knowledge-base/class_10/ch12/0194_geometry.md

---
source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch12
exercise: ex12.1
difficulty: intermediate
---

Class 10 | Ch12: Surface Areas and Volumes | Exercise 12.1

Combined Solids — Surface Area:
When two solids are combined, the surface area of the resulting solid = total surface area of both MINUS the areas that are hidden/joined.

Key Formulas:
Cylinder: CSA=2πrh, TSA=2πr(r+h), V=πr²h
Cone: CSA=πrl, TSA=πr(r+l), V=⅓πr²h, l=√(r²+h²)
Sphere: SA=4πr², V=⅔πr³
Hemisphere: CSA=2πr², TSA=3πr², V=⅔πr³

Q1. Two cubes, each of volume 64cm³, joined end to end. Find SA of resulting cuboid.
Answer: Volume of cube=64 → side=4cm. Cuboid: 8cm×4cm×4cm. SA=2(8×4+4×4+4×8)=2(32+16+32)=160cm². ✅

Q2. Container is hemisphere surmounted by a hollow cylinder. Diameter=14cm, height of cylinder=13cm. Find total CSA.
Answer: r=7cm. CSA of cylinder=2πrh=2×22/7×7×13=572cm². CSA of hemisphere=2πr²=2×22/7×49=308cm². Total=572+308=880cm². ✅

Q3. Toy is cone on hemisphere, both radius 3.5cm. Height of cone=12cm. Find TSA.
Answer: l=√(3.5²+12²)=√(12.25+144)=√156.25=12.5cm. TSA=πrl+2πr²=π×3.5×12.5+2π×3.5²=43.75π+24.5π=68.25π=214.5cm². ✅

Q4. Cuboid dimensions 15cm×10cm×3.5cm. How many silver coins (diameter=1.75cm, thickness=2mm) can be melted to form it?
Answer: V cuboid=525cm³. V coin=π×(0.875)²×0.2=π×0.765625×0.2=0.481cm³. n=525/0.481≈1092. ✅

Q5. Shanta runs ice cream parlour. She supplies ice cream in cones made from scratch — cone+hemisphere combo. Diameter=5cm, height of cone=3.7cm. Find surface area and volume.
Answer: r=2.5, h=3.7, l=√(6.25+13.69)=√19.94≈4.5cm. SA=πrl+2πr²=π(2.5)(4.5)+2π(6.25)=11.25π+12.5π=23.75π≈74.6cm². V=⅓πr²h+⅔πr³=⅓π×6.25×3.7+⅔π×15.625=7.7π+10.4π=18.1π≈56.8cm³. ✅

Q6. Rocket is in shape of cylinder surmounted by cone. Diameter=6cm, cylinder height=12cm, cone height=4cm. Find TSA.
Answer: r=3. l=√(9+16)=5cm. TSA=CSA of cone+CSA of cylinder+base circle=πrl+2πrh+πr²=π(3×5+2×3×12+9)=π(15+72+9)=96π≈301.4cm². ✅