Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 9 Set 9 Surface Area and Volume Solutions

Question 1. If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? \(\left(\pi = \frac{22}{7}\right)\)
Given: For road roller,
diameter (d) = 0.9 m, length (h) = 1.4 m
To find: Area of a field pressed in 500 rotations
Solution:
i. Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller
\(\therefore\) Curved surface area of the road roller = \(2\pi rh\)
= \(\pi dh\),..[: d = 2r]
= \(\frac{22}{7} \times 0.9 \times 1.4\)
= \(22 \times 0.9 \times 0.2\)
= \(3.96\) sq.m.
ii. Area of land pressed in 1 rotation = \(3.96\) sq.m.
\(\therefore\) Area of land pressed in 500 rotations = \(500 \times 3.96\)
= \(1980\) sq.m.
\(\therefore\) \(1980\) sq.m, land will be pressed in \(500\) rotations of the road roller.
In simple words: First, calculate the curved surface area of the road roller, which represents the area covered in one rotation. Then, multiply this single-rotation area by the total number of rotations (500) to find the total area pressed.

🎯 Exam Tip: Remember that the area covered by a road roller in one rotation is equal to its curved surface area. Pay attention to units and ensure correct multiplication for the total area.

Question 2. To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?
Given: Thickness of the glass = 2 mm,
outer length of the tank = 60.4 cm,
outer breadth of the tank = 40.4 cm,
outer height of the tank = 40.2 cm
To find: Volume of water fish tank contains
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक मछली टैंक की संरचना का आरेख है, जिसमें बाहरी आयाम (60.4 सेमी लंबाई, 40.4 सेमी चौड़ाई और 40.2 सेमी ऊँचाई) और कांच की मोटाई (0.2 सेमी) दर्शाई गई है। आरेख आंतरिक आयामों को दर्शाने के लिए मोटाई घटाने की प्रक्रिया को भी दिखाता है, जैसे कि आंतरिक लंबाई (60.4-0.2-0.2) सेमी, आंतरिक चौड़ाई (40.4-0.2-0.2) सेमी, और आंतरिक ऊँचाई (40.2-0.2) सेमी।
i. Thickness of glass = 2 mm.
= \(\frac{2}{10}\) cm
= \(0.2\) cm
Outer length of the tank = 60.4 cm
\(\therefore\) Inner length of the tank (l) = Outer length – thickness of glass on both sides
= \(60.4 - 0.2 - 0.2\)
= \(60\) cm
Outer breadth of the tank = 40.4 cm
\(\therefore\) Inner breadth of the tank (b) = \(40.4 - 0.2 - 0.2\)
= \(40\) cm
Outer height of the tank = 40.2 cm
\(\therefore\) Inner height of the tank (h) = \(40.2 - 0.2\)
= \(40\) cm
ii. Maximum volume of water that can be contained in the tank = volume of the tank
= \(l \times b \times h\)
= \(60 \times 40 \times 40\)
= \(96000\) cubic cm.
\(\therefore\) The fish tank can contain maximum of \(96000\) cubic cm. water in it.
In simple words: To find the maximum volume of water an open fish tank can hold, first calculate the inner dimensions by subtracting the glass thickness from the outer dimensions. Since the tank is open, thickness is subtracted only from the base for height and from both sides for length and breadth. Then, use these inner dimensions to calculate the volume.

🎯 Exam Tip: For open tanks, remember to subtract the glass thickness from both sides for length and breadth, but only from the base (once) for height, as the top is open. Accurate calculation of inner dimensions is crucial.

Question 3. If the ratio of radius of base and height of a cone is 5 : 12 and its volume is 314 cubic metre. Find its perpendicular height and slant height (\(\pi = 3.14\)).
Given: Ratio of radius of base and height of a cone = 5 : 12,
Volume = 314 cubic metre
To find: Perpendicular height (h) and slant height (l)
Solution:
i. The ratio of radius and height of cone is 5 : 12
Let the common multiple be x.
\(\therefore\) Radius of base (r) = 5x
Perpendicular height (h) = 12x
Volume of cone = \(\frac{1}{3} \pi r^2 h\)
\[314 = \frac{1}{3} \times 3.14 \times (5x)^2 \times (12x)\]
\[\frac{314}{1} = \frac{1}{3} \times 3.14 \times 25x^2 \times 12x\]

\[x^3 = \frac{314 \times 3}{3.14 \times 25 \times 12}\]

\[x^3 = \frac{314 \times 3 \times 100}{314 \times 25 \times 12}\]

\[x^3 = \frac{1}{1}\]
\(\therefore x^3 = 1\)
\(\therefore x = 1\) ... [Taking cube root on both sides]
\(\therefore\) r = 5x = 5(1) = 5m
h = 12x = 12(1) = 12 m
ii. Now, \(l^2 = r^2 + h^2\)
= \(5^2 + 12^2\)
= \(25 + 144\)
\(\therefore l^2 = 169\)
\(\therefore l = \sqrt{169}\) ... [Taking square root on both sides]
= \(13\) m
The perpendicular height and slant height of the cone are 12 m and 13 m respectively.
In simple words: Use the given ratio of radius and height to express them in terms of a common multiple, 'x'. Substitute these into the cone's volume formula to solve for 'x'. Once 'x' is known, calculate the actual radius and height. Finally, use the Pythagorean theorem to find the slant height.

🎯 Exam Tip: When given ratios, introduce a common multiple (like 'x') to represent the dimensions. Ensure you use the correct formula for the volume of a cone and the Pythagorean theorem for slant height. Keep track of units.

Question 4. Find the radius of a sphere if its volume is 904.32 cubic cm. (\(\pi = 3.14\))
Given: Volume of sphere = 904.32 cubic cm.
To find: Radius of a sphere
Solution:
Volume of sphere = \(\frac{4}{3} \pi r^3\)
\(\therefore 904.32 = \frac{4}{3} \times 3.14 \times r^3\)

\[r^3 = \frac{904.32 \times 3}{4 \times 3.14}\]

\[r^3 = \frac{90432 \times 3}{4 \times 314}\]

\[r^3 = \frac{288 \times 3}{4}\]
= \(216\)
\(\therefore r = \sqrt[3]{216}\) ... [Taking cube root on both sides]
= \(6\) cm
\(\therefore\) The radius of the sphere is \(6\) cm.
In simple words: To find the radius of a sphere given its volume, rearrange the volume formula to solve for the cube of the radius. Then, calculate the cube root of the result to find the radius.

🎯 Exam Tip: Remember the formula for the volume of a sphere: \(\frac{4}{3} \pi r^3\). Be careful with decimal places during calculations, especially when manipulating the equation to isolate \(r^3\).

Question 5. Total surface area of a cube is 864 sq.cm. Find its volume.
Given: Total surface area of cube = 864 sq. cm
To find: Volume of cube
Solution:
i. Total surface area of cube = \(6l^2\)
\(\therefore 864 = 6l^2\)
\(\therefore l^2 = \frac{864}{6}\)
\(\therefore l^2 = 144\)
\(\therefore l = \sqrt{144}\) ... [Taking square root on both sides]
= \(12\) cm
ii. Volume of cube = \(l^3\)
= \(12^3\)
= \(1728\) cubic cm.
\(\therefore\) The volume of cube is \(1728\) cubic cm.
In simple words: First, use the given total surface area of the cube to find the length of one of its sides by rearranging the surface area formula. Once the side length is determined, calculate the volume of the cube using the formula: side length cubed.

🎯 Exam Tip: Know the formulas for both total surface area (\(6l^2\)) and volume (\(l^3\)) of a cube. Solving for 'l' from the surface area is the critical first step before calculating volume.

Question 6. Find the volume of a sphere, if its surface area is 154 sq.cm.
Given: Surface area of sphere = 154 sq. cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = \(4\pi r^2\)
\[154 = 4 \times \frac{22}{7} \times r^2\]
\[r^2 = \frac{154 \times 7}{4 \times 22}\]
\[r^2 = \frac{49}{4}\]
\[r = \sqrt{\frac{49}{4}}\]
\[r = \frac{7}{2}\] cm ... [Taking square root on both sides]
ii. Volume of sphere = \(\frac{4}{3} \pi r^3\)
\[= \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3\]
\[= \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\]
\[= \frac{1}{3} \times 22 \times 1 \times \frac{7}{2} \times \frac{7}{2}\]
= \(179.67\) cubic cm.
\(\therefore\) The volume of sphere is \(179.67\) cubic cm.
In simple words: First, use the given surface area of the sphere to calculate its radius. Then, substitute this calculated radius into the formula for the volume of a sphere to find the final answer.

🎯 Exam Tip: It's crucial to correctly use the formula for the surface area of a sphere (\(4\pi r^2\)) to find the radius before proceeding to calculate the volume. Be mindful of squaring and cubing the radius correctly.

Question 7. Total surface area of a cone is 616 sq.cm. If the slant height of the cone is three times the radius of its base, find its slant height.
Given: Total surface area of a cone = 616 sq.cm., slant height of the cone is three times the radius of its base
To find: Slant height (l)
Solution:
i. Let the radius of base be r cm.
\(\therefore\) Slant height (l) = 3r cm
Total surface area of cone = \(\pi r (l + r)\)
\(\therefore 616 = \pi r (l + r)\)
\(\therefore 616 = \frac{22}{7} \times r \times (3r + r)\)
\(\therefore 616 = \frac{22}{7} \times 4r^2\)
\[r^2 = \frac{616 \times 7}{22 \times 4}\]
\[r^2 = \frac{28 \times 7}{4}\]
\(\therefore r^2 = 49\)
\(\therefore r = \sqrt{49}\) ... [Taking square root on both sides]
= \(7\)
ii. Slant height (l) = 3r = \(3 \times 7 = 21\) cm
\(\therefore\) The slant height of the cone is \(21\) cm.
In simple words: Express the slant height in terms of the radius using the given relationship. Substitute this into the formula for the total surface area of a cone. Solve the resulting equation to find the radius, and then use the relationship to calculate the slant height.

🎯 Exam Tip: Pay close attention to the relationship between slant height and radius. Using the total surface area formula \(\pi r (l+r)\) is key, and be careful with algebraic manipulation to solve for 'r'.

Question 8. The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs. 52 per sq.m.
Given: Inner diameter (d) = 4.2 m,
To find: depth (h) = 10 m,
rate of plastering = Rs. 52 per sq.m.
Inner surface area and total cost of plastering
Solution:
i. Inner curved surface area of the well = \(2\pi rh\)
= \(\pi dh\) ...[: d = 2r]
= \(\frac{22}{7} \times 4.2 \times 10\)
= \(22 \times 0.6 \times 10\)
= \(22 \times 6\)
= \(132\) sq.m.
ii. Rate of plastering = Rs. 52 per sq.m.
\(\therefore\) Total cost = Curved surface area x Rate of plastering
= \(132 \times 52 = 6864\)
\(\therefore\) The cost of plastering the well from inside is Rs. 6864.
In simple words: First, calculate the inner curved surface area of the cylindrical well using its diameter and depth. Then, multiply this area by the given plastering rate per square meter to find the total cost of plastering.

🎯 Exam Tip: Recognize that the well's inner surface area is its curved surface area. Use the formula \(\pi dh\) (or \(2\pi rh\)). Ensure correct unit conversion and accurate multiplication for the final cost.

Question 9. The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = Rs. 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
Solution:
i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
\(\therefore\) Curved surface area of the road roller = \(2\pi rh\)
= \(\pi dh\) ...[: d = 2r]
= \(\frac{22}{7} \times 1.4 \times 2.1\)
= \(22 \times 0.2 \times 2.1\)
= \(9.24\) sq.m.
ii. Area of ground levelled in 1 rotation = \(9.24\) sq.m.
\(\therefore\) Area of ground levelled in 500 rotations = \(9.24 \times 500\)
= \(4620\) sq.m.
iii. Rate of levelling = Rs. 7 per sq.m.
\(\therefore\) Total cost = Area of ground levelled x Rate of levelling
= \(4620 \times 7\)
= \(32340\)
\(\therefore\) The road roller levels \(4620\) sq.m. land in \(500\) rotation, and the cost of levelling is Rs. 32340.
In simple words: First, calculate the curved surface area of the road roller, which is the area covered in one rotation. Multiply this by the total number of rotations to find the total area levelled. Finally, multiply the total levelled area by the given rate per square meter to determine the total cost.

🎯 Exam Tip: The length of a road roller acts as the height (h) of a cylinder. The area covered in one rotation is its curved surface area. Remember to calculate the total area first, then multiply by the cost per unit area.

Maharashtra Board Class 9 Maths Chapter 9 Surface Area and Volume Practice Set 9 Intext Questions and Activities

Question 1. Curved surface area of cone. (Textbook pg. no. 116)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक शंकु के विभिन्न भागों को दर्शाता है, जिसमें उसके नेट (खुली हुई सतह) भी शामिल हैं। चित्र (a) एक शंकु को, (b) उसकी मुड़ी हुई सतह के नेट को, (c) नेट के टुकड़ों को, और (d) इन टुकड़ों को मिलाकर एक आयत बनने को दर्शाता है, जिसका उपयोग शंकु के वक्र पृष्ठीय क्षेत्रफल को समझने के लिए किया जाता है।
Circumference of base of the cone = \(2\pi r\)
As shown in the figure (c), make pieces of the net as small as possible. Join them as shown in the figure (d). By joining the small pieces of net of the cone, we get a rectangle ABCD approximately.
Total length of AB and CD is \(2\pi r\).
\(\therefore\) length of side AB of rectangle ABCD is \(\pi r\) and length of side CD is also \(\pi r\).
Length of side BC of rectangle = slant height of cone = l.
Curved surface area of cone is equal to the area of the rectangle.
\(\therefore\) curved surface area of cone = Area of rectangle = AB \(\times\) BC = \(\pi r \times l = \pi rl\)
In simple words: To understand the curved surface area of a cone, imagine unrolling it into a flat shape. This shape can be approximated as a rectangle when cut into very small pieces, with its length as the circumference of the cone's base and its width as the slant height, resulting in the formula \(\pi rl\).

🎯 Exam Tip: Visualizing the unrolling of a cone's curved surface into a sector and then approximating it as a rectangle helps understand the \(\pi rl\) formula. This conceptual understanding is key to derivations.

Question 2. Prepare a cylinder of a card sheet, keeping one of its faces open. Prepare an open cone of card sheet which will have the same base-radius and the same height as that of the cylinder. Pour fine sand in the cone till it just fills up the cone. Empty the cone in the cylinder. Repeat the procedure till the cylinder is just filled up with sand. Note how many coneful of sand is required to fill up the cylinder. (Textbook pg. no 117)

ℹ️ चित्र व्याख्या (Diagram Explanation): इस आरेख में एक शंकु और एक बेलन को दर्शाया गया है, दोनों की आधार त्रिज्या 'r' और ऊँचाई 'h' समान हैं। यह गतिविधि यह दर्शाने के लिए है कि एक बेलन को भरने के लिए कितने शंकु-भर रेत की आवश्यकता होती है, जिससे उनके आयतन का संबंध स्पष्ट होता है।
Answer:
To fill the cylinder, three coneful of sand is required.
In simple words: This activity demonstrates that a cylinder with the same base radius and height as a cone can hold exactly three times the volume of the cone. This visually explains the relationship between their volumes.

🎯 Exam Tip: This activity illustrates that the volume of a cone is \(\frac{1}{3}\) the volume of a cylinder with the same base and height. This ratio (\(1:3\)) is a fundamental concept for cone and cylinder volume formulas.

Question 3. Finding total surface area of sphere. (Textbook pg. no 120)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक स्वीट लाइम (मोसंबी) को दो बराबर हिस्सों में काटने की प्रक्रिया को दर्शाता है। यह गतिविधि गोले के कुल पृष्ठीय क्षेत्रफल की अवधारणा को समझाने के लिए उपयोग की जाती है, जहाँ एक गोले का पृष्ठीय क्षेत्रफल उसके चार बराबर वृत्तों के क्षेत्रफल के बराबर होता है।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक स्वीट लाइम के दो हिस्सों को दर्शाता है, जिसमें प्रत्येक अर्ध-भाग का गोलाकार चेहरा है। यह गतिविधि दर्शाती है कि इन हिस्सों को कागज पर रखकर वृत्त बनाए जा सकते हैं, और फिर प्रत्येक अर्ध-भाग को और दो हिस्सों में काटकर कुल चार चौथाई भाग प्राप्त किए जा सकते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक स्वीट लाइम के चार चौथाई भागों को दर्शाता है, जिसमें प्रत्येक भाग की छाल को छोटे टुकड़ों में काटकर एक वृत्त को ढकने का प्रयास किया जाता है। यह गतिविधि यह साबित करती है कि गोले का वक्र पृष्ठीय क्षेत्रफल चार वृत्तों के क्षेत्रफल के बराबर होता है।
i. Take a sweet lime (Mosambe), Cut it into two equal parts.
ii. Take one of the parts. Place its circular face on a paper. Draw its circular border. Copy three more such circles. Again, cut each half of the sweet lime into two equal parts.
iii. Now you get 4 quarters of sweet lime. Separate the peel of a quarter part. Cut it into pieces as small as possible. Try to cover one of the circles drawn, by the small pieces. Observe that the circle gets nearly covered.
The activity suggests that,
Curved surface area of a sphere = \(4\pi r^2\)
\(\therefore\) Curved surface area of a sphere = 4 x Area of a circle
In simple words: This activity demonstrates that the total surface area of a sphere is equivalent to the area of four circles, each with the same radius as the sphere. By peeling a quarter of a sweet lime and attempting to cover a circle, it visually suggests the formula \(4\pi r^2\).

🎯 Exam Tip: This hands-on activity provides an intuitive understanding of why a sphere's surface area is \(4\pi r^2\). Knowing this conceptual derivation can help recall the formula easily during exams.

Question 4. Make a cone and a hemisphere of cardsheet such that radii of cone and hemisphere are equal and height of cone is equal to radius of the hemisphere. Fill the cone with fine sand. Pour the sand in the hemisphere. How many cones are required to fill the hemisphere completely ? (Textbook pg. no. 121)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक शंकु, एक बेलन, और एक अर्धगोले को दर्शाता है, जिनके आयाम संबंधित गतिविधियों को समझने के लिए महत्वपूर्ण हैं। विशेष रूप से, एक शंकु और एक अर्धगोले को समान त्रिज्या और शंकु की ऊँचाई अर्धगोले की त्रिज्या के बराबर होने पर दर्शाया गया है। यह आरेख यह समझने में मदद करता है कि अर्धगोले को भरने के लिए कितने शंकु-भर रेत की आवश्यकता होती है।
Answer:
To fill the hemisphere, two coneful of sand is required.
In simple words: This activity shows that if a cone and a hemisphere have the same radius, and the cone's height is equal to that radius, then two cones full of sand are needed to completely fill the hemisphere. This illustrates the volume relationship between a cone and a hemisphere.

🎯 Exam Tip: This activity visually demonstrates that the volume of a hemisphere is twice the volume of a cone with the same radius and height equal to the radius. This implies the volume of a sphere is four times the volume of such a cone.