Maharashtra Board Class 10 Maths Chapter 7 Mensuration Set 7.2 Solutions

Question 1. The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm³)
Given: Radii (\(r_1\)) = 14 cm, and (\(r_2\)) = 7 cm,
height (\(h\)) = 30 cm
To find: Amount of water the bucket can hold.
Solution:
Volume of frustum = \(\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 \times r_2)\)
\( = \frac{1}{3} \times \frac{22}{7} \times 30 (14^2 + 7^2 + 14 \times 7) \)
\( = \frac{22 \times 10}{7} (196 + 49 + 98) \)
\( = \frac{220}{7} \times 343 \)
\( = 220 \times 49 \)
\( = 10780 \text{ cm}^3 \)
\( = \frac{10780}{1000} \) litres
\([\because 1 \text{ litre} = 1000 \text{ cm}^3]\)
\( = 10.78 \) litres
\(\therefore\) The bucket can hold 10.78 litres of water.
Answer: The bucket can hold 10.78 litres of water.
In simple words: To find the volume a frustum-shaped bucket can hold, we use its given radii and height in the frustum volume formula. The calculated volume in cubic centimeters is then converted to litres using the given conversion factor.

🎯 Exam Tip: Remember to correctly identify \(r_1\), \(r_2\), and \(h\) and use the correct formula for the volume of a frustum. Pay attention to unit conversions, especially cm³ to litres.

Question 2. The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
(i) curved surface area,
(ii) total surface area,
(iii) volume, (\(\pi = 3.14\))
Given: Radii (\(r_1\)) = 14 cm, and (\(r_2\)) = 6 cm,
height (\(h\)) = 6 cm
Solution:
Slant height of frustum \((l) = \sqrt{h^2+(r_1-r_2)^2}\)
\( = \sqrt{6^2+(14-6)^2} \)
\( = \sqrt{6^2+8^2} \)
\( = \sqrt{36+64} \)
\( = \sqrt{100} = 10 \) cm

(i) Curved surface area of frustum
\( = \pi l (r_1 + r_2) \)
\( = 3.14 \times 10(14 + 6) \)
\( = 3.14 \times 10 \times 20 = 628 \text{ cm}^2 \)
\(\therefore\) The curved surface area of the frustum is \(628 \text{ cm}^2\).

(ii) Total surface area of frustum
\( = \pi l (r_1+ r_2) + \pi r_1^2 + \pi r_2^2 \)
\( = 628 + 3.14 \times (14)^2 + 3.14 \times (6)^2 \)
\( = 628 + 3.14 \times 196 + 3.14 \times 36 \)
\( = 628 + 3.14(196 + 36) \)
\( = 628 + 3.14 \times 232 \)
\( = 628 + 728.48 \)
\( = 1356.48 \text{ cm}^2 \)
\(\therefore\) The total surface area of the frustum is \(1356.48 \text{ cm}^2\).

(iii) Volume of frustum
\( = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 \times r_2) \)
\( = \frac{1}{3} \times 3.14 \times 6(14^2 + 6^2 + 14 \times 6) \)
\( = 3.14 \times 2(196 + 36 + 84) \)
\( = 3.14 \times 2 \times 316 \)
\( = 1984.48 \text{ cm}^3 \)
\(\therefore\) The volume of the frustum is \(1984.48 \text{ cm}^3\).
Answer: (i) Curved surface area = 628 cm², (ii) Total surface area = 1356.48 cm², (iii) Volume = 1984.48 cm³.
In simple words: This problem requires calculating three properties of a frustum: its curved surface area, total surface area, and volume, using the given radii and height, and a specific value for pi. The slant height is calculated first as it's needed for the surface area formulas.

🎯 Exam Tip: This question tests multiple formulas for frustums. Ensure you correctly calculate the slant height first, then substitute values accurately into the curved surface area, total surface area, and volume formulas. Using \(\pi = 3.14\) is crucial.

Question 3. The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity.
\(\left(\pi = \frac{22}{7}\right)\)
Solution:
Circumference\(_{1}\) = \(2\pi r_1 = 132\) cm
\(\therefore r_1 = \frac{132}{2\pi} = \frac{132}{2} \times \frac{7}{22} = 21\) cm
Circumference\(_{2}\) = \(2\pi r_2 = 88\) cm
\(\therefore r_2 = \frac{88}{2\pi} = \frac{88}{2} \times \frac{7}{22} = 14\) cm
Slant height of frustum \((l) = \sqrt{h^2+(r_1-r_2)^2}\)
\( = \sqrt{24^2+(21-14)^2} \)
\( = \sqrt{24^2+7^2} \)
\( = \sqrt{576+49} \)
\( = \sqrt{625} \)
\( = 25 \) cm
Curved surface area of frustum = \(\pi (r_1 + r_2) l\)
\( = \pi (21 + 14) \times 25 \)
\( = \pi \times 35 \times 25 \)
\( = \frac{22}{7} \times 35 \times 25 \)
\( = 2750 \text{ cm}^2 \)
Answer: The curved surface area of the frustum is 2750 cm².
In simple words: To complete this activity, first calculate the radii of the frustum's circular faces using their given circumferences. Then, use these radii and the given height to find the slant height. Finally, compute the curved surface area using the slant height and radii.

🎯 Exam Tip: This activity emphasizes step-by-step problem-solving. Carefully calculate the radii from circumferences, then the slant height, and finally the curved surface area using the correct formulas and the specified value of pi.