Question:
The curved surface area of a right circular cone is \(444\pi\text{ cm}^2\) and the diameter of its base is 24 cm. What is the volume of the cone (in \(\text{cm}^3\))?
एक लम्बवृत्तीय शंकु का वक्र पृष्ठीय क्षेत्रफल \(444\pi\text{ cm}^2\) है और इसके आधार का व्यास 24 cm है। शंकु का आयतन (\(\text{cm}^3\) में) क्या है?
Options:
- \(840\pi\)
- \(1120\pi\)
- \(1400\pi\)
- \(1680\pi\)
- \(840\pi\)
- \(1120\pi\)
- \(1400\pi\)
- \(1680\pi\)
Correct Answer: d)
Explanation:
Step 1: Find the slant height \(l\) of the cone:
Base diameter = 24 cm \(\implies\) Base radius \(r = \frac{24}{2} = 12\text{ cm}\).
Curved Surface Area (CSA) of a cone = \(\pi r l\):
\[ \pi r l = 444\pi \]
\[ 12 \times l = 444 \implies l = \frac{444}{12} = 37\text{ cm} \]
Step 2: Calculate the vertical height \(h\) of the cone:
Using Pythagoras’ theorem \(h^2 + r^2 = l^2\):
\[ h = \sqrt{l^2 – r^2} = \sqrt{37^2 – 12^2} = \sqrt{1369 – 144} = \sqrt{1225} = 35\text{ cm} \]
Step 3: Calculate the volume of the cone:
\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 12^2 \times 35 = \frac{1}{3} \pi \times 144 \times 35 = 48 \times 35 \pi = 1680\pi\text{ cm}^3 \]
Hence, the correct answer is option d).
चरण 1: शंकु की तिर्यक ऊँचाई \(l\) ज्ञात करें:
आधार का व्यास = 24 cm \(\implies\) त्रिज्या \(r = 12\text{ cm}\)।
शंकु का वक्र पृष्ठीय क्षेत्रफल = \(\pi r l\):
\[ \pi \times 12 \times l = 444\pi \implies l = \frac{444}{12} = 37\text{ cm} \]
चरण 2: शंकु की ऊँचाई \(h\) ज्ञात करें:
पाइथागोरस प्रमेय से:
\[ h = \sqrt{l^2 – r^2} = \sqrt{37^2 – 12^2} = \sqrt{1369 – 144} = \sqrt{1225} = 35\text{ cm} \]
चरण 3: शंकु का आयतन ज्ञात करें:
\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 12^2 \times 35 = 48 \times 35 \pi = 1680\pi\text{ cm}^3 \]
अतः सही उत्तर विकल्प d) है।