CBSE Class 8 Mathematics Mensuration Practice Paper Worksheet Set 09 with Answers (Old NCERT Syllabus)

This Class 8 Maths Mensuration practice paper carries 40 marks and has 17 questions in five sections (A to E), each with a clear step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The paper tests areas of rhombuses and trapeziums, surface areas and volumes of cuboids, cubes and cylinders, and converting cubic units into litres in real-life problems.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Answer:
A rhombus is a parallelogram, so Area = base × altitude.
Area = 5 × 4.8 = 24 cm².
Also, Area of a rhombus = \( \frac{1}{2} \) × d₁ × d₂.
24 = \( \frac{1}{2} \) × 8 × d₂ = 4 × d₂.
d₂ = 24 ÷ 4 = 6 cm.
So the area is 24 cm² and the other diagonal is 6 cm.
In simple words: Side × height gives the area 24 cm²; half of 8 times the other diagonal must also be 24, so it is 6 cm.

Teacher's Note:
1. A rhombus has two area formulas: base × height and half the product of diagonals.
2. Check: half-diagonals 4 cm and 3 cm give a side of 5 cm (3-4-5 triangle).

Question 8. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Answer:
Volume of one matchbox = l × b × h = 4 × 2.5 × 1.5 = 15 cm³.
Volume of 12 matchboxes = 12 × 15 = 180 cm³.
So the volume of the packet is 180 cm³.
In simple words: One box holds 15 cm³; twelve boxes take 12 × 15 = 180 cm³.

Teacher's Note:
1. Multiply 4 × 2.5 first: it gives 10, which makes the rest easy.
2. The packet's volume is taken as the total volume of the boxes inside it.

Question 9. A carpenter makes a box which has a volume of 13,400 cm³. The base has an area of 670 cm². What is the height of the box?
Answer:
Volume = area of base × height.
13400 = 670 × h.
h = 13400 ÷ 670 = 20 cm.
So the height of the box is 20 cm.
In simple words: Divide the volume by the base area: 13400 ÷ 670 = 20 cm.

Teacher's Note:
1. Volume = base area × height works for every cuboid and cylinder.
2. Check: 670 × 20 = 13400.

Section C: Short Answer Questions (3 Marks Each)

Question 10. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Answer:
Here l = 80 cm, b = 48 cm and h = 24 cm.
Total surface area of one suitcase = 2(lb + bh + hl) = 2(80 × 48 + 48 × 24 + 24 × 80).
= 2(3840 + 1152 + 1920) = 2 × 6912 = 13824 cm².
Area of tarpaulin needed for one suitcase = 13824 cm². Its width is 96 cm, so the length needed = 13824 ÷ 96 = 144 cm.
For 100 suitcases: 144 × 100 = 14400 cm = 144 m.
So 144 m of tarpaulin is required.
In simple words: Each suitcase needs 13824 cm² of cloth, which is a 96 cm wide strip 144 cm long; 100 suitcases need 144 m.

Teacher's Note:
1. Area of the cloth = length × width, so length = area ÷ width.
2. Convert to metres only at the end: 100 cm = 1 m.

Question 11. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Answer:
Radius of the container, r = 14 ÷ 2 = 7 cm.
Height of the label = 20 − 2 − 2 = 16 cm.
The label covers part of the curved surface, so its area = 2πrh.
Area = 2 × \( \frac{22}{7} \) × 7 × 16 = 44 × 16 = 704 cm².
So the area of the label is 704 cm².
In simple words: The label is a band 16 cm tall wrapped around a circle of radius 7 cm, so its area is 44 × 16 = 704 cm².

Teacher's Note:
1. Subtract the 2 cm gaps at both the top and the bottom.
2. Only the curved surface is used; the top and bottom circles are not part of the label.

Question 12. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank.
Answer:
Volume of the cylinder = πr²h.
= \( \frac{22}{7} \) × 1.5 × 1.5 × 7 = 22 × 2.25 = 49.5 m³.
1 m³ = 1000 litres, so 49.5 m³ = 49.5 × 1000 = 49500 litres.
So the tank can store 49500 litres of milk.
In simple words: The tank holds 49.5 cubic metres, and each cubic metre is 1000 litres, so 49500 litres.

Teacher's Note:
1. The length of a lying cylinder plays the role of its height.
2. Remember 1 m³ = 1000 L and 1000 cm³ = 1 L.

Question 13. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.
Answer:
Take l = 2 m, b = 1 m and h = 1.5 m.
Painted area = top + four walls = lb + 2(bh + hl).
= 2 × 1 + 2(1 × 1.5 + 1.5 × 2) = 2 + 2(1.5 + 3) = 2 + 9 = 11 m².
So 11 m² of the box has to be painted.
In simple words: Paint the top (2 m²) and the four sides (9 m²): 11 m² in all.

Teacher's Note:
1. Total surface area minus the bottom: 2(2 + 1.5 + 3) − 2 = 13 − 2 = 11 m² gives the same answer.
2. Read carefully which faces are left out.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Answer:
Volume of the reservoir = 108 m³ = 108 × 1000 = 108000 litres.
Water flows in at 60 litres per minute, i.e. 60 × 60 = 3600 litres per hour.
Time = 108000 ÷ 3600 = 30 hours.
So the reservoir fills in 30 hours.
In simple words: The reservoir holds 108000 litres and gets 3600 litres every hour, so it takes 30 hours.

Teacher's Note:
1. Convert everything to the same units first (litres and hours).
2. Check: 108000 ÷ 60 = 1800 minutes = 30 hours.

Question 15. The circumference of the base of the cylindrical vessel is 132 cm, and its height is 25 cm. How many litres of water can it hold? (Assume π = \( \frac{22}{7} \))
Answer:
Circumference = 2πr = 132, so r = \( \frac{132 × 7}{2 × 22} \) = 21 cm.
Volume = πr²h = \( \frac{22}{7} \) × 21 × 21 × 25 = 22 × 3 × 21 × 25 = 34650 cm³.
1000 cm³ = 1 litre, so 34650 cm³ = 34650 ÷ 1000 = 34.65 litres.
So the vessel can hold 34.65 litres.
In simple words: The radius works out to 21 cm, the volume to 34650 cm³, which is 34.65 litres.

Teacher's Note:
1. Always find the radius first when the circumference is given.
2. Cancel 7 with 21 early: 21 ÷ 7 = 3.

Question 16. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Answer:
Width of the frame = (24 − 16) ÷ 2 = 4 cm, and also (28 − 20) ÷ 2 = 4 cm.
The frame splits into four trapeziums, marked I, II, III and IV below.

Sections I and II (the long sides) have parallel sides 28 cm and 20 cm and height 4 cm.
Area of I = \( \frac{1}{2} \) × (28 + 20) × 4 = \( \frac{1}{2} \) × 48 × 4 = 96 cm². Area of II = 96 cm².
Sections III and IV (the short sides) have parallel sides 24 cm and 16 cm and height 4 cm.
Area of III = \( \frac{1}{2} \) × (24 + 16) × 4 = \( \frac{1}{2} \) × 40 × 4 = 80 cm². Area of IV = 80 cm².
So the side sections are 96 cm² each and the top and bottom sections are 80 cm² each.
Check: 96 + 96 + 80 + 80 = 352 = 24 × 28 − 16 × 20 = 672 − 320.
In simple words: Each piece of the frame is a trapezium 4 cm wide; the long-side pieces are 96 cm² and the short-side pieces 80 cm².

Teacher's Note:
1. The width of each section is the height of each trapezium.
2. Check the total against outer area minus inner area.

Section E: Case Study Based Question (4 Marks)

Question 17. Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see the figure below). She wanted to make a kaleidoscope of length 25 cm with a 7 cm diameter. You may take π = \( \frac{22}{7} \).

Answer the following questions based on the above information:
(i) Find the radius of the kaleidoscope. (1)
(ii) Find the area of chart paper required. (1)
(iii) Find the cost of chart paper if the rate of chart paper is Rs. 2 per 100 cm². (1)
(iv) Find the volume of the kaleidoscope. (1)
Answer:
(i) Radius = diameter ÷ 2 = 7 ÷ 2 = 3.5 cm.
(ii) Chart paper needed = curved surface area = 2πrh = 2 × \( \frac{22}{7} \) × 3.5 × 25 = 22 × 25 = 550 cm².
(iii) Cost = \( \frac{550}{100} \) × Rs 2 = 5.5 × 2 = Rs 11.
(iv) Volume = πr²h = \( \frac{22}{7} \) × 3.5 × 3.5 × 25 = 11 × 3.5 × 25 = 962.5 cm³.
In simple words: Radius 3.5 cm; the paper wrapping it is 550 cm², costing Rs 11; the tube holds 962.5 cm³.

Teacher's Note:
1. The length of the tube is the height h of the cylinder.
2. 2 × \( \frac{22}{7} \) × 3.5 = 22, a handy shortcut for r = 3.5 cm.