---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch11
exercise: ex11.1
difficulty: intermediate
---

Class 9 | Ch11: Surface Areas and Volumes | Exercise 11.1 (Cone)

Key Formulas — Cone (radius=r, height=h, slant height=l=√(r²+h²)):
Curved Surface Area (CSA) = πrl
Total Surface Area (TSA) = πrl + πr² = πr(l+r)
Volume = (1/3)πr²h

Q1. Diameter of base=10.5 cm, slant height=10 cm. Find CSA.
Answer: r=5.25 cm, l=10 cm. CSA=πrl=22/7×5.25×10=165 cm². ✅

Q2. Find TSA of cone with slant height 21 m and diameter 24 m.
Answer: r=12 m, l=21 m. TSA=πr(l+r)=22/7×12×(21+12)=22/7×12×33=1244.57≈1244.57 m². ✅

Q3. CSA=308 cm², slant height=14 cm. Find radius and TSA.
Answer: CSA=πrl=308 → r=308/(π×14)=308×7/(22×14)=7 cm. TSA=πr(l+r)=22/7×7×(14+7)=22×21=462 cm². ✅

Q4. Conical tent 10 m high, radius 24 m. Find slant height and CSA.
Answer: l=√(r²+h²)=√(576+100)=√676=26 m. CSA=πrl=22/7×24×26=1961.14 m². ✅

Q5. Slant height of tent 14 m and CSA=308 m². Find width of tarpaulin needed.
Answer: CSA=308=πrl → r=308/(π×14)=7 m. Width of tarpaulin=1 m (given). Length=308/1=308 m. ✅

Q6. Perimeter of base of cone=44 cm, slant height=25 cm. Find CSA and TSA.
Answer: 2πr=44 → r=7 cm. CSA=πrl=22/7×7×25=550 cm². TSA=πr(l+r)=22/7×7×(25+7)=704 cm². ✅

Q7. Volume of right circular cone=9856 cm³, diameter=28 cm. Find slant height and CSA.
Answer: r=14, V=(1/3)πr²h=9856 → h=9856×3/(π×196)=48 cm. l=√(196+2304)=√2500=50 cm. CSA=πrl=22/7×14×50=2200 cm². ✅

Q8. 50 hollow cones, base diameter=40 cm, height=1 m. Cost of painting outer side=₹12 per m². Find total cost.
Answer: r=20 cm=0.2 m, h=1 m. l=√(0.04+1)=√1.04≈1.02 m. CSA each=πrl=π×0.2×1.02≈0.6408 m². 50 cones: 50×0.6408=32.04 m². Cost=32.04×12=₹384.48. ✅