Class 9th Maths - Surface Case Study Questions and Answers 2022 - 2023
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Surface Areas and Volumes Case Study Questions With Answer Key
9th Standard CBSE
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Reg.No. :
Mathematics
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Mathematics teacher of a school took her 9th standard students to show Gol Gumbaz. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Gol Gumbaz to students. Gol Gumbaz is the tomb of king Muhammad Adil Shah, Adil Shah Dynasty. Construction of the tomb, located in Vijayapura , Karnataka, India, was started in 1626 and completed in 1656. It reaches up to 51 meters in height while the giant dome has an external diameter of 44 meters, making it one of the largest domes ever built. At each of the four corners of the cube is a dome shaped octagonal tower seven stories high with a staircase inside.
(a) What is the total surface area of a cuboid?(i) lb + bh + hl (ii) 2(lb + bh + hl) (iii) 2(lb + bh) (iv) I2+ b2+ h2 (b) What is the curved surface area of hemispherical dome ?
(i) 908π m2 (ii) 968π m2 (iii) 340π m2 iv) 780π m2 (c) What is the height of the cubodial part ?
(i) 14 m (ii) 7 m (iii) 29 m (iv) 18 m d) What is the circumference of the base of the dome ?
(i) 34π (ii) 22π (iii) 44π (iv) 55π (v) The total surface area of a hemispherical dome having radius 7 cm is:
(a) 462 cm2 (b) 294 cm2 (c) 588 cm2 (d) 154 cm2 -
Mathematics teacher of a school took his 10th standard students to show Taj Mahal. It was a part of their Educational trip. The teacher had interest in history as well. He narrated the facts of Taj Mahal to the students. Then the teacher said in this monument one can find combination of solid figures. There are 4 pillars which are cylindrical in shape. Also, 2 domes at the back side which are hemispherical. 1 big domes at the centre. It is the finest example of the symmetry. (Use π = 22/7)
(i)How much cloth material will be required to cover 2 small domes each of radius 4.2 metres?
(a) 52.08 cm2 (b) 52.8 m2 (c) 52 m2 (d) none of these (ii) Write the formula to find the volume of one pillar (including hemispherical dome) :
(a) πr2(l + r) (b) πr2(2/3 r + h) (c) 2πr2h (d) none of these (iii) The volume of the hemispherical dome at the centre if base radius is 7 m is :
(a) 718.66 cm3 (b) 152.8 m3 (c) 718.66 m3 (d) 56 m3 (iv) What is the lateral surface area of all 4 pillars if height of the each pillar is 14 m and base radius is 1.4 m (without dome)?
(a) 508 m2 (b) 591.36 m2 (c) 52 m2 (d) none of these (v) The cost of polishing all the four pillars if the cost of 1 m2 is Rs. 270, will be :
(a) Rs. 1,59,667.20 (b) Rs. 2,00,000 (c) Rs. 1,52,567.50 (d) none of these -
Mathematics teacher of a school took his 10th standard students to show Taj Mahal. It was a part of their Educational trip. The teacher had interest in history as well. He narrated the facts of Taj Mahal to the students. Then the teacher said in this monument one can find combination of solid figures. There are 4 pillars which are cylindrical in shape. Also, 2 domes at the back side which are hemispherical. 1 big domes at the centre. It is the finest example of the symmetry. (Use π = 22/7)
(i)How much cloth material will be required to cover 2 small domes each of radius 4.2 metres?(a) 52.08 cm2 (b) 52.8 m2 (c) 52 m2 (d) none of these (ii) Write the formula to find the volume of one pillar (including hemispherical dome) :
(a) πr2(l + r) (b) πr2(2/3 r + h) (c) 2πr2h (d) none of these (iii) The volume of the hemispherical dome at the centre if base radius is 7 m is :
(a) 718.66 cm3 (b) 152.8 m3 (c) 718.66 m3 (d) 56 m3 (iv) What is the lateral surface area of all 4 pillars if height of the each pillar is 14 m and base radius is 1.4 m (without dome)?
(a) 508 m2 (b) 591.36 m2 (c) 52 m2 (d) none of these (v) The cost of polishing all the four pillars if the cost of 1 m2 is Rs. 270, will be :
(a) Rs. 1,59,667.20 (b) Rs. 2,00,000 (c) Rs. 1,52,567.50 (d) none of these -
Mathematics teacher of a school took her 9th standard students to show Red fort. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Red fort to students. Then the teacher said in this monument one can find combination of solid figures. There are 2 pillars which are cylindrical in shape. Also 2 domes at the corners which are hemispherical.7 smaller domes at the centre. Flag hoisting ceremony on Independence Day takes place near these domes.
(i) How much cloth material will be required to cover 2 big domes each of radius 2.5 metres?(a) 75 m2 (b) 78.57 m2 (c) 87.47 m2 (d) 25.8 m2 (ii) Write the formula to find the volume of a cylindrical pillar:
(a) πr2h (b) πrl (c) πr(l + r) (d) 2πr (iii) Find the lateral surface area of two pillars if height of the pillar is 7 m and radius of the base is 1.4 m.
(a) 112.3 cm2 (b) 12.2 m2 (c) 90m2 (d) 345.2cm2 (iv) How much is the volume of a hemisphere if the radius of the base is 3.5 m?
(a) 85.9 m3 (b) 80 m3 (c) 98 m3 (d) 89.83 m3 (v) What is the ratio of sum of volumes of two hemispheres of radius 1 cm each to the volume of a sphere of radius 2 cm?
(a) 1:1 (b) 1:8 (c) 8:1 (d) 1:16 -
An architect's planned design for a room with dimensions of 8.6 m, 5.4 m and 4 m respectively. He also planned to make 4 windows with blue colour and 2 doors with brown wood colour. The room needs to be painted with Asian paint of Green colour except for the floor and square tiles were used for flooring as shown in the below figure:
.JPG)
(i) The total area of the four walls is :(a) 112 sq m (b) 212 sq m (c) 312 sq m (d) 412 sq m (ii) If the area of windows and doors is 22 sq m. The area of the walls to be painted
(a) 100 sq m (a) 100 sq m (a) 100 sq m (d) 132 sq m (iii) What is the area of the tiles to be used for flooring?
(a) 64.6 sq m (b) 46.44 sq m (c) 66.4 sq m (d) 44.6 sq m (iv) The total area of the room is (including windows and doors):
(a) 48.02 sq m (b) 840.4 sq m (c) 402.8 sq m (c) 402.8 sq m (v) What is the volume of the air in the room?
(a) 157.68 cu. m (a) 157.68 cu. m (a) 157.68 cu. m (d) 185.76 cu. m -
Mohan lives in Hyderabad in Telangana. Those were very hot days of May. He thought that if we human beings need so much of water to drink, won’t the birds also be thirsty. He decided to prepare a vessel to provide water for birds. He found a flexible blue coloured plastic rectangular sheet 44 cm × 15 cm. He rolled it along its length and joined the two opposite ends using a tape. He wanted to have a circular base for this cylinder and searched for another sheet. He found a square sheet 15 cm × 15 cm. He got a circular sheet just equal to the base of the cylinder cut from it. 15 cm 44 cm 15 cm 15 cm
(a) The curved surface area of the cylinder formed is(i) 540 cm² (ii) 560 cm² (iii) 640 cm² (iv) 660 cm² (b) The radius of the base of the cylinder is
(i) 2 cm (ii) 3.5 cm (iii) 7 cm (iv) 10 cm (c) The area of the circular base required for the cylinder is
(i) 154 cm² (ii) 216 cm² (iii) 260 cm² (iv) 308 cm² (d) How much will be the area of square sheet left unused after removing the circular base of the cylinder from it?
(i) 69 cm² (ii) 71 cm² (iii) 83 cm² (iv) 91 cm² (e) Find the volume of water that can be filled in the cylinder.
(i) 1410 ml (ii) 1730 ml (iii) 2170 ml (iv) 2310 ml -
Mr. Kumar, a Mathematics teacher brings some green coloured clay in the classroom to teach the topic 'mensuration'. First, he forms a cylinder of radius 6 cm and height 8 cm with the clay. Then, he moulds that cylinder into a sphere similarly, he moulds the sphere in other different shapes. Answer the following questions:
(i) Which of the following is not a 3D shape?
(a) cone (b) cuboid (c) rectangle (d) sphere (ii) What is the volume of cylindrical shape?
(a) 268π cm3 (b) 288π cm3 (c) 36π cm2 (d) 48π cm3 (iii) What is the formula of volume of sphere?
(a) \(\frac{2}{3}\) πr3 (b) πr3 (c) \(\frac{1}{3}\) πr3 (d) \(\frac{4}{3}\) πr3 (iv) When clay changes into on shape to other which of the following remains same.
(a) Volume (b) Area (c) CSA (c) Radius (v) The radius of the sphere formed is :
(a) 4 cm (b) 5 cm (c) 6 cm (c) 3 cm -
In a newly constructed park which is situated in the heart of a city Hyderbad, an architect has form a structure in the given shape. The shape has a cuboid, which is standing on the two cylindrical beams. The dimensions of the cuboid are 1.5 m, 3 m and 0.5 m. The dimensions of the cylinders are of height 2 m and diameter 0.6 m.
(i) As the structure is made from the concrete, how much volume of concrete is reqired to make the cuboidal shape?(a) 1.75 m3 (b) 2.20 m3 (c) 2.25 m3 (d) 1.25 m3 (ii) What is formula for calculating the lateral surface area of the cylinder?
(a) πr2 (b) 2πrh (c) πr2h (d) 2πr3 (iii) What is the volume of two cylinders?
(a) 1.20 m3 (b) 1.134 m3 (c) 3 m3 (d) 2.2 m3 (iv) If the cuboid needs to be painted red, how much area need to be painted?
(a) 5.2 m2 (b) 13 m2 (c) 6.75 m2 (d) 5.7 m2 (v) If a cloth is needed to cover the cylindrical part, how much cloth is needed?
(a) 8.25m2 (b) 1.25 m2 (c) 4.50 m2 (d) 7.536 m2 -
One day teacaher planned to take all the Class X students to the milk factory (Chilling plant) and asked the students to observe it carefully. Refer to this plant, its machinery is shown below. It is related to some solid shapes, which we study in our curriculum.
(i) Refer to cylindrical container, calculate the quantity of milk it can store.(a) 200 m3 (b) 300 m3 (c) 942.477 m3 (d) 1000 m3 (ii) What is the formula for calculating the total surface area of the cylindrical container i.e., milk container?
(a) 2πrh (b) 2πr (r + h) (c) 2πr2 (d)2πrh (iii) If the cube shown in the picture is of dimension 6 cm each. Find the capacity of this cubic container.
(a) 186 (b) 200 (c) 216 (d) 185 (iv) Find the ratio between the volume and C.S.A. of the cylindrical container.
(a) \(\frac{2}{1}\) (b) \(\frac{5}{2}\) (c) \(\frac{4}{9}\) (d) \(\frac{3}{7}\) (v) What is the formula for calculating the total surface area of the hemisphere?
(a) 2πr2 (b) 4πr2 (c) 3πr2 (d) \(\frac{1}{2}\) πr2
Case Study
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Answers
Surface Areas and Volumes Case Study Questions With Answer Key Answer Keys
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(a) (ii) 2(lb + bh + hl)
(b) (ii) 968π m2
Diameter = 44 m
Radius = 22 m
Curved surface area of hemispherical dome = 2πr2
= 2π(22)2 = 968π m2
(c) (iii) 29 m
Total height of monument = 51 m
Radius of hemispherical dome part
= height of hemispherical dome = 22 m
Height of the cuboid part = 51 m - 22 m = 29 m
(d) (iii) 44π
Circumference of the base of the dome = 2πr
= 2π(22) = 44π
(v) The total surface of a hemispherical dome having radius 7 cm is:
(v) (a) 462 cm2
TSA of hemispherical dome = 3πr2
= 3 x \(\begin{equation} \frac{22}{7} \end{equation}\) x 7 x 7
= 462 cm2 -
(i) (d) none of these
C.SA of hemispheres = 2 x 2πr2
= 4 x \(\frac{22}{7}\) x \(\frac{42}{10}\) x \(\frac{42}{10}\) = \(\frac{88 \times 252}{100}\)
= 221.76 m2
(ii) (b) πr2(2/3 r + h)
Volume of pillar = volume of cylinder + volume of hemisphere
= πr2 + \(\frac{2}{3}\)πr3 = πr2\(\left(h+\frac{2}{3} r\right)\)
(iii) (c) 718.66 m3
Volume of the hemispherical dome = \(\frac{2}{3}\)πr3
= \(\frac{2}{3}\) x \(\frac{22}{7}\) x 7 x 7 x7 = 718. 66 m3
(iv) (b) 591.36 m2
Lateral surface are of 4 pillars = 4 x 2πrh
= 4 x 2 x \(\frac{22}{7}\) x 14 x 14
= 591.36 m2
(v) (a) Rs. 1,59,667.20
The cost of polishing all the four pillars = 591.36 x 270
= 159,667.20 -
(i) (d) none of these
C.SA of hemispheres = 2 x 2πr2
= 4 x \(\frac{22}{7}\) x \(\frac{42}{10}\) x \(\frac{42}{10}\) = \(\frac{88 \times 252}{100}\)
= 221.76 m2
(ii) (b) πr2(2/3 r + h)
Volume of pillar = volume of cylinder + volume of hemisphere
= πr2 + \(\frac{2}{3}\)πr3 = πr2\(\left(h+\frac{2}{3} r\right)\)
(iii) (c) 718.66 m3
Volume of the hemispherical dome = \(\frac{2}{3}\)πr3
= \(\frac{2}{3}\) x \(\frac{22}{7}\) x 7 x 7 x7 = 718. 66 m3
(iv) (b) 591.36 m2
Lateral surface are of 4 pillars = 4 x 2πrh
= 4 x 2 x \(\frac{22}{7}\) x 14 x 14
= 591.36 m2
(v) (a) Rs. 1,59,667.20
The cost of polishing all the four pillars = 591.36 x 270
= 159,667.20 -
(i) (b) 78.57 m2
Radius of a dome, r = 2.5 cm
The dome is hemispherical in shape.
Then, cloth material required = 2 x Surface area of hemisphere
= 2 x 2πr2
= 4 x \(\frac{22}{7}\) x 2.5 x 2.5 = 78.57 m2
(ii) (a) πr2h
(iii) (b) 123.2m2
Height of each pillar, h = 7 m
Radius of base, r = 1.4 m
Lateral surface area or curved surface area of 2 pillars = 2 x 2πrh
= 4 x \(\frac{22}{7}\) x 1.4 x 7
= 123.2 m2
(iv) (d) 89.83 m3
Radius of hemisphere, r = 3.5 m
Then, volume of a hemisphere, V = \(\frac{2}{3}\)πr3
= \(\frac{2}{3} \times \frac{22}{7}\) x (3.5)3
= 89.83 m3
(v) (b) 1:8
Volume of 2 hemispheres of radius 1 cm = 2 x \(\frac{2}{3}\)πr3 = \(\frac{4}{3}\)π(1)3 = \(\frac{4}{3}\)π cm3
Volume of 1 shpere of radius 2cm = \(\frac{4}{3}\)π(r3) = \(\frac{4}{3}\)π(2)3 =\(\frac{32}{3}\)π cm3
Then, required ratio = \(\frac{\frac{4}{3} \pi}{\frac{32}{3} \pi}\) = \(\frac{1}{8}\) -
(i) (a) 112 sq m
Total area of four walls = 2 (l + b)h
= 2 (8.6 + 5.4) x 4
= 112 sq m
(ii) (c) 90 sq m
Wall painted area = 112 sq m - 22 sq m
= 90 sq m
(iii) (b) 46.44 sq m
Area of tiles = 1 x b
= 8.6 m x 5.4 m
= 46.44 sq m.
(iv) (d) 204.88 sq m
Total area of room = 2(lb + bh + hl)
= 2 (8.6 x 5.4 + 5.4 x 4 + 8.6)
= 2 (46.44 + 21.6 + 34.4)
= 204.88 sq m
(v) (d) 204.88 sq m
Volume of the air in the room = lbh
= 8.6 x 5.4 x 4
= 185.76 cu.m -
(a) (iv) 660 cm²
The curved surface area of the cylider formed
= area of the plastic rectangular sheet = (44 x 15) cm²
= 660 cm²
(b) (iii) 7 cm
circumference of the base
= length of the rectangular sheet = 44 cm
∴ 2πr = 44 ⇒ 2 x \(\frac{22}{7}\) x r = 44
⇒ r = 7.
(c) (i) 154 cm²
= \(\left(\frac{22}{7} \times 7 \times 7\right)\) cm² = 154 cm² .
(d) (ii) 71 cm²
Area of the square sheet = a2 = (15 x 15) cm² = 225 cm².
∴ Area of square sheet left unused
= Area of square - Area of the base of the cylinder
= (225 - 154) cm² = 71 cm².
(e) (iv) 2310 ml
Volume of water that can be filled in the cylinder
= Volume of the cylinder = πr2h
\(\left(\frac{22}{7} \times 7 \times 7 \times 15\right)\) cm3
= 2310 cm3
= 2310 ml. [∵ 1 ml = 1 cm3] -
(i) (c) rectangle
(ii) (b) 288π cm3
Volume of cylindrical shape = πr2h
= π (6)2 x 8
= 288 π cm3
(iii) (d) \(\frac{4}{3}\) πr3
(iv) (a) Volume
(v) (c) 6 cm
Volume of Sphere = Volume of cylindrical shape
⇒ \(\frac{4}{3}\)πR3 = πr2h
⇒ R3 = \(\frac{3}{4}\) (6)2 x 8
⇒ R3 = 216
⇒ R = 6 cm -
(i) (c) 2.25 m3
Here, l = 1.5 m, b = 3 m, h = 0.5 m
volume of cuboid, lbh = 1.5 x 3 x 0.5 = 4.5 x 0.5
= 2.25 m3
(ii) (b) 2πrh
C.S.A. of cylinder = 2πrh
(iii) (b) 1.134 m3
Volume of two cylinders = 2 x πr2h = 2x3.14x\(\begin{equation} \left(\frac{0.6}{2}\right)^{2} \end{equation}\) x2
= 3.14 x 0.6 x 0.6
= 1.1304 m3
(iv) (b) 13 m2
Area to be painted = T.S.A. of cuboid - 2 x base circular area = 2(lb + bh + hl) - 2πr2|
= 2(1.5 x 3 + 3 x 0.5 + 0.5 x 1.5) - 2 x 3.14 x \(\begin{equation} \left(\frac{0.6}{2}\right)^{2} \end{equation}\)
= 2(4.5 + 1.5 + 0.75 - 3.14 x 0.3 x 0.6)
= 2 x 6.75 - 0.57 = 12.93 ~ 13
(v) (d) 7.536 m2
C.S.A. of cylinder = 2 x 2πrh
= 4 x 3.14 x \(\left(\frac{0.6}{2}\right)\) x 2
= 4 x 3.14 x 0.6
= 7.536 m2 -
(i) (c) 942.477 m3
Quantity of milk = Volume of cotainer
= πr2h
= π x (5)2 x (12)
= 942.477 m3
(ii) (b) 2πr (r + h)
(iii) (c) 216
Volume of cube = a3 = 6 x 6 x 6
= 216 cm3
(iv) (b) \(\frac{5}{2}\)
Ratio = \(\frac{\text { Volume }}{\text { C.S.A. }}\)
= \(\frac{\pi r^{2} h}{2 \pi r h}\) = \(\frac{r}{2}\) = \(\frac{5}{2}\)
(v) (c) 3πr2
2πr2 + πr2 = 3πr2
