## MENSURATION

### Solved Question Papers

Q.
In a right angled triangle, find the hypotenuse if base and perpendicular are respectively 36015 cm and 48020 cm.

1.69125 cm
2.60025 cm
3.391025 cm
4.60125 cm

ANS.2

Explanation :
Let hypotenuse = x cm
Then, by Pythagoras theorem:
x2 = (48020)2 + (36015)2
x so 60025 cm

Q.
The perimeter of an equilateral triangle is 723–√ cm. Find its height.

1.63
2.24
3.18
4.36

ANS.4

Explanation :
Let one side of the △ be = a Perimeter of equilateral triangle = 3a 3a = 723–√ = a = 243–√ cm Height = AC; by Pythagoras theorem AC2 = a2 – (a/2)2 AC = 36 cm

Q.
The inner circumference of a circular track is 440 cm. The track is 14 cm wide. Find the diameter of the outer circle of the track.

1.84 cm
2.168 cm
3.336 cm
4.77 cm

ANS.2

Explanation :
Let inner radius = A; then 2pr = 440 so p = 70
Radius of outer circle = 70 + 14 = 84 cm
Outer diameter = 2 × Radius = 2 × 84 = 168

Q.
A race track is in the form of a ring whose inner and outer circumference are 352 metre and 396 metre respectively. Find the width of the track.

1.7
2.14
3.14pi
4.7pi

ANS.1

Explanation :
Width = R – r = 396/2π – 352/2π
so (R – r) = 44/2π = 7 meters

Q.
The outer circumference of a circular track is 220 metre. The track is 7 metre wide everywhere. Calculate the cost of levelling the track at the rate of 50 paise per square metre.

1.1556.5
2.3113
3.593
4.693

ANS.4

Explanation :
Let outer radius = R; then inner radius = r = R – 7
2pR = 220 so 35m;
r = 35 – 7 = 28 m
Area of torch = pR2 – pr2 p(R2 – r2) = 1386 m2
Cost of traveling it = 1386 × = ` 693

Q.
Find the area of a quadrant of a circle in cm2 whose circumference is 44 cm

1.77
2.38.5
3.19.25
4.19.25pi

ANS.2

Explanation :
Circumference of circle = 2pr = 44
= r = 7 cm
Area of a quadrant = π * r2 / 4 = 38.5 cm2

Q.
A pit 7.5 metre long, 6 metre wide and 1.5 metre deep is dug in a field. Find the volume of soil in m3 removed in cubic metres

1.135
2.101.25
3.50.625
4.67.5

ANS.4

Explanation :
Volume of soil removed = l × b × h

Q.
The length, breadth and height of a room are in the ratio of 3 : 2 : 1. If its volume be 1296 m3 , find its breadth

1.18 metres
2.18 met
3.16 metres
4.12 metres

ANS.4

Explanation :
(d)
Let the common ratio be = x
Then; length = 3x, breadth = 2x and height = x
Then; as per question 3x * 2x * x = 1296 so 6×3 = 1296
fi x = 6 m
Breadth = 2x = 12 m

Q.
The volume of a cube is 216 cm3 . Part of this cube is then melted to form a cylinder of length 8 cm. Find the volume of the cylinder in cm3

1.342
2.216
3.36
4.data insufficient

ANS.4

Explanation :
Data is inadequate as it’s not mentioned that what part of the cube is melted to form cylinder

Q.
The whole surface of a rectangular block is 8788 square cm. If length, breadth and height are in the ratio of 4 : 3 : 2, find length. in cm

1.26 cm
2.52 cm
3.104 cm
4.13 cm

ANS.2

Explanation :
(b)
Let the common ratio be = x
Then, length = 4x, breadth = 3x and height = 2x
As per question;
2(4x * 3x + 3x * 2x + 2x * 4x) = 8788
2(12×2 + 6×2 + 8×2) = 8788 fi 52×2 = 8788
fi x = 13
Length = 4x = 52 cm

Q.
Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the side of the resulting cube in cm

1.11
2.12
3.13
4.24

ANS.12

Explanation :
The total volume will remain the same, let the side of the resulting cube be = a. Then,
63 + 83 + 103 = a3
fi a = cube root of 1728 = 12 cm

Q.
Find curved and total surface area of a conical flask of radius 6 cm and height 8 cm.

1.60p, 96p
2.20p, 96p
3.60p, 48p
4.30p, 48p

ANS.2

Explanation :
Slant length = l = 62+82−−−−−−√= 10 cm Then curved surface area = prl = p × 6 × 10 so 60p And total surface area = prl + pr2 so p((6 × 10) + 62) = 96p

Q.
The volume of a right circular cone is 100p cm3 and its height is 12 cm. Find its curved surface area in cm2

1.130 p
2.65 p
3.204 p
4.65

ANS.2

Explanation :
Volume of a cone = π * r2 * h / 3
Then;100p = π * r2 * 12 / 3 = fi r = 5 cm
Curved surface area = prl
l = √ h2 + r2 = √ 52 + 122 = 13
then,prl = p × 13 × 5 = 65p cm2

Q.
The diameters of two cones are equal. If their slant height be in the ratio 5 : 7, find the ratio of their curved surface areas

1.25 : 7
2.25 : 49
3.5 : 49
4.5 : 7

ANS.4

Explanation :
Let the radius of the two cones be = x cm
Let slant height of 1st cone = 5 cm and
Slant height of 2nd cone = 7 cm
Then ratio of covered surface area = 5π / 7π = 5 : 7

Q.
The curved surface area of a cone is 2376 square cm and its slant height is 18 cm. Find the diameter.

1.6 cm
2.18 cm
3.84 cm
4.12 cm

ANS.3

Explanation :
Radius = π * r * l / π * l = 2376 / 3.14 * 18 = 42 cm
Diameter = 2 × Radius = 2 × 42 = 84 cm

Q.
The ratio of radii of a cylinder to a that of a cone is 1 : 2. If their heights are equal, find the ratio of their volumes?

1.1 : 3
2.2 : 3
3.3 : 4
4.3 : 1

ANS.3

Explanation :
Let the radius of cylinder = 1(r)
Then the radius of cone be = 2(R)
Then as per question = π * r2*h / π * R2*h / 3 = 3 * π * r2*h / π * R2*h
so 3 : 4

Q.
A silver wire when bent in the form of a square, encloses an area of 484 cm2 . Now if the same wire is bent to form a circle, the area of enclosed by it would be

1.308
2.196
3.616
4.88

ANS.3

Explanation :
The perimeter would remain the same in any case.
Let one side of a square be = a cm
Then a2 = 484 fi a = 22 cm \ perimeter = 4a = 88 cm
Let the radius of the circle be = r cm
Then 2pr = 88 fi r = 14 cm
Then area = pr2 = 616 cm2

Q.
The circumference of a circle exceeds its diameter by 16.8 cm. Find the circumference of the circle

1.12.32 cm
2.49.28 cm
3.58.64 cm
4.24.64 cm

ANS.4

Explanation :
Let the radius of the circle be = p
Then 2pr – 2r = 16.8 fi r = 3.92 cm
Then 2pr = 24.6 cm

Q.
A bicycle wheel makes 5000 revolutions in moving 11 km. What is the radius of the wheel?

1.70 cm
2.135 cm
3.17.5 cm
4.35 cm

ANS.4

Explanation :
Let the radius of the wheel be = p
Then 5000 × 2pr = 1100000 cm fi r = 35 cm

Q.
The volume of a right circular cone is 100p cm3 and its height is 12 cm. Find its slant height

1.13 cm
2.16 cm
3.9 cm
4.26 cm

ANS.1

Explanation :
Let the slant height be = l
Then v = π * r2 * h/3 = so r = √ 3v/πh = fi = 5 cm
l = √(h2 + l2) = √(122 + 52) = 13 cm

Q.
The short and the long hands of a clock are 4 cm and 6 cm long respectively. What will be sum of distances travelled by their tips in 4 days? (Take p = 3.14)

1.954.56 cm
2.3818.24 cm
3.2909.12 cm
4.2703.56 cm

ANS.2

Explanation :
In 4 days, the short hand covers its circumference
4 × 2 = 8 times long hand covers its circumference
4 × 24 = 96 times
Then they will cover a total distance of:-
(2 × p × 4)8 + (2 × p × 6)96 fi 3818.24 cm

Q.
The outer and inner diameters of a spherical shell are 10 cm and 9 cm respectively. Find the volume of the metal contained in the shell. (Use p = 22/7) in cm3

1.6956
2.141.95
3.283.9
4.478.3

ANS.2

Explanation :
Inner radius(p) = 9/2 = 4.5 cm
Outer radius (R) = 10/2 = 5 cm
Volume of metal contained in the shell = 4πR3 – 4πr3 = 141.9
fi 141.9 cm3

Q.
The radii of two spheres are in the ratio of 1 : 2. Find the ratio of their surface areas

1.1 : 3
2.2 : 3
3.1 : 4
4.3 : 4

ANS.3

Explanation :
Let smaller radius (r) = 1
Then bigger radius (R) = 2
Then, as per question
4πR2 / 4πr2 = (1/2)2 = 1 : 4

Q.
A sphere of radius r has the same volume as that of a cone with a circular base of radius r. Find the height of cone

1.2r
2.r/3
3.4r
4.(2/3)r

ANS.3

Explanation :
As per question 4πr3/3 = πr2h / 3 = 4r

Q.
Find the number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm, required to construct a wall 12 m long, 5 m high and 0.25 m thick, while the sand and cement mixture occupies 5% of the total volume of wall.

1.6080
2.3040
3.1520
4.12160

ANS.1

Explanation :
Volume of wall = 1200 × 500 × 25 = 15000000 cm3
Volume of cement = 5% of 15000000 = 750000 cm3
Remaining volume = 15000000 – 750000 = 14250000 cm3
Volume of a brick = 25 × 12.5 × 7.5 = 2343.75 cm3
Number of bricks used = 14250000 / 2343.75 = 6080