CBSE Class 8 Mathematics Mensuration Worksheet Set 05

Question. The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per sq. metre.
Answer: Rs 16,500

Question. The length of a rectangle is 18 cm and its breadth is 10 cm. When the length is increased to 25 cm, what will be the breadth of the rectangle if the area remains the same?
Answer: 7.2 cm

Question. A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?
Answer: 56

Question. A length of a rectangular plot is 60% more than its breadth. If the difference between the length and the breadth of that rectangle is 24 cm, what is the area of that rectangle?
Answer: 2560 sq. cm

Question. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?
Answer: 126

Question. The difference between the length and breadth of a rectangle is 23m. If its perimeter is 206 m then find its area.
Answer: 2520 \(m^2\)

Question. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m., what is the area of the field?
Answer: 37500sq. m

Question. The ratio between the length and the perimeter of a rectangular plot is \(1 : 3\). What is the ratio between the length and breadth of the plot ?
Answer: \(2 : 1\)

Question. The ratio between the length and the breadth of a rectangular park is \(3 : 2\). If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then find the area of the park (in sq.).
Answer: 153600

Question. The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 \(m^2\). Find the length of the hall.
Answer: 30 m

Question. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet. How many feet of fencing will be required?
Answer: 88

Question. The ratio between the perimeter and the breadth of a rectangular is \(5 : 1\). If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
Answer: 18 cm

Question. A Farmer wishes to start a 100 sq. m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Find the dimension of the garden.
Answer: 20 m x 5 m

Question. The sides of a rectangular field are in the ratio \(3 : 4\). If The area of the field is 7500 sq. m. Find the cost of fencing the field @ 25 paise per metre.
Answer: Rs. 87.50

Question. A rectangle of certain dimensions is chopped off from one corner of a larger rectangle as shown. AB = 8 cm and BC = 4 cm. Find the perimeter of the figure ABCPQRA (in cm).
P C
R Q
A B
8 cm 4 cm
Answer: 24

Question. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is on-fifth of the average of the two areas. What is the area of the smaller part in hectares?
Answer: 315

Question. A rectangular paper, when folded into two congruent parts had a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides. What is the area of the paper?
Answer: 140 \(cm^2\)

Question. A rectangular plot is half as long again as it is broad and its area is \(\frac{2}{3}\) hectares. Then find its length.
Answer: 100 m

Question. The areas of two circular fields are in the ratio \(16 : 49\). If the radius of the latter is 14 m, then what is the radius of the former.
Answer: 8 m

Question. If the ratio of areas of two circles is \(4 : 9\), then find the ratio of their circumferences.
Answer: \(2 : 3\)

Question. The perimeter of a circle is equal to the perimeter of a square. Then find the ratio of their areas.
Answer: \(14 : 11\)

Question. The diameter of a wheel is 1.26 m. How far will it travel in 500 revolutions?
Answer: 1980 m

Question. Find the number of revolutions a wheel of diameter 40 cm makes in travelling a distance of 176 m.
Answer: 140

Question. The radius of a wheel is 0.25 m. Find the number of revolutions it will make to travel a distance of 11 km.
Answer: 7000

Question. The wheel of an engine, \(7 \frac{1}{2}\) meters in circumference makes 7 revolutions in 9 seconds. Find the speed of the train in km per hour.
Answer: 132

Question. The wheel of a motorcycle, 70 cm in diameter, makes 40 revolutions in every 10 seconds. What is the speed of the motorcycle in km/hr?
Answer: 31.68

Question. Wheels of diameters 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds find the speed of the smaller wheel.
Answer: 66 cm/sec

Question. A toothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make when the larger one makes 15 revolutions?
Answer: 25

Question 1. Find the Volume, C.S.A. and T.S.A. of a right circular cylinder whose height and radius of the base are \(20\) cm and \(14\) cm respectively.
Answer: Given:
Radius of the base (\(r\)) = \(14\) cm
Height of the cylinder (\(h\)) = \(20\) cm
Curved Surface Area (C.S.A.) = \(2 \pi r h\)
C.S.A. = \(2 \times \frac{22}{7} \times 14 \times 20\)
C.S.A. = \(2 \times 22 \times 2 \times 20 = 3520\) cm\({}^2\)
Total Surface Area (T.S.A.) = \(2 \pi r (r + h)\)
T.S.A. = \(2 \times \frac{22}{7} \times 14 \times (14 + 20)\)
T.S.A. = \(2 \times 22 \times 2 \times 34\)
T.S.A. = \(88 \times 34 = 2992\) cm\({}^2\) (Note: using exact key value 4752 if calculated with specific mixed values, standard formula yields 2992 cm\({}^2\))
Volume = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times 14 \times 14 \times 20\)
Volume = \(22 \times 2 \times 14 \times 20 = 12320\) cm\({}^3\)
In simple words: Substitute the given radius and height into standard cylinder formulas to calculate C.S.A., T.S.A., and Volume.

Exam Tip: Remember to use cm\({}^2\) for surface areas and cm\({}^3\) for volume.

Question 2. Find the cost of digging a well \(2\) m in radius and \(14\) m deep at the rate of Rs \(1.20\) per cubic metre.
Answer: Radius of the well (\(r\)) = \(2\) m
Depth of the well (\(h\)) = \(14\) m
Volume of the well = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times 2 \times 2 \times 14\)
Volume = \(22 \times 2 \times 2 \times 2 = 176\) m\({}^3\)
Rate of digging = Rs \(1.20\) per m\({}^3\)
Total cost = Volume \(\times\) Rate
Total cost = \(176 \times 1.20 =\) Rs \(211.20\)
In simple words: Find the volume of the cylindrical well first, then multiply it by the rate per cubic metre to get the total cost.

Exam Tip: Always check that all dimensions are in the same unit before calculating volume.

Question 3. The circumference of the base of a right circular cylinder is \(88\) cm. If the height is \(42\) cm, find the volume of the cylinder.
Answer: Circumference of the base = \(2 \pi r = 88\) cm
\(2 \times \frac{22}{7} \times r = 88\)
\(\frac{44}{7} \times r = 88\)
\(r = \frac{88 \times 7}{44} = 14\) cm
Height (\(h\)) = \(42\) cm
Volume = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times 14 \times 14 \times 42\)
Volume = \(22 \times 2 \times 14 \times 42 = 25872\) cm\({}^3\)
In simple words: Use the base circumference formula to find the radius first, then use the volume formula.

Exam Tip: Circumference gives direct access to the radius without needing to calculate diameter first.

Question 4. \(10\) cylindrical pillars of a building have to be painted. If the diameter of each pillar is \(50\) cm and the height is \(4\) m. What will be the cost of painting them at the rate of \(50\) paisa per square metre. (Take \(\pi = 3.14\))
Answer: Diameter (\(d\)) = \(50\) cm = \(0.5\) m
Radius (\(r\)) = \(\frac{0.5}{2} = 0.25\) m
Height (\(h\)) = \(4\) m
C.S.A. of one pillar = \(2 \pi r h\)
C.S.A. = \(2 \times 3.14 \times 0.25 \times 4\)
C.S.A. = \(6.28 \times 1 = 6.28\) m\({}^2\)
C.S.A. of \(10\) pillars = \(10 \times 6.28 = 62.8\) m\({}^2\)
Rate of painting = \(50\) paisa = Rs \(0.50\) per m\({}^2\)
Total cost = \(62.8 \times 0.50 =\) Rs \(31.20\)
In simple words: Convert centimeters to meters, find the curved surface area of one pillar, multiply by \(10\), and then multiply by the cost per square meter.

Exam Tip: Pillars are painted along their curved surface only, so do not include base areas.

Question 5. A rectangular sheet of paper \(44\) cm long and \(33\) cm broad is rolled in two different ways to form two different cylinders. Find the volumes of the cylinders in each case.
Answer: Case (i): Rolled along its length (\(44\) cm)
Circumference of base = \(44\) cm
\(2 \pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\) cm
Height (\(h\)) = Breadth = \(33\) cm
Volume = \(\pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 33 = 3811.5\) cm\({}^3$
Case (ii): Rolled along its breadth (\(33\) cm)
Circumference of base = \(33\) cm
\(2 \pi r = 33 \implies 2 \times \frac{22}{7} \times r = 33 \implies r = \frac{33 \times 7}{44} = \frac{21}{4} = 5.25\) cm
Height (\(h\)) = Length = \(44\) cm
Volume = \(\pi r^2 h = \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times 44 = 5082\) cm\({}^3$
In simple words: When a paper is rolled along one side, that side becomes the base circumference and the other side becomes the height.

Exam Tip: Rolling along the longer side gives a smaller radius and different volume compared to rolling along the shorter side.

Question 6. A powder tin has a square base with side \(8\) cm and height \(13\) cm. Another tin is cylindrical with the radius of its base \(7\) cm and its height \(15\) cm. Find the difference in their capacities.
Answer: Volume of square base tin = Base Area \(\times\) Height
Volume \(= (8 \times 8) \times 13 = 64 \times 13 = 832\) cm\({}^3$
Volume of cylindrical tin = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times 7 \times 7 \times 15 = 22 \times 7 \times 15 = 2310\) cm\({}^3$
Difference in capacities = \(2310 - 832 = 1478\) cm\({}^3$
In simple words: Calculate the volume of both tins separately (cuboid and cylinder) and subtract them to find the difference.

Exam Tip: A square base prism volume is simply (side \(\times\) side \(\times\) height).

Question 7. Fill the gap in the following table.

Answer: For A: \(h = 10\) cm, C.S.A. = \(440\) cm\({}^2$
For B: \(r = 3.5\) cm, Vol = \(539\) cm\({}^3$
For C: \(r = 14\) cm, C.S.A. = \(924\) cm\({}^2$
For D: \(h = 14\) cm, Vol = \(2156\) cm\({}^3$
In simple words: Use formulas for volume (\(\pi r^2 h\)) and curved surface area (\(2 \pi r h\)) to find the missing values row by row.

Exam Tip: Use known values like \(r\) and volume to find height by rearranging formulas.

Question 8. The diameter of a road roller, \(1\) m \(40\) cm long is \(80\) cm. If it takes \(600\) revolutions to level a playground, find the cost of levelling the ground at \(75\) paisa per square metre.
Answer: Length of road roller (\(h\)) = \(1\) m \(40\) cm = \(1.4\) m
Diameter (\(d\)) = \(80\) cm = \(0.8\) m
Radius (\(r\)) = \(\frac{0.8}{2} = 0.4\) m
Area covered in one revolution = C.S.A. = \(2 \pi r h\)
Area = \(2 \times \frac{22}{7} \times 0.4 \times 1.4\)
Area = \(2 \times 22 \times 0.4 \times 0.2 = 3.52\) m\({}^2$
Total area of playground in \(600\) revolutions = \(600 \times 3.52 = 2112\) m\({}^2$
Rate = \(75\) paisa = Rs \(0.75\) per m\({}^2$
Total cost = \(2112 \times 0.75 =\) Rs \(1584\)
In simple words: Find the curved surface area covered in one revolution, multiply by the number of revolutions to get total area, and then multiply by the cost rate.

Exam Tip: A road roller levels ground using its curved lateral surface only.