 360°

---> x = 20°

Hence, the

Given,

And,

Now,

= 180° â€“ (40° +80°)

= 180° â€“ 120°

=

Hence, the

Given, ratio of angles of quadrilateral = 2 : 4 : 7 : 5

Then, 2x + 4x + 7x + 5x = 360°

---> 18x = 360°

---> x = 20°

Hence, the

**Smallest angle of quadrilateral**= 2x = 2 * 20° =**40°**Given,

**Smallest angle of a triangle**= Smallest angle of quadrilateral =**40°**And,

**Second angle of triangle**= 2 * 40° =**80°**Now,

**Third angle of triangle**= 180° â€“ (Sum of the other two angles of the triangle)= 180° â€“ (40° +80°)

= 180° â€“ 120°

=

**60°**Hence, the

**second largest angle of the triangle = 60°**
Workspace

43. If ratio of volume of sphere to volume of cylinder is 2 : 3 and radius of cylinder is equal to the side of square whose perimeter is 84 cm then find the curved surface area of cylinder(in cmÂ²). Given that radius of sphere is equal to radius of cylinder.

SHOW ANSWER

Correct Ans:5544

Explanation:

Volume of sphere = 4/3(πr

Volume of the cylinder = πr

Volume of sphere/Volume of the cylinder = 2 : 3

[(4/3(πr

2r = h

Perimeter of square = 4 * side = 84 cm

Side of square = 84/4 = 21 cm

Radius of cylinder = side of square = 21 cm

Curved surface area of cylinder = 2πrh

= 2 * (22/7) * 21 * 42 = 5544 cm

^{3})Volume of the cylinder = πr

^{2}hVolume of sphere/Volume of the cylinder = 2 : 3

[(4/3(πr

^{3})] / (πr^{2}h) = 2/32r = h

Perimeter of square = 4 * side = 84 cm

Side of square = 84/4 = 21 cm

Radius of cylinder = side of square = 21 cm

Curved surface area of cylinder = 2πrh

= 2 * (22/7) * 21 * 42 = 5544 cm

^{2}
Workspace

44. If length of a rectangle is decreased by 6 cm we get a square and the area of square formed is 252 cm

^{2}less than the area of square formed when breadth of the original rectangle is increased by 6 cm. Find the perimeter of the rectangle.SHOW ANSWER

Correct Ans:84 cm

Explanation:

Let the length of the rectangle be 'l cm' and breadth of the rectangle be 'b cm'.

As per the question,

lb + 6l - bl + 6b = 252

6(l + b) = 252

2(l + b) = 84 cm

Therefore, perimeter of rectangle is 84 cm.

As per the question,

**Area of rectangle when breadth increased - Area of reactangle when length decreased = 252****[l x (b + 6)] - [b x (l - 6)] = 252**lb + 6l - bl + 6b = 252

6(l + b) = 252

2(l + b) = 84 cm

Therefore, perimeter of rectangle is 84 cm.

Workspace

45. The barrel of a fountain pen is cylindrical in shape which radius of base as 0.7 cm and is 5 cm long. One such barrel in the pen can be used to write 300 words. A barrel full of ink which has a capacity of 14 cu.cm can be used to write how many words approximately?

SHOW ANSWER

Correct Ans:545

Explanation:

Given:

Radius of base = 0.7 cm; Height = 5 cm

WKT,

Volume of barrel in pen = (22/7)*(0.7)

= 7.7 cu.cm

So, capacity of 7.7 cu.cm in pen can write 300 words.

Therefore, capacity of 14 cu.cm can be used to write = (300 x 14)/7.7 =

Radius of base = 0.7 cm; Height = 5 cm

WKT,

**Volume of cylinder = πr**^{2}hVolume of barrel in pen = (22/7)*(0.7)

^{2}*5= 7.7 cu.cm

So, capacity of 7.7 cu.cm in pen can write 300 words.

Therefore, capacity of 14 cu.cm can be used to write = (300 x 14)/7.7 =

**545 words.**
Workspace

46. The radius of the circle is equal to seven - ninth of the side of the square.The area of the circle is 9856 sq cm. Find the perimeter of a square?

SHOW ANSWER

Correct Ans:288 cm

Explanation:

Given: Radius of Circle = (7/9) * Side of Square

Area of circle = πRÂ²

9856 = (22/7) * RÂ²

RÂ² = 9856 / (22/7)

= 448 * 7

R = √(448 *7)

R = 56 cm

And Side of Square = Radius of Circle * (9/7)

= 56*(9/7)

Side = 72 cm

Perimeter of the Square = 4 * Side of Square.

= 72*4

= 288 cm.

Area of circle = πRÂ²

9856 = (22/7) * RÂ²

RÂ² = 9856 / (22/7)

= 448 * 7

R = √(448 *7)

R = 56 cm

And Side of Square = Radius of Circle * (9/7)

= 56*(9/7)

Side = 72 cm

Perimeter of the Square = 4 * Side of Square.

= 72*4

= 288 cm.

Workspace

47. The side of a rhombus is 13m. If one of its diagonal is 24m. Find the area of the rhombus.

SHOW ANSWER

Correct Ans:120 m

^{2}Explanation:

Given: âˆ†BCE

BE² + EC² = BC²

BD = 24 m, BE = BD/2 = 12 m

EC = √[(13)² -(12)²] = 5 m

Diagonal AC = 2(EC) = 2*5 = 10 m

Required area = (Product of diagonals)/2

= (10*24)/2

= 120 m².

Workspace

48. A ground is square in shape has 12 m wide road inside it running along its sides. The area occupied by the road is 5290 sq m. What is the perimeter along the outer edge of the road?

SHOW ANSWER

Correct Ans:488

Explanation:

Let the side of outer square of the road = x meter.

So the side of inner square of the road = (x - 2(12)) = (x - 24) m.

According to the question:

We know that

Area of outer square of the road - Area of inner square of the road = Area occupied by the road

x

x

48x = 5866

x = 122.2 m

Therefore side of outer square of the road = x = 122 m

Perimeter of outer square of the road= 4*a = 4*122 = 488 m.

So the side of inner square of the road = (x - 2(12)) = (x - 24) m.

According to the question:

We know that

**Area of square = (side)**^{2}Area of outer square of the road - Area of inner square of the road = Area occupied by the road

x

^{2}- (x - 24)^{2}= 5290 m^{2}x

^{2}- x^{2}- 576 + 48x = 529048x = 5866

x = 122.2 m

Therefore side of outer square of the road = x = 122 m

**Perimeter of square = 4a**,where a is area.Perimeter of outer square of the road= 4*a = 4*122 = 488 m.

Workspace

49. The total area of a circle and a square is equal to 5450 sq.cm. The diameter of the circle is 70 cm. What is the sum of the circumference of the circle and the perimeter of the square?

SHOW ANSWER

Correct Ans:380 cm

Explanation:

Given: Diameter of circle = 70 cm

Radius of circle = 35 cm

WKT,

a

a

a

a

a = 40

As per the question,

=

=

= 2 x (22/7) x 35 + 4(40)

= 220 + 160

= 380 cm

Radius of circle = 35 cm

WKT,

**Area of square = a**^{2}**Area of circle = πr**^{2}**Area of square + Area of circle = 5450**a

^{2}+ (22/7) x 35 x 35 = 5450a

^{2}+ 3850 = 5450a

^{2}= 5450 - 3850a

^{2}= 1600a = 40

As per the question,

=

**Circumference of the circle + Perimeter of the square**=

**2πr + 4a**= 2 x (22/7) x 35 + 4(40)

= 220 + 160

= 380 cm

Workspace

50. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively 144 cm and 100 cm?

SHOW ANSWER

Correct Ans:240

Explanation:

Total area of the rectangular region = Length * Breadth

= 144 * 100

= 14400 cm

Area of one tile = 12 * 5 = 60 cm

= 144 * 100

= 14400 cm

^{2}Area of one tile = 12 * 5 = 60 cm

^{2}**Number of tiles required**= 14400 / 60 =**240**
Workspace

51. A rectangular garden has a four-metre-wide road along all the four sides. The area of the road is 1104 sq metre. What is the sum of the length and the breadth of the garden?

SHOW ANSWER

Correct Ans:130m

Explanation:

Formula:

Area of Rectangle = l*b

Let the length and breadth of the rectangular garden be l and b

The 4m wide road covers the inner garden.

Therefore, the length and breadth of the outer road is (l+8)*(b+8)

Given:

Area of rectangular garden including the 4m wide road = Area of the road

(l+8)*(b+8)-lb = 1104

lb+8l+8b+64-lb = 1104

8l+8b = 1104-64

l+b = 130

Area of Rectangle = l*b

Let the length and breadth of the rectangular garden be l and b

The 4m wide road covers the inner garden.

Therefore, the length and breadth of the outer road is (l+8)*(b+8)

Given:

Area of rectangular garden including the 4m wide road = Area of the road

(l+8)*(b+8)-lb = 1104

lb+8l+8b+64-lb = 1104

8l+8b = 1104-64

l+b = 130

Workspace

52. The circumference of a circle is thrice the perimeter of a rectangle. The area of the circle is 616 sq. m. What is the area of the rectangle if the breadth of the rectangle is 60 m?

SHOW ANSWER

Correct Ans:4320 sq m

Explanation:

Area of a circle= πr

616 = 22r

616*(7/22) = r

r

r = 14 m

Circumference of a circle = 2πr

= 2*22/7 * 14

= 88 m

It is given that the circumference of a circle is thrice the perimeter of a rectangle

Circumference of a circle= 3Perimeter of the rectangle

Circumference of a circle = 88 m

= 3*88

Circumference of a circle= 264 m

264 = 2 (l + 60)

132 = l + 60

l = 132 - 60

l = 72

Area of the rectangle = l*b

= 72*60

Area of the rectangle = 4320 m

^{2}616 = 22r

^{2}/7616*(7/22) = r

^{2}r

^{2}= 196r = 14 m

Circumference of a circle = 2πr

= 2*22/7 * 14

= 88 m

^{2}It is given that the circumference of a circle is thrice the perimeter of a rectangle

Circumference of a circle= 3Perimeter of the rectangle

Circumference of a circle = 88 m

^{2}= 3*88

Circumference of a circle= 264 m

^{2}264 = 2 (l + 60)

132 = l + 60

l = 132 - 60

l = 72

Area of the rectangle = l*b

= 72*60

Area of the rectangle = 4320 m

^{2}
Workspace

53. The dimension of an open container are 100cm, 80cm, and 46cm. Its thickness is 6cm. If 1 cubic cm of metal used in the box weight 1 gram, find the weight of the container.

SHOW ANSWER

Correct Ans:128.64kg

Explanation:

Volume of a Cube (V) = a

External volume = (100*80*46) cm

= 368000 cm

Internal volume = (88* 68* 40) cm

= 239360 cm

Volume of the metal used in the container = 368000 - 239360

= 128640 cm

Converting into g to Kg:

Weight of the metal = 128640 / 1000

= 128.64 kg

^{3}External volume = (100*80*46) cm

^{3}= 368000 cm

^{3}Internal volume = (88* 68* 40) cm

^{3}= 239360 cm

^{3}Volume of the metal used in the container = 368000 - 239360

= 128640 cm

^{3}Converting into g to Kg:

Weight of the metal = 128640 / 1000

= 128.64 kg

Workspace

54. The number of sperical bullets that can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being of 4 cm diameter, is (take π = 22/7)

SHOW ANSWER

Correct Ans:2541

Explanation:

Given:

Bullet diameter - 4 cm; Bullet radius - 2cm

Cube edges - 44 cm

WKT, Volume of cube = a

Volume of sphere = (4/3)πr

Total number of spherical bullets = Volume of solid cube/Volume of 1 bullet

= (44 x 44 x 44)/[(4/3)x (22/7) x 2 x 2 x 2]

= (44 x 44 x 44 x 7 x 3)/(22 x 4 x 2 x 2 x 2)

= 2541.

Bullet diameter - 4 cm; Bullet radius - 2cm

Cube edges - 44 cm

WKT, Volume of cube = a

^{3}Volume of sphere = (4/3)πr

^{3}Total number of spherical bullets = Volume of solid cube/Volume of 1 bullet

= (44 x 44 x 44)/[(4/3)x (22/7) x 2 x 2 x 2]

= (44 x 44 x 44 x 7 x 3)/(22 x 4 x 2 x 2 x 2)

= 2541.

Workspace

55. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope stretched and describes 88 m when it has traced out 72° at the center the length of the rope is (take π = 22/7)

SHOW ANSWER

Correct Ans:70 m

Explanation:

WKT, Length of the arc = 2πrθ/360°

As per the question,

2πrθ/360° = 88

2πr x 72/360 = 88

r = (88 x 5)/(2 x π)

r = (88 x 5 x 7)/(2 x 22)

r = 70 m

Here, length of the rope is nothing but radius.

Therefore, length ogf the radius is 70 m.

Workspace

56. A square lawn has a path of 4m wide around it. If the area of the path is 196 m

^{2}, then each side of the lawn isSHOW ANSWER

Correct Ans:8.25 m

Explanation:

Let each side of the square lawn be 'X' m.

As per the question,

Area of lawn and path - Area of lawn = 196

(X + 4 + 4)

^{2}- (X)

^{2}= 196

X

^{2}+ 64 + 16X - X

^{2}= 196

16X = 196 - 64

X = 132/16

X = 8.25 m.

Workspace

57. The diagonal of a square equals the side of an equilateral triangle. If the area of the square is 6√3 sq.cm, what is the area of the equilateral triangle?

SHOW ANSWER

Correct Ans:9 sq.cm

Explanation:

Given:

Area of square = 6√3 sq.cm

a

a = √(6√3) cm

WKT, Diagonal of square = a√2

Diagonal of square = √(12√3) cm

Area of equilateral triangle = (√3/4)a

= (√3/4) x [√(12√3)]

= (√3/4) x (12√3)

= 9 sq.cm

Area of square = 6√3 sq.cm

a

^{2}= 6√3 sq.cma = √(6√3) cm

WKT, Diagonal of square = a√2

Diagonal of square = √(12√3) cm

Area of equilateral triangle = (√3/4)a

^{2}= (√3/4) x [√(12√3)]

^{2}= (√3/4) x (12√3)

= 9 sq.cm

Workspace

58. A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the liquid rises in the cylindrical vessel.

SHOW ANSWER

Correct Ans:24 cm

Explanation:

Let the height of the liquid in the conical vessel be 'h'.

Given:

Internal radius of cone =12 cm

Internal height of cone = 50 cm

Internal radius of cylinder = 10 cm

WKT, Volume of cylinder = πr

Volume of cone = 1/3(πr

Volume of the liquid in the cylindrical vessel = Volume of liquid in the conical vessel

πr

[(22/7) x 10 x 10 x h] = [(22 x 12 x 12 x 50)/(3 x 7)]

[(22/7) x 10 x 10 x h] = [(22 x 4 x 12 x 50)/7]

h = [(4 x 12 x 50)/(10 x 10)]

h = 24 cm.

Given:

Internal radius of cone =12 cm

Internal height of cone = 50 cm

Internal radius of cylinder = 10 cm

WKT, Volume of cylinder = πr

^{2}h cm^{3}Volume of cone = 1/3(πr

^{2}h)cm^{3}Volume of the liquid in the cylindrical vessel = Volume of liquid in the conical vessel

πr

^{2}h = 1/3(πr^{2}h)[(22/7) x 10 x 10 x h] = [(22 x 12 x 12 x 50)/(3 x 7)]

[(22/7) x 10 x 10 x h] = [(22 x 4 x 12 x 50)/7]

h = [(4 x 12 x 50)/(10 x 10)]

h = 24 cm.

Workspace

59. The area of a square is 1296 sq cm. Find the perimeter of the rectangle, if length is (1/3) the side of the square and breadth is 28 cm less than the side of the square?

SHOW ANSWER

Correct Ans:40 cm

Explanation:

Given: Area of a square = 1296 sq.cm.

Area of square = a

a = √1296

a = 36 cm

Length of the rectangle = (1/3) x 36 = 12 cm

Breadth of the rectangle = 36 -28 = 8 cm

Perimeter of the rectangle = 2(l + b)

= 2(12 + 8)

= 2(20)

= 40 cm.

Area of square = a

^{2}sq.cma = √1296

a = 36 cm

Length of the rectangle = (1/3) x 36 = 12 cm

Breadth of the rectangle = 36 -28 = 8 cm

Perimeter of the rectangle = 2(l + b)

= 2(12 + 8)

= 2(20)

= 40 cm.

Workspace

60. Some bricks are arranged in an area measuring 50 m

^{3}. If the length, breadth and height of each brick is 25 cm, 12.5 cm and 4 cm respectively. Find the number of bricks that are used supposing there is no gap between two bricks.SHOW ANSWER

Correct Ans:40,000

Explanation:

Given: Volume of area = 50m

Length of brick = 25 cm = 25/100 m

Breadth of brick = 12.5 cm = 12.5/100 m

Width of brick = 4 cm = 4/100 m

WKT, Volume of each brick = l x b x h

Volume of that area = Number of bricks x Volume of each brick

50 = N x [(25 x 12.5 x 4)/(100 x 100 x 100)]

N = 50/0.00125

N = 40,000

^{3}Length of brick = 25 cm = 25/100 m

Breadth of brick = 12.5 cm = 12.5/100 m

Width of brick = 4 cm = 4/100 m

WKT, Volume of each brick = l x b x h

Volume of that area = Number of bricks x Volume of each brick

50 = N x [(25 x 12.5 x 4)/(100 x 100 x 100)]

N = 50/0.00125

N = 40,000

Workspace

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