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Unit 14 Brain Teaser | Mensuration | Class-8 DAV Secondary Mathematics
A1. Tick (✓) the correct option.
(i) The volume of a cube of side 0.01 m in cm2 is
(a) 0.001 (b) 1 (c) 0.0001 (d) 0.000001
(ii) If the height of a cylinder is halved. its volume will be
(a) 1/2 times (b) 1/3 times (c) 2 times (d) 3 times
(iii) The area of a trapezium having two parallel sides as 10 cm and 12 cm and height as 4 cm is.
(a) 42 cm2 (b) 44 cm2 (c) 46 cm2 (d) 48 cm2
(iv) The number of faces a tetrahedron has
(a) 14 (b) 12 (c) 8 (d) 4
(v) If the length of side of a cube doubled, then the ratio of volumes of new cube and original cube is
(a) 1: 2 (b) 2 : 1
(c) 4 : 1 (d) 8 : 1
QB. Answer the following questions.
(i) Find the number of edges of a polyhedron having 20 faces and 12 vertices.
(ii) Find the surface area of a cube whose volume is 729 m3.
(iii) Volume of a cylinder is 2,376 cm3. If the diameter of its base is 12cm, then find its height.
(iv) Draw the solid represented by the views given below:
(iv) Find the total surface area of a cuboid of length 10cm, breadth 8 cm and height 6 cm.
Q2. Find the area of given quadrilateral.
Q3. A field is in the shape of a trapezium. One of its parallel sides is twice the other parallel side and distance between them is 100 m. If area of the field is 10,500 m2, find the length of the parallel sides.
Q4. Two parallel sides DC and AB of a trapezium are 12 cm and 36 cm respectively. If non-parallel sides are each equal to 15 cm, find the area of the trapezium.
Q5. The length, breadth and height of a cuboidal box are 2 m 10 cm, 1 m and 80 cm respectively. Find the area of canvas required to cover this box.
Answer 5:
l = 2 m 10 cm = 2.1 m, b = 1 m, h = 80 cm = 0.8 m.
Canvas = total surface area = 2(lb + bh + hl) = 2(2.1 × 1 + 1 × 0.8 + 0.8 × 2.1)
= 2(2.1 + 0.8 + 1.68) = 2 × 4.58 = 9.16 m².
Q6. A metallic pipe is 0.7 cm thick. The inner radius of the pipe is 3.5 cm and length is 5 dm. Find its total surface area.
Q7. A cylinder of curved surface area 1250 m2 is formed from a rectangular metallic sheet. Find the dimensions of the rectangular sheet if its length is double its breadth.
Q8. Diameter and length of a roller are 84 cm and 120 cm respectively. In how many revolutions, can the roller level the playground of area 1584 m2?
Q9. Water is pouring into reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Q10. An iron pipe is 21 m long and its exterior diameter is 8 cm. If the thickness of the pipe is 1cm and iron weight is 8 g/cm3, find the weight of the pipe.
Q11. A well is dug 20 m deep and it has a diameter 7 m. The earth which is so dug out is spread evenly on a rectangular plot 22m long and 14 m broad. What is the height of the platform formed?
Q12. The volume of a right circular cylinder is 448π cm3 and height 7 cm. Find the lateral surface area total surface area.
Answer 12: (radius is 8 cm, not 7 cm)
Volume = πr²h ⇒ πr² × 7 = 448π ⇒ r² = 64 ⇒ r = 8 cm.
Lateral (curved) surface area = 2πrh = 2 × 22/7 × 8 × 7 = 352 cm².
Total surface area = 2πr(r + h) = 2 × 22/7 × 8 × 15 = 5280/7 ≈ 754.29 cm² (= 240π cm²).
Q13. Verify Euler’s formula for
(i) Square pyramid
(ii) Triangular prism
(iii) Rectangular prism.
HOTS (Higher Order Thinking Skills) Questions
HOTS 1: A rectangular metal sheet of dimensions 44 cm × 18 cm is rolled along its length without overlapping to form a right circular cylinder. Find the volume of the cylinder formed.
Solution:
When rolled along length (44 cm):
Circumference of circular base = 44 cm
2πr = 44 → 2 × (22/7) × r = 44 → r = (44 × 7) / 44 = 7 cm.
Height of cylinder h = width of sheet = 18 cm.
Volume of cylinder = π r2 h
= (22/7) × 7 × 7 × 18
= 22 × 7 × 18 = 154 × 18 = 2,772 cm3.
HOTS 2: The diagonal of a quadrilateral shaped plot is 24 m and the perpendicular offsets dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the plot.
Solution:
Area of general quadrilateral = (1/2) × diagonal × (h1 + h2)
= (1/2) × 24 × (8 + 13)
= 12 × 21 = 252 m2.
HOTS 3: A circular well of diameter 14 m is dug 15 m deep. The earth dug out of it is spread evenly all around the well to a width of 7 m to form an elevated circular embankment. Find the height of the embankment.
Solution:
Radius of well r = 14 / 2 = 7 m.
Depth of well h = 15 m.
Volume of earth taken out = π r2 h = (22/7) × 7 × 7 × 15 = 22 × 105 = 2,310 m3.
The embankment forms a circular ring around the well:
Inner radius r1 = 7 m.
Outer radius r2 = 7 + 7 = 14 m.
Area of embankment ring = π(r22 – r12) = (22/7) × (142 – 72)
= (22/7) × (196 – 49) = (22/7) × 147 = 22 × 21 = 462 m2.
Height of embankment H = Volume of earth / Area of ring
H = 2,310 / 462 = 5 metres.
Value Based Questions (VBQs)
Question: A residential society constructs a cylindrical rainwater harvesting well of diameter 2 m and depth 7 m to store rainwater runoff from rooftop catchments.
(a) Calculate the water holding capacity of the harvesting well in litres (1 m3 = 1000 litres).
Solution:
Radius r = 2 / 2 = 1 m, Height h = 7 m.
Volume = π r2 h = (22/7) × 1 × 1 × 7 = 22 m3.
Capacity in litres = 22 × 1,000 = 22,000 litres.
(b) Why is rainwater harvesting crucial for sustainable urban communities?
Solution:
Recharges falling groundwater levels, mitigates seasonal urban water shortages, reduces flood runoff, and fosters eco-friendly living.
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