UP Board Solutions Class 9 Maths Chapter 18 Surface Area and Volume 18.1

Question 1. एक छड़ की अधिकतम लम्बाई ज्ञात कीजिए जिसे 10 मीx 8 मीx 6 मी विमा वाले एक कमरे में रखा जा सकता है।
Answer: हलः छड़ की अधिकतम लम्बाई = घनाभ का विकर्ण
In simple words: The maximum length of a rod that can be placed in a room is equivalent to the length of the room's diagonal, calculated using the dimensions of the cuboid-shaped room.

🎯 Exam Tip: Remember that the longest object that can fit in a cuboid is its space diagonal, found using the formula \( \sqrt{l^2 + b^2 + h^2} \).

Question 2. यदि एक घनाभ की लम्बाई, चौड़ाई तथा गहराई का योग 19 सेमी है तथा इसका विकर्ण \(5\sqrt{5}\) है तब इसका पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः
घनाभ की \(l + b + h = 19\) ... (1)
विकर्ण \( \sqrt{l^2 + b^2 + h^2} = 5\sqrt{5} \)
वर्ग करने पर \( l^2 + b^2 + h^2 = 125 \) ... (2)
समीकरण (1) का वर्ग करने पर
\( (l + b + h)^2 = (19)^2 \)
\( l^2 + b^2 + h^2 + 2(lb + bh + hl) = 361 \)
\( 125 + 2(lb + bh + hl) = 361 \)
\( 2(lb + bh + hl) = 361 - 125 \)
\( 2(lb + bh + hl) = 236 \)
घनाभ का पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl) = 236 \text{ सेमी}^2\)
In simple words: Given the sum of the cuboid's dimensions and its diagonal length, we use the algebraic identity \( (l+b+h)^2 \) to find the sum of the squares of the dimensions, which helps in calculating the total surface area.

🎯 Exam Tip: Problems involving the sum of dimensions and diagonal of a cuboid often require using the identity \( (l+b+h)^2 = l^2+b^2+h^2+2(lb+bh+hl) \), where \( 2(lb+bh+hl) \) is the surface area.

Question 3. यदि ऊँचाई \(h\) तथा त्रिज्या \(r\) वाले एक ठोस लम्ब वृत्तीय बेलन का वक्र पृष्ठीय क्षेत्रफल, इसके कुल पृष्ठीय क्षेत्रफल का \( \frac{1}{3} \) है तब \(h\) ज्ञात कीजिए।
Answer: हलः लम्ब वृत्तीय बेलन का वक्र पृष्ठीय क्षेत्रफल = \(2\pi rh\)
कुल पृष्ठीय क्षेत्रफल = \(2\pi r(h + r)\)
\( 2\pi rh = \frac{2\pi r(h + r)}{3} \)
\( \implies 3h = h + r \)
\( \implies 2h = r \)
\( \implies h = \frac{r}{2} \)
In simple words: By setting the curved surface area of a cylinder equal to one-third of its total surface area, we can algebraically solve for the height \(h\) in terms of its radius \(r\).

🎯 Exam Tip: Understanding the formulas for curved surface area \( (2\pi rh) \) and total surface area \( (2\pi r(h+r)) \) of a cylinder is crucial for solving such ratio-based problems.

Question 4. एक बेलन में, त्रिज्या दोगुनी है तथा ऊँचाई आधी है तो वक्र पृष्ठीय क्षेत्रफल होगाः (a) समान (b) आधा (c) दोगुना (d) इनमें से कोई नहीं
Answer: (a) समान
हलः बेलन का वक्र पृष्ठीय क्षेत्रफल = \(2\pi rh\)
त्रिज्या दो गुनी तथा ऊँचाई आधी करने पर वक्र पृष्ठीय क्षेत्रफल समान रहेगा।
In simple words: If the radius of a cylinder is doubled and its height is halved, the curved surface area remains the same because the changes in radius and height cancel each other out in the formula \(2\pi rh\).

🎯 Exam Tip: To assess changes in geometric properties, substitute the new dimensions into the relevant formula. Here, \(2\pi (2r) (\frac{h}{2}) = 2\pi rh\), showing no change.

Question 5. एक घनाभ 12 सेमी लम्बा, 9 सेमी चौड़ा तथा 8 सेमी ऊँचा है इसका कुल पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घनाभ की \(l = 12\) सेमी, \(b = 9\) सेमी, \(h = 8\) सेमी
कुल पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl) = 2(12 \times 9 + 9 \times 8 + 8 \times 12) = 2(108 + 72 + 96) = 2(276) = 552 \text{ सेमी}^2\)
In simple words: To find the total surface area of a cuboid, you simply plug its given length, breadth, and height into the formula \(2(lb + bh + hl)\) and perform the calculation.

🎯 Exam Tip: Ensure correct substitution of length, breadth, and height into the total surface area formula for a cuboid, and pay attention to units in the final answer.

Question 6. यदि एक घन के एक विकर्ण की लम्बाई \(8\sqrt{3}\) सेमी है तब इसका पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः
घन का विकर्ण = \(a\sqrt{3}\)
\(a\sqrt{3} = 8\sqrt{3}\)
\( \implies a = 8 \)
पृष्ठीय क्षेत्रफल = \(6a^2 = 6 \times (8)^2 = 6 \times 64 = 384 \text{ सेमी}^2\)
In simple words: Given the diagonal of a cube, first find the side length using the diagonal formula \(a\sqrt{3}\), and then calculate the total surface area using the formula \(6a^2\).

🎯 Exam Tip: Remember that the diagonal of a cube with side 'a' is \(a\sqrt{3}\). This relationship is key to finding the side length before calculating the surface area.

Question 7. तीन बराबर घन, एक आसन्नीय क्रम में रखे गये हैं तब घनाभ के कुल पृष्ठीय क्षेत्रफल से तीनों घनों के पृष्ठीय क्षेत्रफलों के योग का अनुपात ज्ञात कीजिए।
Answer: हलः
तीन घनों का वक्र पृष्ठीय क्षेत्रफल = \(3 \times 6a^2 = 18a^2\)
बने घनाभ का वक्र पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl)\)
(\(l = 3a, b = a, h = a\))
= \(2[3a^2 + a^2 + 3a^2] = 14a^2\)
अनुपात = \(14a^2 : 18a^2 = 7 : 9\)
In simple words: When three identical cubes are joined end-to-end, they form a cuboid. The problem asks for the ratio of the surface area of this new cuboid to the sum of the surface areas of the three individual cubes, which can be found by calculating each and simplifying.

🎯 Exam Tip: When combining identical cubes, identify the new dimensions (length, breadth, height) of the resulting cuboid carefully. The exposed surface area of the individual cubes changes when joined.

Exercise 18.1 Surface Area And Volume Of A Cube, Cuboid And Right Circular Cylinder लघु उत्तरीय प्रश्न (Short Answer Type Questions)

Question 8. एक घनाभ का पृष्ठीय क्षेत्रफल 1300 सेमी है। यदि इसकी चौड़ाई 10 सेमी है तथा ऊँचाई 20 सेमी है तो इसकी लम्बाई ज्ञात कीजिए।
Answer: हलः
\(b = 10\) सेमी, \(h = 20\) सेमी
घनाभ का पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl)\)
\(1300 = 2(l \times 10 + 10 \times 20 + 20 \times l)\)
\(1300 = 2(10l + 200 + 20l)\)
\( \frac{1300}{2} = 30l + 200 \)
\( 650 - 200 = 30l \)
\( 450 = 30l \)
\( l = \frac{450}{30} \)
\( l = 15 \text{ सेमी} \)
In simple words: To find the unknown length of a cuboid when its total surface area, breadth, and height are given, rearrange the surface area formula to solve for the length.

🎯 Exam Tip: Always write down the given values and the formula for surface area. Substitute the knowns and then isolate the unknown variable through algebraic manipulation.

Question 9. तीन घनों जिनमें प्रत्येक की भुजा 4 सेमी है की बाह्य सतहों को आपस में जोड़ दिया गया है तो प्राप्त घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः तीन घनों को आपस में जोड़ने पर प्राप्त घनाभ की लम्बाई \(l = 4 + 4 + 4 = 12\) सेमी
\(b = 4\) सेमी
\(h = 4\) सेमी
अतः घनाभ का पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl) = 2(12 \times 4 + 4 \times 4 + 4 \times 12)\)
= \(2(48 + 16 + 48) = 2 \times 112 = 224 \text{ सेमी}^2\)
In simple words: When three cubes of equal side length are joined in a row, the new length of the resulting cuboid is three times the side of a single cube, while the breadth and height remain the same. Calculate its surface area using these new dimensions.

🎯 Exam Tip: For problems involving joining cubes, accurately determine the dimensions (length, breadth, height) of the new composite solid before applying the appropriate surface area formula.

Question 10. एक घनाभ का कुल पृष्ठीय क्षेत्रफल 372 सेमी है तथा इसका पार्श्व पृष्ठीय क्षेत्रफल 180 सेमी है। इसके आधार का क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घनाभ का कुल पृष्ठीय क्षेत्रफल = पार्श्व पृष्ठीय क्षेत्रफल + \(2 \times\) (आधार का क्षेत्रफल)
\(372 = 180 + 2 \times\) (आधार का क्षेत्रफल)
\(372 - 180 = 2 \times\) (आधार का क्षेत्रफल)
\(192 = 2 \times\) आधार का क्षेत्रफल
\( \frac{192}{2} = \) आधार का क्षेत्रफल
\( \implies \) आधार का क्षेत्रफल = \(96 \text{ सेमी}^2\)
In simple words: The total surface area of a cuboid is the sum of its lateral surface area and twice the area of its base. By subtracting the lateral surface area from the total surface area, we can find twice the base area and then determine the base area.

🎯 Exam Tip: Remember the relationship: Total Surface Area = Lateral Surface Area + 2 * (Base Area). This formula allows you to find any one component if the others are known.

Question 11. 8 सेमी भुजा वाले एक घन का पार्श्व पृष्ठीय क्षेत्रफल तथा कुल पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घन की भुजा = 8 सेमी
पार्श्व पृष्ठीय क्षेत्रफल = \(4 \times (\text{भुजा})^2 = 4 \times (8)^2\)
= \(4 \times 64 = 256 \text{ सेमी}^2\)
कुल पृष्ठीय क्षेत्रफल = \(6 \times (\text{भुजा})^2 = 6 \times (8)^2\)
= \(6 \times 64 = 384 \text{ सेमी}^2\)
In simple words: For a cube with a given side length, its lateral surface area (area of four walls) is \(4a^2\), and its total surface area (area of all six faces) is \(6a^2\).

🎯 Exam Tip: Clearly distinguish between lateral surface area \( (4a^2) \) and total surface area \( (6a^2) \) for a cube; a common error is confusing the two.

Question 12. यदि एक घनाभ की लम्बाई, चौड़ाई और ऊँचाई क्रमशः 15 सेमी, 10 सेमी तथा 20 सेमी है तो इसका पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घनाभ की \(l = 15\) सेमी, \(b = 10\) सेमी, \(h = 20\) सेमी
कुल पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl) = 2(15 \times 10 + 10 \times 20 + 20 \times 15)\)
= \(2(150 + 200 + 300) = 2(650) = 1300 \text{ सेमी}^2\)
In simple words: To calculate the total surface area of a cuboid, use its length, breadth, and height in the standard formula \(2(lb + bh + hl)\).

🎯 Exam Tip: Practice the formula for the surface area of a cuboid thoroughly. Be careful with calculations, especially when dealing with three sets of multiplications and additions.

Question 13. एक आयताकार हॉल के फर्श का परिमाप 250 मी है तथा ऊँचाई 6 मी है। चारों दीवारों का क्षेत्रफल ज्ञात कीजिए।
Answer: हलः हाँल की चारों दीवारों का क्षेत्रफल = \(2(l + b) \times h = \text{परिमाप} \times \text{ऊँचाई}\)
= \(250 \times 6 = 1500 \text{ मी}^2\)
In simple words: The area of the four walls of a rectangular hall is equivalent to the perimeter of its base multiplied by its height.

🎯 Exam Tip: Remember that the area of four walls (lateral surface area) is \(2(l+b)h\). If the perimeter of the floor \( (2(l+b)) \) is given directly, it simplifies the calculation.

Question 14. 24 सेमी × 12 सेमी x 5 सेमी आकार के एक डिब्बे को बनाने के लिए आवश्यक गत्ते का क्षेत्रफल ज्ञात कीजिए।
Answer: हलः डिब्बे का क्षेत्रफल = \(2(lb + bh + hl) = 2(24 \times 12 + 12 \times 5 + 5 \times 24)\)
= \(2(288 + 60 + 120) = 2(468) = 936 \text{ सेमी}^2\)
In simple words: The area of cardboard required to make a box is equal to the total surface area of the box, which can be calculated using its given length, breadth, and height.

🎯 Exam Tip: When making a box, "area of cardboard needed" refers to the total surface area of the cuboid. Ensure all dimensions are in the same unit.

Question 15. एक घनाभ की लम्बाई, चौड़ाई और गहराई का योग 19 सेमी है तथा इसके विकर्ण की लम्बाई 11 सेमी है। घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः
घनाभ की \(l + b + h = 19\) ... (1)
घनाभ का विकर्ण \( \sqrt{l^2 + b^2 + h^2} = 11 \)
समीकरण (1) का वर्ग करने पर
\( (l + b + h)^2 = 19^2 \)
\( l^2 + b^2 + h^2 + 2(lb + bh + hl) = 361 \)
(: \( l^2 + b^2 + h^2 = 121 \) क्योंकि \( \text{विकर्ण}^2 = 11^2 = 121 \))
\( 121 + 2(lb + bh + hl) = 361 \)
\( 2(lb + bh + hl) = 361 - 121 \)
\( 2(lb + bh + hl) = 240 \)
घनाभ का पृष्ठीय क्षेत्रफल = \(240 \text{ सेमी}^2\)
In simple words: Given the sum of a cuboid's dimensions and the length of its diagonal, its surface area can be found by squaring the sum of dimensions and using the diagonal squared value.

🎯 Exam Tip: Similar to Question 2, this problem uses the algebraic identity relating the sum of dimensions, sum of squares of dimensions, and surface area of a cuboid. Square the diagonal to get \(l^2+b^2+h^2\).

Question 16. एक घनाभ की विमाएँ 1 : 2 :3 के अनुपात में हैं तथा इसका कुल पृष्ठीय क्षेत्रफल 88 मी\(^2\) है। विमाएँ ज्ञात कीजिए ।
Answer: हलः
माना घनाभ की विमायें = \(x, 2x, 3x\)
\(2(lb + bh + hl) = 88\)
\(2(x \times 2x + 2x \times 3x + 3x \times x) = 88\)
\(2(2x^2 + 6x^2 + 3x^2) = 88\)
\(2(11x^2) = 88\)
\(11x^2 = \frac{88}{2}\)
\(11x^2 = 44\)
\(x^2 = \frac{44}{11}\)
\(x^2 = 4\)
\(x = \sqrt{4}\)
\(x = 2\)
घनाभ की विमायें = 2,4,6
In simple words: When the dimensions of a cuboid are given in a ratio and its total surface area is known, represent the dimensions as multiples of \(x\), form an equation using the surface area formula, solve for \(x\), and then find the actual dimensions.

🎯 Exam Tip: For problems with dimensions in a ratio, always introduce a common multiplier, e.g., \(x\), to represent the actual dimensions. Careful algebraic manipulation is key to solving for \(x\).

Question 17. तीन घनों जिनमें प्रत्येक की भुजा 5 सेमी है की बाह्य सतहों को आपस में जोड़ दिया गया है। परिणामी घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घनों को जोड़ने पर प्राप्त घनाभ की लम्बाई \(l = 5 + 5 + 5 = 15\) सेमी,
\(b = 5\) सेमी,
\(h = 5\) सेमी
परिणामी घनाभ का पृष्ठीय क्षेत्रफल = \(2(lb + bh + hl) = 2(15 \times 5 + 5 \times 5 + 5 \times 15)\)
= \(2(75 + 25 + 75) = 2 \times 175 = 350 \text{ सेमी}^2\)
In simple words: By joining three identical cubes, we form a cuboid whose length is three times the side of a single cube, while its breadth and height remain the same. Calculate the total surface area using these derived dimensions.

🎯 Exam Tip: Similar to Question 9, clearly define the new length, breadth, and height of the combined solid before applying the cuboid surface area formula. Visualizing the combination helps prevent errors.

Question 18. एक आयताकार हॉल के फर्श का परिमाप 250 मी है। इसकी ऊँचाई 6 मी है। Rs. 6 प्रति मीटर की दर से इसकी चारों दीवारों पर पुताई कराने का खर्च ज्ञात कीजिए।
Answer: हलः हॉल की चारों दीवारो का क्षेत्रफल = \(2(l + b) \times h = \text{परिमाप} \times \text{ऊँचाई}\)
= \(250 \times 6 = 1500 \text{ मी}^2\)
कुल खर्च = \(1500 \times 6 = \text{Rs. } 9000\)
In simple words: To find the cost of painting the four walls of a rectangular hall, first calculate the area of the four walls by multiplying the perimeter of the floor by the height, then multiply this area by the given cost per square meter.

🎯 Exam Tip: For cost-related problems, first calculate the total area to be painted. Then, multiply this area by the rate per unit area to find the total cost. Remember to use the correct units.

Question 19. एक घनाभ का कुल पृष्ठीय क्षेत्रफल 40 मी\(^2\) है तथा इसका पार्श्व पृष्ठीय क्षेत्रफल 26 मी\(^2\) है। इसके आधार का क्षेत्रफल ज्ञात कीजिए।
Answer: हलः घनाभ का कुल पृष्ठीय क्षेत्रफल = पार्श्व पृष्ठीय क्षेत्रफल + \(2 \times\) आधार का क्षेत्रफल
\(40 = 26 + 2 \times\) आधार का क्षेत्रफल
\(40 - 26 = 2 \times\) आधार का क्षेत्रफल
\(14 = 2 \times\) आधार का क्षेत्रफल
\( \frac{14}{2} = \) आधार का क्षेत्रफल
\( \implies \) आधार का क्षेत्रफल = \(7 \text{ मी}^2\)
In simple words: The base area of a cuboid can be found by subtracting its lateral surface area from its total surface area, and then dividing the result by two.

🎯 Exam Tip: This problem reinforces the relationship between total surface area, lateral surface area, and base area. It's a direct application of the formula: TSA = LSA + 2(Base Area).

Question 20. एक शीतगृह की लम्बाई, इसकी चौड़ाई की दोगुनी है। इसकी ऊँचाई 3 मी है। इसकी चारों दीवारों (दरवाजों सहित) का क्षेत्रफल 108 मी\(^2\) है। इसका आयतन ज्ञात कीजिए ।
Answer: हलः
\(l = 2b\) ... (1)
\(h = 3\) ... (2)
चारो दीवारों का क्षेत्रफल = \(2(l + b) \times h\)
\(108 = 2(2b + b) \times 3\)
\(108 = 6b \times 3\)
\(108 = 18b\)
\( b = \frac{108}{18} \)
\( b = 6 \text{ मी} \)
समीकरण (1) से
\( l = 2 \times 6 = 12 \text{ मी} \)
शीत गृह का आयतन = \(lbh\)
= \(12 \times 6 \times 3 = 216 \text{ मी}^3\)
In simple words: Given the relationship between length and breadth, the height, and the lateral surface area of a cold storage, we can first find the breadth and length, and then calculate its volume.

🎯 Exam Tip: Break down complex problems into steps. First, use the lateral surface area and height to find the base dimensions. Then, apply the volume formula \( (lbh) \) using the calculated dimensions.