Class 10 CBSE Maths Chapter 13: Surface Areas and Volumes, including:

✅ Key Concepts & Formulas

📘 All NCERT Exercise Questions with detailed solutions

📝 CBSE Previous Year Questions from this chapter

📚 Chapter 13: Surface Areas and Volumes

This chapter explores the curved surface area (CSA), total surface area (TSA), and volume of 3D solids, including cylinders, cones, spheres, hemispheres, frustums, and composite solids.

🧠 Key Concepts and Formulas

Let $r$ = radius, $h$ = height

CSA = $2\pi rh$

TSA = $2\pi r(h + r)$

Volume = $\pi r^2 h$

Let $r$ = radius, $h$ = height, $l$ = slant height
Where $l = \sqrt{r^2 + h^2}$

CSA = $\pi r l$

TSA = $\pi r(l + r)$

Volume = $\frac{1}{3} \pi r^2 h$

Let $r$ = radius

Surface Area = $4\pi r^2$

Volume = $\frac{4}{3} \pi r^3$

CSA = $2\pi r^2$

TSA = $3\pi r^2$

Volume = $\frac{2}{3} \pi r^3$

Let $R$ = larger radius, $r$ = smaller radius, $h$ = height, $l = \sqrt{h^2 + (R - r)^2}$

CSA = $\pi (R + r)l$

TSA = $\pi (R + r)l + \pi R^2 + \pi r^2$

Volume = $\frac{1}{3} \pi h (R^2 + r^2 + Rr)$

Add volumes or surface areas of individual components as needed. Visualize how solids are joined (e.g., cone on hemisphere, hemisphere on cylinder, etc.)

📘 NCERT Exercise Solutions

$\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$

$\text{TSA} = 2\pi r (r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 10) = 2112 \text{ cm}^2$

$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = 1232 \text{ cm}^3$

$V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 10.5^3 = 4851 \text{ cm}^3$

$\text{Total Volume} = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \pi r^2\left(\frac{2r + h}{3}\right) = \frac{22}{7} \times 3.5^2 \times \frac{2 \times 3.5 + 12}{3} \approx 268.5 \text{ cm}^3$

Use volume conservation:
$\text{Volume of sphere} = n \times \text{Volume of one cone} \Rightarrow n = \frac{V_{\text{sphere}}}{V_{\text{cone}}}$

$V = \frac{1}{3} \pi h(R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 4 (49 + 9 + 21) = 1069.71 \text{ cm}^3$

📝 CBSE Previous Year Questions (PYQs)

Year Question Marks

2024 A hemispherical bowl of radius 7 cm is filled with liquid and poured into cones. Find number of cones formed. 3

2023 A toy made of a cone mounted on a hemisphere. Find surface area. 4

2022 Frustum-based question on surface area 3

2020 Volume of metal cube melted into a cylinder 3

2019 Ice-cream cone with hemisphere — find total surface area 4

2018 Cone and cylinder combination — find volume 4

2017 Frustum volume and CSA 4

2016 Sphere converted into smaller balls — find how many 3

Previous Year Questions (PYQs) from Class 10 Maths Chapter 13: Surface Areas and Volumes, covering a wide variety of patterns asked in board exams.

📝 CBSE PYQ Solutions – Chapter 13: Surface Areas and Volumes

Q. A hemispherical bowl of radius 7 cm is filled with liquid and poured into cylindrical cones each of radius 3.5 cm and height 6 cm. Find the number of cones filled.

Solution:

Volume of hemisphere:
$V_1 = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = 716.67 \text{ cm}^3$
Volume of 1 cone:
$V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 6 = \frac{22}{7} \times \frac{12.25 \times 6}{3} = 77 \text{ cm}^3$
Number of cones =
$\frac{716.67}{77} = \boxed{9.3} \Rightarrow \boxed{9 \text{ cones (completely)}}$

Q. A toy is in the shape of a cone mounted on a hemisphere. Radius = 3.5 cm, height of cone = 5 cm. Find total surface area.

Solution:

Radius $r = 3.5$ cm

Cone height $h = 5$ cm

Slant height $l = \sqrt{r^2 + h^2} = \sqrt{12.25 + 25} = \sqrt{37.25} ≈ 6.1 \text{ cm}$

CSA of cone = $\pi r l = \frac{22}{7} \times 3.5 \times 6.1 ≈ 67.1 \text{ cm}^2$
CSA of hemisphere = $2\pi r^2 = 2 \times \frac{22}{7} \times 3.5^2 = 77 \text{ cm}^2$

Total Surface Area = $67.1 + 77 = \boxed{144.1 \text{ cm}^2}$

Q. A frustum has height 10 cm, radii of bases are 4 cm and 2 cm. Find total surface area.

Solution:

Given:

$R = 4$ , $r = 2$ , $h = 10$

Slant height $l$ =
$\sqrt{(R - r)^2 + h^2} = \sqrt{(2)^2 + (10)^2} = \sqrt{4 + 100} = \sqrt{104} ≈ 10.2 \text{ cm}$
CSA = $\pi(R + r)l = \frac{22}{7}(6)(10.2) ≈ 193.7 \text{ cm}^2$
Area of both bases = $\pi R^2 + \pi r^2 = \frac{22}{7}(16 + 4) = \frac{22}{7} \times 20 = 62.86$

TSA = CSA + base areas
$≈ 193.7 + 62.86 = \boxed{256.56 \text{ cm}^2}$

Q. A cube of side 10 cm is melted and recast into a cylinder of radius 5 cm. Find height.

Solution:

Volume of cube = $a^3 = 10^3 = 1000 \text{ cm}^3$
Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times 25 \times h$

Equating:
$\frac{22}{7} \times 25 \times h = 1000 \Rightarrow h = \frac{1000 \times 7}{22 \times 25} = \boxed{12.73 \text{ cm}}$

Q. A toy is made by attaching a hemisphere to the base of a cone. Radius = 3.5 cm, height = 10.5 cm. Find TSA.

Solution:

Radius $r = 3.5$ , height $h = 10.5$

Slant height $l = \sqrt{r^2 + h^2} = \sqrt{12.25 + 110.25} = \sqrt{122.5} ≈ 11.07$

CSA of cone = $\pi r l = \frac{22}{7} \times 3.5 \times 11.07 ≈ 121.15$
CSA of hemisphere = $2\pi r^2 = 77$

TSA = 121.15 + 77 = \boxed{198.15 \text{ cm}^2}

Q. A solid is composed of a cylinder of radius 4 cm, height 10 cm and a cone of same base radius and height 5 cm mounted on top. Find volume.

Solution:

Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times 16 \times 10 = 502.86 \text{ cm}^3$
Volume of cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 16 \times 5 = 120.0 \text{ cm}^3$

Total Volume = $502.86 + 120 = \boxed{622.86 \text{ cm}^3}$

Q. A frustum of cone has radii 6 cm and 3 cm, height 4 cm. Find volume and CSA.

Solution:

$R = 6$ , $r = 3$ , $h = 4$

Volume:
$V = \frac{1}{3} \pi h(R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 4(36 + 9 + 18) = \boxed{369.14 \text{ cm}^3}$
Slant height:
$l = \sqrt{(6 - 3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
CSA = $\pi(R + r)l = \frac{22}{7} \times 9 \times 5 = \boxed{141.43 \text{ cm}^2}$

Q. A spherical ball of radius 7 cm is melted and recast into smaller balls of radius 1 cm. Find number of balls.

Solution:

Volume of original sphere = $\frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times 343 = 1436.19$

Volume of one small ball = $\frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 1 = 4.19$

Number of balls =
$\frac{1436.19}{4.19} ≈ \boxed{343 \text{ balls}}$

📎 Summary Table of Formulas

Solid Surface Area Volume

Cylinder TSA: $2\pi r(h + r)$ $\pi r^2 h$

Cone TSA: $\pi r(l + r)$ $\frac{1}{3} \pi r^2 h$

Sphere $4\pi r^2$ $\frac{4}{3} \pi r^3$

Hemisphere TSA: $3\pi r^2$ $\frac{2}{3} \pi r^3$

Frustum TSA: $\pi(R + r)l + \pi R^2 + \pi r^2$ $\frac{1}{3} \pi h(R^2 + r^2 + Rr)$

Tags:
#Class10Maths, #SurfaceAreasAndVolumes, #CBSEMaths, #NCERTSolutions, #MathsMadeEasy, #GeometryClass10, #BoardExamPrep, #VolumeAndArea, #MensurationFormulas, #FrustumProblems

Class 10 CBSE Maths Chapter 13: Surface Areas and Volumes, including: ✅ Key Concepts & Formulas 📘 All NCERT Exercise Questions with detailed solutions 📝 CBSE Previous Year Questions from this chapter

📚 Chapter 13: Surface Areas and Volumes

This chapter explores the curved surface area (CSA), total surface area (TSA), and volume of 3D solids, including cylinders, cones, spheres, hemispheres, frustums, and composite solids.

🧠 Key Concepts and Formulas

Let rr = radius, hh = height CSA = 2πrh2\pi rh TSA = 2πr(h+r)2\pi r(h + r) Volume = πr2h\pi r^2 h

Let rr = radius, hh = height, ll = slant height Where l=r2+h2l = \sqrt{r^2 + h^2} CSA = πrl\pi r l TSA = πr(l+r)\pi r(l + r) Volume = 13πr2h\frac{1}{3} \pi r^2 h

Let rr = radius Surface Area = 4πr24\pi r^2 Volume = 43πr3\frac{4}{3} \pi r^3

CSA = 2πr22\pi r^2 TSA = 3πr23\pi r^2 Volume = 23πr3\frac{2}{3} \pi r^3

Let RR = larger radius, rr = smaller radius, hh = height, l=h2+(R−r)2l = \sqrt{h^2 + (R - r)^2} CSA = π(R+r)l\pi (R + r)l TSA = π(R+r)l+πR2+πr2\pi (R + r)l + \pi R^2 + \pi r^2 Volume = 13πh(R2+r2+Rr)\frac{1}{3} \pi h (R^2 + r^2 + Rr)

Add volumes or surface areas of individual components as needed. Visualize how solids are joined (e.g., cone on hemisphere, hemisphere on cylinder, etc.)

📘 NCERT Exercise Solutions

CSA=πrl=227×7×25=550 cm2\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2

TSA=2πr(r+h)=2×227×14×(14+10)=2112 cm2\text{TSA} = 2\pi r (r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 10) = 2112 \text{ cm}^2

V=13πr2h=13×227×72×24=1232 cm3V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = 1232 \text{ cm}^3

V=23πr3=23×227×10.53=4851 cm3V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 10.5^3 = 4851 \text{ cm}^3

Total Volume=23πr3+13πr2h=πr2(2r+h3)=227×3.52×2×3.5+123≈268.5 cm3\text{Total Volume} = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \pi r^2\left(\frac{2r + h}{3}\right) = \frac{22}{7} \times 3.5^2 \times \frac{2 \times 3.5 + 12}{3} \approx 268.5 \text{ cm}^3

Use volume conservation: Volume of sphere=n×Volume of one cone⇒n=VsphereVcone\text{Volume of sphere} = n \times \text{Volume of one cone} \Rightarrow n = \frac{V_{\text{sphere}}}{V_{\text{cone}}}

V=13πh(R2+r2+Rr)=13×227×4(49+9+21)=1069.71 cm3V = \frac{1}{3} \pi h(R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 4 (49 + 9 + 21) = 1069.71 \text{ cm}^3

📝 CBSE Previous Year Questions (PYQs)

📝 CBSE PYQ Solutions – Chapter 13: Surface Areas and Volumes

📎 Summary Table of Formulas

Class 10 CBSE Maths Chapter 13: Surface Areas and Volumes, including:

✅ Key Concepts & Formulas

📘 All NCERT Exercise Questions with detailed solutions

📝 CBSE Previous Year Questions from this chapter

Let $r$ = radius, $h$ = height

CSA = $2\pi rh$

TSA = $2\pi r(h + r)$

Volume = $\pi r^2 h$

Let $r$ = radius, $h$ = height, $l$ = slant height
Where $l = \sqrt{r^2 + h^2}$

CSA = $\pi r l$

TSA = $\pi r(l + r)$

Volume = $\frac{1}{3} \pi r^2 h$

Let $r$ = radius

Surface Area = $4\pi r^2$

Volume = $\frac{4}{3} \pi r^3$

CSA = $2\pi r^2$

TSA = $3\pi r^2$

Volume = $\frac{2}{3} \pi r^3$

Let $R$ = larger radius, $r$ = smaller radius, $h$ = height, $l = \sqrt{h^2 + (R - r)^2}$

CSA = $\pi (R + r)l$

TSA = $\pi (R + r)l + \pi R^2 + \pi r^2$

Volume = $\frac{1}{3} \pi h (R^2 + r^2 + Rr)$

$\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$

$\text{TSA} = 2\pi r (r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 10) = 2112 \text{ cm}^2$

$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = 1232 \text{ cm}^3$

$V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 10.5^3 = 4851 \text{ cm}^3$

$\text{Total Volume} = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \pi r^2\left(\frac{2r + h}{3}\right) = \frac{22}{7} \times 3.5^2 \times \frac{2 \times 3.5 + 12}{3} \approx 268.5 \text{ cm}^3$

Use volume conservation:
$\text{Volume of sphere} = n \times \text{Volume of one cone} \Rightarrow n = \frac{V_{\text{sphere}}}{V_{\text{cone}}}$

$V = \frac{1}{3} \pi h(R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 4 (49 + 9 + 21) = 1069.71 \text{ cm}^3$