Surface Areas and Volumes

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CLASS X Mathematics ~5 marks/year Ch 12 of 14
Surface Areas and Volumes

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Real objects (toys, capsules, tents, bird-baths) are combinations of the five basic solids — cuboid, cube, cylinder, cone, sphere, hemisphere.
  • Surface area of a combination = sum of the EXPOSED (visible) curved/flat surfaces — the joined faces vanish, so you do NOT just add the full TSAs.
  • Volume of a combination = sum of the volumes of the parts (nothing disappears here — for a cavity you subtract).
  • Master one formula table for $r,h,l$ of each solid; almost every question is "identify the parts, then add/subtract".
  • For a cone, the slant height $l=\sqrt{r^{2}+h^{2}}$ is needed for CSA — a very common slip-up.
  • Board weightage: ~5 marks/year — usually one combined-solid surface-area or volume problem (3–5 marks), often as a real-life word problem.
Detailed notes

1. What this chapter is really about

In Class 9 you learnt to find surface areas and volumes of the five basic solids: cuboid (and cube), cylinder, cone, sphere and hemisphere. In real life we rarely meet a single clean solid — instead we meet combinations:

  • An oil tanker = a cylinder with a hemisphere stuck at each end.
  • A test tube or medicine capsule = a cylinder with hemispherical end(s).
  • A spinning top (lattu) or ice-cream cone = a cone joined to a hemisphere.
  • A tent = a cylinder topped by a cone.

This chapter teaches just two skills: (a) find the surface area of such a combination, and (b) find its volume / capacity. The whole trick is to break the object into known basic solids, handle each piece with formulas you already know, and then combine them carefully.

Note (rationalised syllabus): the frustum of a cone and "conversion of one solid into another" (melting/recasting) have been trimmed from the current NCERT chapter. We base these notes strictly on the present chapter — combinations of solids.

2. The formula table — your single most useful tool

Memorise this. Here $r$ = radius, $h$ = height, $l$ = slant height of a cone. CSA = curved surface area (the rounded/side part only), TSA = total surface area (everything, including flat ends).

SolidCSA / LSATSAVolume
Cuboid ($l,b,h$)$2h(l+b)$$2(lb+bh+hl)$$lbh$
Cube (edge $a$)$4a^{2}$$6a^{2}$$a^{3}$
Cylinder$2\pi r h$$2\pi r(h+r)$$\pi r^{2}h$
Cone$\pi r l$$\pi r(l+r)$$\dfrac{1}{3}\pi r^{2}h$
Sphere$4\pi r^{2}$$4\pi r^{2}$$\dfrac{4}{3}\pi r^{3}$
Hemisphere$2\pi r^{2}$$3\pi r^{2}$$\dfrac{2}{3}\pi r^{3}$

Two relationships you will use again and again:

$$l=\sqrt{r^{2}+h^{2}}\qquad\text{(cone slant height)},\qquad r=\frac{d}{2}\quad(\text{radius from diameter}).$$

For a hemisphere: CSA (the curved dome) $=2\pi r^{2}$, while TSA $=3\pi r^{2}$ because it adds the flat circular base $\pi r^{2}$. For a solid sphere there is no flat face, so CSA = TSA $=4\pi r^{2}$.

3. Surface area of a combination — the golden rule

When two solids are joined, the surfaces where they touch disappear from the outside — you can no longer see them. So:

$$\text{TSA of combination}=\text{sum of the EXPOSED surface areas of all parts}.$$

You almost never add the full TSAs of the parts. Instead, walk over the new object and ask, "what can I actually see?" Typical patterns:

  • Cylinder + hemisphere on each end (tanker, capsule): you see CSA of cylinder + CSA of two hemispheres. The flat circular ends of the cylinder are covered.
  • Cone on a hemisphere (top/toy): you see CSA of cone + CSA of hemisphere. (Their joined flat circles vanish.)
  • Hemisphere fixed on a cube: TSA of cube, but minus the circle where it sits (a base of radius $r$), plus the curved dome $2\pi r^{2}$.
  • Hemisphere scooped OUT of a solid (depression): TSA of the block, minus the flat circle removed, plus the curved hollow $2\pi r^{2}$.

The key idea: when a flat circular face of area $\pi r^{2}$ is hidden or replaced, you subtract that $\pi r^{2}$ and add whatever new curved surface appears.

NCERT Example 1 — area to paint a spinning top (cone + hemisphere)

The top is a cone surmounted by a hemisphere. Total height $=5$ cm, diameter $=3.5$ cm so $r=\dfrac{3.5}{2}=1.75$ cm. Take $\pi=\dfrac{22}{7}$.

Hemisphere CSA $=2\pi r^{2}=2\times\dfrac{22}{7}\times\dfrac{3.5}{2}\times\dfrac{3.5}{2}$.

Height of cone $=$ total height $-$ radius of hemisphere $=5-\dfrac{3.5}{2}=5-1.75=3.25$ cm.

Slant height $l=\sqrt{r^{2}+h^{2}}=\sqrt{(1.75)^{2}+(3.25)^{2}}\approx3.7$ cm.

Cone CSA $=\pi r l=\dfrac{22}{7}\times\dfrac{3.5}{2}\times3.7$.

Area to colour $=$ CSA of hemisphere $+$ CSA of cone $=\dfrac{22}{7}\times\dfrac{3.5}{2}(3.5+3.7)=\dfrac{11}{2}(3.5+3.7)=\dfrac{11}{2}\times7.2\approx39.6\ \text{cm}^{2}$. Note this is not the sum of the full TSAs of the two pieces.

4. Hemisphere sitting on a cube (Example 2)

NCERT Example 2 — decorative block (cube + hemisphere)

A cube of edge $5$ cm carries a hemisphere of diameter $4.2$ cm on top, so $r=2.1$ cm. Take $\pi=\dfrac{22}{7}$.

TSA of cube $=6\times(\text{edge})^{2}=6\times5\times5=150\ \text{cm}^{2}$.

The hemisphere hides a circle of area $\pi r^{2}$ on the top face, and adds its curved dome $2\pi r^{2}$. So:

Surface area $=$ (TSA of cube) $-\pi r^{2}+2\pi r^{2}=150+\pi r^{2}$.

$\pi r^{2}=\dfrac{22}{7}\times2.1\times2.1=13.86\ \text{cm}^{2}$. Therefore surface area $=150+13.86=163.86\ \text{cm}^{2}$.

5. When the bases differ — exposed rings (Example 3)

If a cone sits on a cylinder but the cone's base is wider than the cylinder's, the leftover ring of the cone's base is exposed and must be counted.

NCERT Example 3 — toy rocket (cone on cylinder), $\pi=3.14$

Cone: $r=2.5$ cm, $h=6$ cm $\Rightarrow l=\sqrt{2.5^{2}+6^{2}}=\sqrt{6.25+36}=\sqrt{42.25}=6.5$ cm. Cylinder: $r'=1.5$ cm, height $=26-6=20$ cm.

Orange (cone) area $=$ CSA of cone $+$ base ring $=\pi r l+\pi r^{2}-\pi r'^{2}=\pi[(2.5\times6.5)+2.5^{2}-1.5^{2}]=\pi[16.25+6.25-2.25]=\pi\times20.25=3.14\times20.25=63.585\ \text{cm}^{2}$.

Yellow (cylinder) area $=$ CSA of cylinder $+$ its one base $=2\pi r'h'+\pi r'^{2}=\pi r'(2h'+r')=3.14\times1.5\times(40+1.5)=4.71\times41.5=195.465\ \text{cm}^{2}$.

6. Depression (cavity) on the surface (Example 4)

NCERT Example 4 — bird-bath (cylinder with hemispherical depression)

Cylinder height $h=1.45$ m $=145$ cm, radius $r=30$ cm. The hemispherical scoop at the top adds curved area but the visible surface is simply CSA of cylinder $+$ CSA of hemisphere.

TSA $=2\pi r h+2\pi r^{2}=2\pi r(h+r)=2\times\dfrac{22}{7}\times30\times(145+30)=2\times\dfrac{22}{7}\times30\times175=33000\ \text{cm}^{2}=3.3\ \text{m}^{2}$.

7. Volume of a combination — simply add

Volume behaves more kindly than surface area. When two solids are joined, no volume disappears, so:

$$\text{Volume of combination}=\text{sum of the volumes of the parts.}$$

If material is removed (a hemispherical scoop, a conical cavity, a sphere dropped in), you subtract the removed volume. "Capacity" means the volume of liquid/air it can hold.

NCERT Example 5 — shed (cuboid + half-cylinder), $\pi=\dfrac{22}{7}$

Cuboid: $15\times7\times8$. Half-cylinder on top: diameter $7$ m so $r=3.5$ m, length $15$ m.

Volume $=lbh+\dfrac{1}{2}\pi r^{2}\,(\text{length})=15\times7\times8+\dfrac12\times\dfrac{22}{7}\times3.5\times3.5\times15=840+288.75=1128.75\ \text{m}^{3}$.

Machinery $=300$ m³; workers $=20\times0.08=1.6$ m³. Air left $=1128.75-(300+1.6)=827.15\ \text{m}^{3}$.

NCERT Example 6 — juice glass with raised hemispherical bottom, $\pi=3.14$

Glass is a cylinder: diameter $5$ cm $\Rightarrow r=2.5$ cm, height $10$ cm.

Apparent capacity $=\pi r^{2}h=3.14\times2.5\times2.5\times10=196.25\ \text{cm}^{3}$.

The raised hemisphere reduces it by $\dfrac{2}{3}\pi r^{3}=\dfrac{2}{3}\times3.14\times2.5^{3}=32.71\ \text{cm}^{3}$.

Actual capacity $=196.25-32.71=163.54\ \text{cm}^{3}$.

NCERT Example 7 — toy (hemisphere + cone) and circumscribing cylinder, $\pi=3.14$

Diameter $4$ cm $\Rightarrow r=2$ cm; cone height $h=2$ cm.

Volume of toy $=\dfrac{2}{3}\pi r^{3}+\dfrac{1}{3}\pi r^{2}h=\dfrac{2}{3}\times3.14\times8+\dfrac{1}{3}\times3.14\times4\times2=16.747\ldots+8.373\ldots=25.12\ \text{cm}^{3}$.

Circumscribing cylinder: radius $2$ cm, height $=AO+OP=2+2=4$ cm. Its volume $=\pi r^{2}\times4=3.14\times4\times4=50.24\ \text{cm}^{3}$.

Difference $=50.24-25.12=25.12\ \text{cm}^{3}$.

8. NCERT Exercise 12.1 — fully solved

Unless stated otherwise, $\pi=\dfrac{22}{7}$.

Q1. Two cubes of volume 64 cm³ joined end to end — surface area of the cuboid. Edge $=\sqrt[3]{64}=4$ cm. Joined cuboid: $l=8,\ b=4,\ h=4$. Surface area $=2(lb+bh+hl)=2(8\times4+4\times4+4\times8)=2(32+16+32)=2\times80=160\ \text{cm}^{2}$.

Q2. Hollow hemisphere mounted by a hollow cylinder; diameter 14 cm, total height 13 cm — inner surface area. $r=7$ cm. Cylinder height $=13-7=6$ cm. Inner SA $=2\pi r h+2\pi r^{2}=2\pi r(h+r)=2\times\dfrac{22}{7}\times7\times(6+7)=44\times13=572\ \text{cm}^{2}$.

Q3. Cone on a hemisphere, both radius 3.5 cm, total height 15.5 cm — TSA. $r=3.5$ cm. Cone height $=15.5-3.5=12$ cm; $l=\sqrt{3.5^{2}+12^{2}}=\sqrt{12.25+144}=\sqrt{156.25}=12.5$ cm. TSA $=\pi r l+2\pi r^{2}=\dfrac{22}{7}\times3.5\times12.5+2\times\dfrac{22}{7}\times3.5\times3.5=137.5+77=214.5\ \text{cm}^{2}$.

Q4. Cube of side 7 cm surmounted by a hemisphere — greatest diameter and surface area. Greatest diameter $=$ side $=7$ cm, so $r=3.5$ cm. Surface area $=6a^{2}-\pi r^{2}+2\pi r^{2}=6a^{2}+\pi r^{2}=6\times49+\dfrac{22}{7}\times3.5\times3.5=294+38.5=332.5\ \text{cm}^{2}$.

Q5. Hemispherical depression cut from one face of a cube, diameter $l$ = edge — surface area of remaining solid. $r=\dfrac{l}{2}$. SA $=6l^{2}-\pi r^{2}+2\pi r^{2}=6l^{2}+\pi r^{2}=6l^{2}+\pi\dfrac{l^{2}}{4}=\dfrac{l^{2}}{4}(24+\pi)$ square units.

Q6. Medicine capsule: cylinder with a hemisphere at each end; length 14 mm, diameter 5 mm — surface area. $r=2.5$ mm. Cylinder length $=14-2(2.5)=14-5=9$ mm. SA $=2\pi r h+2(2\pi r^{2})=2\pi r(h+2r)=2\times\dfrac{22}{7}\times2.5\times(9+5)=\dfrac{110}{7}\times14=220\ \text{mm}^{2}$.

Q7. Tent: cylinder + conical top; cylinder height 2.1 m, diameter 4 m, slant height of cone 2.8 m — canvas area and cost at ₹500/m². $r=2$ m. Canvas $=$ CSA cylinder $+$ CSA cone $=2\pi r h+\pi r l=\pi r(2h+l)=\dfrac{22}{7}\times2\times(4.2+2.8)=\dfrac{44}{7}\times7=44\ \text{m}^{2}$. Cost $=44\times500=₹22000$.

Q8. Cone hollowed out of a cylinder, same height 2.4 cm and diameter 1.4 cm — TSA of remaining solid (nearest cm²). $r=0.7$ cm, $h=2.4$ cm; $l=\sqrt{0.7^{2}+2.4^{2}}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5$ cm. Remaining solid shows: CSA of cylinder $+$ one base of cylinder $+$ CSA of the conical cavity $=2\pi r h+\pi r^{2}+\pi r l=\pi r(2h+r+l)=\dfrac{22}{7}\times0.7\times(4.8+0.7+2.5)=2.2\times8.0=17.6\approx18\ \text{cm}^{2}$.

Q9. Hemisphere scooped from each end of a cylinder; height 10 cm, radius 3.5 cm — TSA of the article. $r=3.5$ cm. Visible: CSA of cylinder $+$ CSA of two hemispheres $=2\pi r h+2(2\pi r^{2})=2\pi r(h+2r)=2\times\dfrac{22}{7}\times3.5\times(10+7)=22\times17=374\ \text{cm}^{2}$.

9. NCERT Exercise 12.2 — fully solved

Unless stated otherwise, $\pi=\dfrac{22}{7}$.

Q1. Cone on a hemisphere, both radius 1 cm, cone height = radius — volume in terms of $\pi$. $r=1,\ h=1$. Volume $=\dfrac{1}{3}\pi r^{2}h+\dfrac{2}{3}\pi r^{3}=\dfrac{1}{3}\pi(1)+\dfrac{2}{3}\pi(1)=\pi\ \text{cm}^{3}$.

Q2. Cylinder with a cone at each end; diameter 3 cm, total length 12 cm, each cone height 2 cm — volume of air. $r=1.5$ cm. Cylinder length $=12-2(2)=8$ cm. Volume $=\pi r^{2}h_{\text{cyl}}+2\!\left(\dfrac{1}{3}\pi r^{2}h_{\text{cone}}\right)=\pi(1.5)^{2}\!\left(8+\dfrac{2\times2}{3}\right)=\dfrac{22}{7}\times2.25\times\dfrac{28}{3}\approx66\ \text{cm}^{3}$.

Q3. 45 gulab jamuns (cylinder + two hemispherical ends), length 5 cm, diameter 2.8 cm; syrup ≈ 30% of volume. $r=1.4$ cm. Cylinder length $=5-2(1.4)=2.2$ cm. One jamun $=\pi r^{2}h+\dfrac{4}{3}\pi r^{3}=\dfrac{22}{7}(1.4)^{2}\!\left(2.2+\dfrac{4\times1.4}{3}\right)=\dfrac{22}{7}\times1.96\times\dfrac{12.2}{3}\approx25.05\ \text{cm}^{3}$. For 45: $\approx1127.3\ \text{cm}^{3}$. Syrup $=30\%\approx338\ \text{cm}^{3}$.

Q4. Wood cuboid 15×10×3.5 cm with 4 conical depressions (radius 0.5 cm, depth 1.4 cm) — volume of wood. Cuboid $=15\times10\times3.5=525\ \text{cm}^{3}$. One cone $=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\times\dfrac{22}{7}\times0.25\times1.4=\dfrac{11}{30}\approx0.3667\ \text{cm}^{3}$; four cones $=4\times0.3667=1.4667\ \text{cm}^{3}$. Wood $=525-1.4667\approx523.53\ \text{cm}^{3}$.

Q5. Inverted cone (height 8 cm, top radius 5 cm) full of water; lead spheres radius 0.5 cm dropped, ¼ of water flows out — number of shots. Cone volume $=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\pi\times25\times8=\dfrac{200\pi}{3}$. Water out $=\dfrac{1}{4}\times\dfrac{200\pi}{3}=\dfrac{50\pi}{3}$. One shot $=\dfrac{4}{3}\pi(0.5)^{3}=\dfrac{4}{3}\pi\times0.125=\dfrac{\pi}{6}$. Number $=\dfrac{50\pi/3}{\pi/6}=\dfrac{50}{3}\times6=100$ lead shots.

Q6. Iron pole: cylinder (height 220 cm, base diameter 24 cm) + cylinder (height 60 cm, radius 8 cm); mass at 8 g/cm³, $\pi=3.14$. Lower cylinder $r=12$: $\pi\times144\times220=99475.2\ \text{cm}^{3}$. Upper cylinder $r=8$: $\pi\times64\times60=12057.6\ \text{cm}^{3}$. Total volume $=111532.8\ \text{cm}^{3}$. Mass $=111532.8\times8=892262.4\ \text{g}\approx892.26\ \text{kg}$.

Q7. Cone (height 120 cm, radius 60 cm) on a hemisphere (radius 60 cm) placed in a cylinder (radius 60 cm, height 180 cm) full of water — volume of water left. Cylinder $=\pi r^{2}H=\pi\times3600\times180=648000\pi$. Solid $=\dfrac{1}{3}\pi r^{2}h+\dfrac{2}{3}\pi r^{3}=\dfrac{1}{3}\pi\times3600\times120+\dfrac{2}{3}\pi\times216000=144000\pi+144000\pi=288000\pi$. Water left $=648000\pi-288000\pi=360000\pi=360000\times\dfrac{22}{7}\approx1131428.6\ \text{cm}^{3}\approx1.131\ \text{m}^{3}$.

Q8. Spherical vessel with cylindrical neck (8 cm long, 2 cm diameter), spherical part diameter 8.5 cm; child claims volume 345 cm³, $\pi=3.14$. Neck: $r=1$, $\pi r^{2}h=3.14\times1\times8=25.12$. Sphere: $r=4.25$, $\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times3.14\times76.765625\approx321.39$. Total $=25.12+321.39=346.51\ \text{cm}^{3}$. The child's value 345 cm³ is not correct (correct value ≈ 346.51 cm³).

10. Common mistakes to avoid

  • Adding full TSAs of the parts for a surface-area question — only the exposed (visible) surfaces count; joined faces vanish.
  • Forgetting the slant height: CSA of a cone is $\pi r l$, not $\pi r h$. Always compute $l=\sqrt{r^{2}+h^{2}}$ first.
  • Confusing diameter with radius — halve the diameter before substituting.
  • Using TSA of a hemisphere ($3\pi r^{2}$) when only the curved dome ($2\pi r^{2}$) is exposed in a combination.
  • For a combined-solid length like a capsule, subtract the hemispheres' radii to get the true cylinder length (e.g. cylinder $=$ total $-2r$).
  • Forgetting to subtract the removed volume for a cavity/depression, or to add the exposed ring when bases differ.
  • Mixing units (cm and m) in one calculation — convert everything first.

11. Quick revision checklist

  • Identify the basic solids, their shared radius, and each height.
  • Cone slant height $l=\sqrt{r^{2}+h^{2}}$ before any CSA of a cone.
  • Surface area: add only exposed curved/flat areas; subtract a $\pi r^{2}$ wherever a flat face is hidden/replaced.
  • Volume: add the parts; subtract for cavities. "Capacity" = volume held.
  • Hemisphere: CSA $=2\pi r^{2}$, volume $=\dfrac{2}{3}\pi r^{3}$. Sphere: SA $=4\pi r^{2}$, volume $=\dfrac{4}{3}\pi r^{3}$.
  • Convert all measurements to a single unit; report area in (unit)² and volume in (unit)³.
Practice MCQs
1. The curved surface area of a hemisphere of radius $r$ is:
  1. $4\pi r^{2}$
  2. $3\pi r^{2}$
  3. $2\pi r^{2}$
  4. $\pi r^{2}$
Answer: (C) $2\pi r^{2}$ — the curved dome only; TSA would be $3\pi r^{2}$.
2. Two cubes of edge $a$ are joined end to end. The surface area of the resulting cuboid is:
  1. $12a^{2}$
  2. $10a^{2}$
  3. $8a^{2}$
  4. $6a^{2}$
Answer: (B) cuboid $2a\times a\times a$: $2(2a\cdot a+a\cdot a+a\cdot2a)=2(2a^{2}+a^{2}+2a^{2})=10a^{2}$.
3. A cone of base radius $r$, height $h$ has slant height:
  1. $r+h$
  2. $\sqrt{r^{2}+h^{2}}$
  3. $\sqrt{r^{2}-h^{2}}$
  4. $rh$
Answer: (B) $l=\sqrt{r^{2}+h^{2}}$ (Pythagoras on the right triangle of the cone).
4. When a solid hemisphere of radius $r$ is fixed on the flat top of a cube, the surface area changes by:
  1. $+2\pi r^{2}$
  2. $+\pi r^{2}$
  3. $+3\pi r^{2}$
  4. $-\pi r^{2}$
Answer: (B) lose base $\pi r^{2}$, gain dome $2\pi r^{2}$, net $+\pi r^{2}$.
5. The volume of a sphere of radius $r$ is:
  1. $\dfrac{2}{3}\pi r^{3}$
  2. $\dfrac{4}{3}\pi r^{3}$
  3. $\dfrac{1}{3}\pi r^{3}$
  4. $4\pi r^{3}$
Answer: (B) $\dfrac{4}{3}\pi r^{3}$; a hemisphere is half of this, $\dfrac{2}{3}\pi r^{3}$.
6. A capsule is a cylinder with a hemisphere at each end; total length $14$ mm, diameter $5$ mm. Its surface area is:
  1. $110\ \text{mm}^{2}$
  2. $220\ \text{mm}^{2}$
  3. $330\ \text{mm}^{2}$
  4. $440\ \text{mm}^{2}$
Answer: (B) $r=2.5$, cylinder $=9$ mm; $2\pi r(h+2r)=2\times\dfrac{22}{7}\times2.5\times14=220\ \text{mm}^{2}$.
7. When two basic solids are joined, the total volume of the combination equals:
  1. sum of the volumes
  2. sum of the volumes minus the joined faces
  3. sum of the surface areas
  4. difference of the volumes
Answer: (A) volume is conserved on joining; nothing disappears.
8. A cylinder, a cone and a hemisphere have the same radius and the same height $r$. The ratio of their volumes is:
  1. $1:2:3$
  2. $3:1:2$
  3. $2:1:3$
  4. $1:3:2$
Answer: (B) $\pi r^{3}:\dfrac{1}{3}\pi r^{3}:\dfrac{2}{3}\pi r^{3}=3:1:2$.
9. The total surface area of a solid hemisphere of radius $r$ is:
  1. $2\pi r^{2}$
  2. $3\pi r^{2}$
  3. $4\pi r^{2}$
  4. $\pi r^{2}$
Answer: (B) curved $2\pi r^{2}$ + flat base $\pi r^{2}=3\pi r^{2}$.
10. A hemispherical depression is scooped from one face of a cube of edge $a$ (diameter $=a$). The remaining surface area is:
  1. $6a^{2}$
  2. $6a^{2}-\dfrac{\pi a^{2}}{4}$
  3. $6a^{2}+\dfrac{\pi a^{2}}{2}$
  4. $6a^{2}+\dfrac{\pi a^{2}}{4}$
Answer: (D) $6a^{2}-\pi r^{2}+2\pi r^{2}=6a^{2}+\pi r^{2}=6a^{2}+\dfrac{\pi a^{2}}{4}$ (since $r=\tfrac{a}{2}$).
Assertion–Reason
A: The surface area of a combination of solids is generally less than the sum of the total surface areas of its parts.   R: When solids are joined, the surfaces at the join are no longer exposed.
Answer: Both A and R are true, and R is the correct explanation of A — hidden joined faces are dropped from the count.
A: The volume of a toy made of a cone on a hemisphere equals the sum of their volumes.   R: Volume is lost at the surface where two solids are joined.
Answer: A is true, R is false — no volume is lost on joining; only surface area is affected.
Previous-year questions
Q1. A toy is in the form of a cone of radius $3.5$ cm mounted on a hemisphere of the same radius. The total height of the toy is $15.5$ cm. Find its total surface area. (CBSE, 3 marks)
Answer: $r=3.5$, cone height $=12$, $l=12.5$. TSA $=\pi r l+2\pi r^{2}=\dfrac{22}{7}\times3.5\times12.5+2\times\dfrac{22}{7}\times3.5^{2}=137.5+77=214.5\ \text{cm}^{2}$.
Q2. A tent is a cylinder (height $2.1$ m, diameter $4$ m) surmounted by a cone of slant height $2.8$ m. Find the canvas area and its cost at ₹500/m². (CBSE, 3 marks)
Answer: $r=2$. Canvas $=2\pi r h+\pi r l=\pi r(2h+l)=\dfrac{22}{7}\times2\times7=44\ \text{m}^{2}$. Cost $=44\times500=₹22000$.
Q3. From a solid cylinder of height $2.4$ cm and diameter $1.4$ cm, a conical cavity of the same height and diameter is hollowed out. Find the TSA of the remaining solid (nearest cm²). (CBSE, 4 marks)
Answer: $r=0.7$, $h=2.4$, $l=2.5$. TSA $=2\pi r h+\pi r^{2}+\pi r l=\pi r(2h+r+l)=\dfrac{22}{7}\times0.7\times8=17.6\approx18\ \text{cm}^{2}$.
Q4. A solid is a cone of height $120$ cm and radius $60$ cm standing on a hemisphere of radius $60$ cm, placed in a cylinder (radius $60$ cm, height $180$ cm) full of water. Find the volume of water left. (CBSE, 4 marks)
Answer: Cylinder $=648000\pi$, solid (cone + hemisphere) $=144000\pi+144000\pi=288000\pi$. Water left $=360000\pi\approx1.131\ \text{m}^{3}$.
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