Class 9 Maths Chapter 11 Surface Areas and Volumes – Exercise 11.4 NCERT Solutions

Introduction

Exercise 11.4 focuses on surface areas and volumes of cones. Students learn how to apply formulas for curved surface area, total surface area, and volume of a cone. This exercise is important for understanding 3D geometry and mensuration.

Key Formulas

1. Curved Surface Area (CSA) of Cone:

$$\text{CSA}=\pi rl$$

where $r$ = radius, $l$ = slant height.

2. Total Surface Area (TSA) of Cone:

$$\text{TSA}=\pi r(l+r)$$

3. Volume of Cone:

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

where $h$ = height.

Common Mistakes

• Confusing slant height with vertical height.

• Forgetting to use Pythagoras theorem: $l=\sqrt{r^2+h^2}$.

• Mixing CSA with TSA.

• Not converting units properly.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Find CSA of cone with radius 7 cm and slant height 25 cm.

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

Q2. Find TSA of cone with radius 7 cm and slant height 25 cm.

$$\text{TSA}=\pi r(l+r)=\frac{22}{7}\times 7\times (25+7)=704{\text{cm}}^2$$

Q3. Find volume of cone with radius 7 cm and height 24 cm.

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times 49\times 24=1232{\text{cm}}^3$$

Q4. Find CSA of cone with radius 3.5 cm and slant height 12 cm.

$$\text{CSA}=\pi rl=\frac{22}{7}\times 3.5\times 12=132{\text{cm}}^2$$

Q5. Find TSA of cone with radius 3.5 cm and slant height 12 cm.

$$\text{TSA}=\pi r(l+r)=\frac{22}{7}\times 3.5\times (12+3.5)=170.5{\text{cm}}^2$$

Q6. Find volume of cone with radius 3.5 cm and height 12 cm.

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times (3.5{)}^2\times 12=154{\text{cm}}^3$$

Q7. Find CSA of cone with radius 14 cm and slant height 30 cm.

$$\text{CSA}=\pi rl=\frac{22}{7}\times 14\times 30=1320{\text{cm}}^2$$

Q8. Find TSA of cone with radius 14 cm and slant height 30 cm.

$$\text{TSA}=\pi r(l+r)=\frac{22}{7}\times 14\times (30+14)=1936{\text{cm}}^2$$

Q9. Find volume of cone with radius 14 cm and height 24 cm.

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times 196\times 24=12352{\text{cm}}^3$$

Q10. Find CSA, TSA, and volume of cone with radius 5 cm and height 12 cm. Slant height:

$$l=\sqrt{r^2+h^2}=\sqrt{25+144}=13$$

CSA:

$$\pi rl=\frac{22}{7}\times 5\times 13=204.7{\text{cm}}^2$$

TSA:

$$\pi r(l+r)=\frac{22}{7}\times 5\times (13+5)=282.7{\text{cm}}^2$$

Volume:

$$\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times 25\times 12=314.3{\text{cm}}^3$$

FAQs (10)

FAQ1. What is CSA of cone? Curved surface area = $\pi rl$.

FAQ2. What is TSA of cone? Total surface area = $\pi r(l+r)$.

FAQ3. What is volume of cone? $\frac{1}{3}\pi r^{2}h$.

FAQ4. How to find slant height? Use Pythagoras theorem: $l=\sqrt{r^2+h^2}$.

FAQ5. What is difference between CSA and TSA? CSA excludes base, TSA includes base.

FAQ6. What is unit of volume? Cubic units (cm${}^3$, m${}^3$).

FAQ7. What is unit of surface area? Square units (cm${}^2$, m${}^2$).

FAQ8. What is real‑life example of cone? Ice‑cream cone, funnel.

FAQ9. Why cube radius in volume formula? Volume depends on three dimensions.

FAQ10. Why is Exercise 11.4 important? It builds foundation for 3D mensuration and applications.

Conclusion

Exercise 11.4 covers surface areas and volumes of cones with solved examples and FAQs. Mastering these problems helps students in geometry, mensuration, and practical applications.

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