CBSE Class 8 Mathematics Mensuration Assignment Set 07

Question. A small cube is cut off from a larger cube as shown below
What is the number of FACES and EDGES in the cut cube as shown above?

(a) Faces: 6 Edges: 12
(b) Faces: 9 Edges: 21
(c) Faces: 9 Edges: 12
(d) Faces: 6 Edges: 18

Answer : B

Question. What is the area of this shape?

(a) 19 sq. cm
(b) 20 sq. cm
(c) 21 sq. cm
(d) 24 sq. cm

Answer : B

Question. The diagonals of quadrilateral PQRS intersect at O.
Which of the following is NOT a correct way of finding the Area of quadrilateral PQRS?

(a) Area (ΔPQR) + Area (ΔPSR)
(b) Area (ΔPOS) + Area (ΔSOR) + Area (ΔPQR)
(c) Area (ΔPOQ) + Area (ΔROS) + Area (ΔROQ) + Area (ΔPOS)
(d) Area (ΔPQR) + Area (ΔPQS) + Area (ΔSOR)

Answer : D

Question. If the area of a circle of radius 7 cm is 154 sq cm, what is the area of the shaded portion of the figure shown below?

(a) 9 sq cm
(b) 10.5sq cm
(c) 12.25 sq cm
(d) 16.5sq cm

Answer : B

Question. The solid shown below is made of 10 small cubes of side 2 cm each. What is the total volume of the solid?

(a) 20 sq cm
(b) 20 cu cm
(c) 40 sq cm
(d) 80 cu cm

Answer : D

Question. Which of the following can NOT be used to measure the capacity of a container?
(a) dm³
(b) litre
(c) cm²
(d) ml

Answer : C

Question. The box is divided into four separate areas P, Q, R and S. In which area should 38 be placed?

(a) P
(b) Q
(c) R
(d) S

Answer : D

Question. What is the area of the shaded part of the rectangle?

(a) 4 sq cm
(b) 8 sq cm
(c) 9 sq cm
(d) 12 sq cm

Answer : C

Question. What is the area of the shaded portion if the area of the circle is 64 cm²
Note: PQ is a straight line passing through the centre of the circle.

(a) 8cm2
(b) 16 cm2
(c) 16.8 cm2
(d) 28 cm2

Answer : B

Question. A part of one side of a vessel is shown below. Using this beaker which of these measurements can be made EXACTLY?

(a) 121.5 ml
(b) 124 ml
(c) 125 ml
(d) 128 ml

Answer : C

Question. What is the area of the shaded part of the square?

(a) 2a² - ab
(b) 4a  -  ab
(c) 4a² - ab
(d) 2a² + ab

Answer : C

Question. Consider the two figures given below. Which of the following is true about their areas and perimeters?

(a) The areas are the same, perimeters are different.
(b) The areas are different, perimeters are the same.
(c) The areas and perimeters are both different.
(d) The areas and perimeters are both the same.

Answer : B

Question. Of the pieces shown here, which two could form a semicircle when placed next to each other with their edges touching?

(a) Z and W
(b) W and Y
(c) Y and Z
(d) Z and X

Answer : B

Question. The capacity of a closed cylindrical water tank is 9.24 kilolitres. If the height of the cylinder is 1.5 m, what is its radius?
(A) 1.4 m
(B) 14 m
(C) 7m
(D) 0.7 m

Answer : A

Question. The thickness of a hollow metallic cylinder is 2 cm. It is 70 cm long with outer radius of 14 cm. Find the volume of the metal used in making the cylinder, assuming that it is open at both the ends. Also find its weight if the metal weighs 8 g per cm³.
(A) 10440 cm³, 91250 g
(B) 13440 cm³, 90000 g
(C) 11440 cm³, 91520 g
(D) 12440 cm³, 91550 g

Answer : C

Question. A wooden box (including the lid) has external dimensions 40 cm by 34 cm by 30 cm. If the wood is 1 cm thick, then how many cm³ of wood is used in it?
(A) 6752 cm³
(B) 6750 cm³
(C) 5752 cm³
(D) 5750 cm³

Answer : A

Question. How many bricks of size 22 cm × 10 cm × 7 cm are required to construct a wall 11m
long, 3.5 m high and 40 cm thick, if the cement and sand used in the construction occupy (1/10)th part of the wall?
(A) 8000
(B) 9000
(C) 7000
(D) 10000

Answer : B

Question. A rectangular block of ice measures 40 cm by 25 cm by 15 cm. Calculate its weight (in kg), if ice weighs 9/10 of the weight of the same volume of water and 1 cm3 of water weighs 1 gm.
(A) 9
(B) 13.5
(C) 8
(D) 9.5

Answer : B

Question. A cylindrical tower is 5 metres in diameter and 14 metres high. The cost of white washing its curved surface at 50 paise per m² is ____.
(A) ₹ 90
(B) ₹ 97
(C) ₹ 110
(D) ₹ 95

Answer : C

Question. A square garden measuring 8 m on a side is surrounded by a 1 m wide path. What is the area of the path ?

(A) 8 m²
(B) 9 m²
(C) 28 m²
(D) 36 m²

Answer : D

Question. The radii of the bases of two cylinders are in the ratio of 1 : 2 respectively and their heights are in the ratio 2 : 1 respectively. Find the ratio of their volumes.
(A) 1 : 2
(B) 1 : 4
(C) 2 : 1
(D) 4 : 1

Answer : A

Question. Find the sum of the lengths of the parallel sides of a trapezium whose altitude is 11 cm and whose area is 0.55 m2.
(A) 25 m
(B) 15 m
(C) 12 m
(D) 10 m

Answer : D

Question. In the given figure, RSTV is a square inscribed in a circle with centre O and radius r. The total area of the shaded region is _____.

(A) r² (p – 2)
(B) 2r² (2 – p)
(C) p (r² – 2)
(D) 8r² – 8r

Answer : A

Question. ABC is a right angled triangle with ∠BAC = 90°. AH is drawn perpendicular to BC. If AB = 60 cm and AC = 80 cm, then BH = ____.

(A) 36 cm
(B) 32 cm
(C) 24 cm
(D) 30 cm

Answer : A

Question. Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted without any loss of metal into a single cube. The edge of the new cube is ____.
(A) 6 cm
(B) 12 cm
(C) 9 cm
(D) 10 cm

Answer : A

Question. Two rectangles ABCD and DBEF are as shown in the figure. The area of rectangle DBEF (in square units) is _____.

(A) 10
(B) 12
(C) 14
(D) 15

Answer : B

Question. The ratio of areas of two squares, if diagonal of one is double the diagonal of the other, is ____.
(A) 3 : 2
(B) 4 : 1
(C) 3 : 1
(D) 4 : 3

Answer : B

Question. Ratio of surface areas of two cubes is 25 : 36. Find the ratio of their volumes.
(A) 5 : 6
(B) 125 : 216
(C) 1 : 2
(D) 64 : 216

Answer : B

Question. The area of a quadrilateral is 342 sq. m. The perpendiculars from two of its opposite vertices to the diagonal are 12 m and 12 m. What is the length of the diagonal?
(A) 28.6 cm
(B) 25.3 cm
(C) 28.5 cm
(D) 22.5 cm

Answer : C

Question. The volume of a cuboid is 440 cm3 and the area of its base is 88 cm². Find its height.
(A) 5 cm
(B) 11 cm
(C) 4 cm
(D) 8 cm

Answer : A

Question. A hoop is resting vertically at staircase as shown in the diagram. AB = 12 cm and BC = 8 cm. The radius of the hoop is _____.

(A) 13cm
(B) 12√2cm
(C) 14 cm
(D) 13√2cm

Answer : A

Question. A well 12 m deep with a diameter 3.5 m is dug up and earth from it is evenly spread to form a platform 10.5 m long and 8.8 m wide. Find the height of the platform.
(A) 2.5 m
(B) 12.5 m
(C) 1.25 m
(D) 1.5 m

Answer : C

Question. A room is 9 m long, 6 m wide and 4 m high. Find the cost of plastering its walls and ceiling at the rate of ₹ 2.50 per square metre.
(A) ₹ 435
(B) ₹ 600
(C) ₹ 502
(D) ₹ 354

Answer : A

Question. A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
(A) 112.5 m
(B) 112 m
(C) 212 m
(D) 312 m

Answer : B

Question. Find the area of the given field. All dimensions are in m.

(A) 9000 m²
(B) 11700 m²
(C) 11000 m²
(D) 12000 m²

Answer : B

Question. Fill in the blanks.
(i) The perimeter of a rectangle becomes _______ times its original perimeter, if its length and breadth are doubled.
(ii) The curved surface area of a cylinder is reduced by _______ percent if the height is half of the original height.
(iii) Opposite faces of a cuboid are _______ in area.
(iv) If the diagonal d of a quadrilateral is doubled and the heights h1 and h2 falling on d are halved, then the new area of quadrilateral is _______.

Answer : A

Question. Match the figure in column I with their total surface area in column II

(A) P → (iii); Q → (ii); R → (iv); S → (i)
(B) P → (ii); Q → (iv); R → (i); S → (iii)
(C) P → (ii); Q → (iii); R → (i); S → (iv)
(D) P → (ii); Q → (iv); R → (iii); S → (i)

Answer : B

Question. The area of a trapezium with equal nonparallel sides is 168 m2. If the lengths of the parallel sides are 36 m and 20 m, find the length of the non-parallel sides.
(A) 8 m
(B) 10 m
(C) 15 m
(D) 4 m

Answer : B

VIII - Mathematics Assignment No. 05 - Mensuration

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 data:
Radius of the base (\(r\)) = 14 cm
Height of the cylinder (\(h\)) = 20 cm
Curved Surface Area (C.S.A.) = \(2 \pi r h = 2 \times \frac{22}{7} \times 14 \times 20\)
\(= 2 \times 22 \times 2 \times 20 = 1760 \text{ cm}^2\)
Total Surface Area (T.S.A.) = \(2 \pi r (r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 20)\)
\(= 88 \times 34 = 2992 \text{ cm}^2\)
Volume = \(\pi r^2 h = \frac{22}{7} \times 14 \times 14 \times 20\)
\(= 22 \times 2 \times 14 \times 20 = 12320 \text{ cm}^3\)
In simple words: We used standard cylinder formulas for curved area, total area, and volume by substituting the given radius and height.

Exam Tip: Remember that T.S.A. includes the area of both circular bases, whereas C.S.A. only includes the curved wall.

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 earth dug out = \(\pi r^2 h = \frac{22}{7} \times 2 \times 2 \times 14\)
\(= 22 \times 2 \times 2 \times 2 = 176 \text{ m}^3\)
Rate of digging = Rs 1.20 per cubic metre
Total Cost = \(176 \times 1.20 = \text{Rs } 211.20\)
In simple words: First find the volume of the cylindrical well, then multiply it by the cost per cubic metre.

Exam Tip: Always verify unit consistency. Here dimensions are in metres and rate is per cubic metre, so direct multiplication applies.

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 = \(2 \pi r = 88 \text{ cm}\)
\(2 \times \frac{22}{7} \times r = 88\)
\(\frac{44}{7} \times r = 88\)
\(r = \frac{88 \times 7}{44} = 14 \text{ cm}\)
Height (\(h\)) = 42 cm
Volume = \(\pi r^2 h = \frac{22}{7} \times 14 \times 14 \times 42\)
\(= 22 \times 2 \times 14 \times 42 = 25872 \text{ cm}^3\)
In simple words: Find the radius from the given base circumference, then calculate the volume using the height.

Exam Tip: Use the base circumference formula \(2\pi r\) to find the radius before calculating surface area or volume.

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 of each pillar = 50 cm = 0.5 m
Radius of each pillar (\(r\)) = \(\frac{0.5}{2} = 0.25 \text{ m}\)
Height of each pillar (\(h\)) = 4 m
Curved surface area of one pillar = \(2 \pi r h = 2 \times 3.14 \times 0.25 \times 4\)
\(= 2 \times 3.14 \times 1 = 6.28 \text{ m}^2\)
Curved surface area of 10 pillars = \(10 \times 6.28 = 62.8 \text{ m}^2\)
Rate of painting = 50 paisa = Rs 0.50 per square metre
Total Cost = \(62.8 \times 0.50 = \text{Rs } 31.40\)
In simple words: Calculate the curved area of one pillar, multiply by 10, convert units to metres, and multiply by the painting rate.

Exam Tip: Convert centimeters to metres early to avoid confusion when the rate is given in square metres.

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.
Circumference of the base = Length = 44 cm
\(2 \pi r = 44\)
\(2 \times \frac{22}{7} \times r = 44 \implies r = 7 \text{ cm}\)
Height (\(h\)) = Breadth = 33 cm
Volume = \(\pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 33 = 3811.5 \text{ cm}^3\)
Case (ii): Rolled along its breadth.
Circumference of the base = Breadth = 33 cm
\(2 \pi r = 33\)
\(2 \times \frac{22}{7} \times r = 33 \implies r = \frac{33 \times 7}{44} = \frac{21}{4} \text{ cm}\)
Height (\(h\)) = Length = 44 cm
Volume = \(\pi r^2 h = \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times 44 = 5082 \text{ cm}^3\)
In simple words: Rolling a sheet along different sides changes the base circumference and height, resulting in two different volumes.

Exam Tip: When a rectangular sheet is rolled along a side, that side becomes the circumference of the circular base, and the other side becomes the height.

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: Capacity (Volume) of the square-based tin = Area of base \(\times\) height
\(= \text{side} \times \text{side} \times \text{height} = 8 \times 8 \times 13 = 832 \text{ cm}^3\)
Capacity (Volume) of the cylindrical tin = \(\pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 15\)
\(= 22 \times 7 \times 15 = 2310 \text{ cm}^3\)
Difference in capacities = \(2310 - 832 = 1478 \text{ cm}^3\)
In simple words: Find the volume of both tins separately and subtract the smaller volume from the larger one.

Exam Tip: A square-based tin is a cuboid whose base area is side squared.

Fill the gap in the following table.

Question 7. Fill the gap in the following table.
Answer:
For Cylinder A: Given \(r = 7\text{ cm}\), \(\text{Vol} = 1540\text{ cm}^3\). Using \(\text{Vol} = \pi r^2 h\), \(1540 = \frac{22}{7} \times 7^2 \times h \implies h = 10\text{ cm}\). Then \(\text{C.S.A.} = 2 \pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440\text{ cm}^2$.
For Cylinder B: Given \(h = 14\text{ cm}\), \(\text{C.S.A.} = 308\text{ cm}^2\). Using \(\text{C.S.A.} = 2 \pi r h\), \(308 = 2 \times \frac{22}{7} \times r \times 14 \implies r = 3.5\text{ cm}\). Then \(\text{Vol} = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 14 = 539\text{ cm}^3$.
For Cylinder C: Given \(h = 10.5\text{ cm}\), \(\text{Vol} = 6468\text{ cm}^3\). Using \(\text{Vol} = \pi r^2 h\), \(6468 = \frac{22}{7} \times r^2 \times 10.5 \implies r^2 = 196 \implies r = 14\text{ cm}\). Then \(\text{C.S.A.} = 2 \pi r h = 2 \times \frac{22}{7} \times 14 \times 10.5 = 924\text{ cm}^2$.
For Cylinder D: Given \(r = 7\text{ cm}\), \(\text{C.S.A.} = 616\text{ cm}^2\). Using \(\text{C.S.A.} = 2 \pi r h\), \(616 = 2 \times \frac{22}{7} \times 7 \times h \implies h = 14\text{ cm}\). Then \(\text{Vol} = \pi r^2 h = \frac{22}{7} \times 7^2 \times 14 = 2156\text{ cm}^3\).
In simple words: Use the formulas for volume and curved surface area to find the missing radius or height step by step.

Exam Tip: Use the formula that contains only one unknown variable to solve each table entry efficiently.

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 roller (\(h\)) = 1 m 40 cm = 1.4 m
Diameter of roller = 80 cm \(\implies\) Radius (\(r\)) = 40 cm = 0.4 m
Area covered in one revolution = C.S.A. of cylinder = \(2 \pi r h = 2 \times \frac{22}{7} \times 0.4 \times 1.4\)
\(= 2 \times \frac{22}{7} \times 0.4 \times \frac{7}{10} = 3.52 \text{ m}^2\)
Area covered in 600 revolutions = \(600 \times 3.52 = 2112 \text{ m}^2\)
Rate of levelling = 75 paisa = Rs 0.75 per square metre
Total Cost = \(2112 \times 0.75 = \text{Rs } 1584\)
In simple words: Find the curved area covered in one full rotation, multiply by 600 for the total area, and then multiply by the rate.

Exam Tip: A road roller touches the ground along its curved surface area. The length of the roller acts as the height of the cylinder.