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NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes Exercise 11.1

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NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 - FREE PDF Download

NCERT Solutions for Maths Chapter 11 - Surface Areas and Volumes, Exercise 11.1 Class 9, is designed to help students understand and master the calculations of surface areas for different shapes. This exercise includes practical problems that enhance students' understanding and application of these concepts, aiding in their overall mathematical proficiency. Vedantu provides comprehensive solutions prepared by experienced teachers, ensuring that students can easily grasp the concepts. These solutions offer step-by-step explanations and illustrations, making it simpler for students to follow along. Practicing these exercises boosts confidence and helps students achieve high marks in their Class 9 Maths exams.

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Table of Content
1. NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 11 Exercise 11.1 Class 9 | Vedantu
3. Formulas Used in Class 9 Chapter 11 Exercise 11.1
4. Access NCERT Solutions for Maths Class 9 Chapter 11 - Surface Areas and Volumes
5. Class 9 Maths Chapter 11: Exercises Breakdown
6. CBSE Class 9 Maths Chapter 11 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 11 Exercise 11.1 Class 9 | Vedantu

  • Maths Chapter 11 Exercise 11.1 Class 9 explains about 1 Surface Area of a Right Circular Cone, right circular cone, curved surface area of a cone.

  • Understanding these concepts helps in solving geometric problems related to the surface areas of cones.

  • A right circular cone is a 3D shape with a circular base and an apex perpendicular to the base.

  • The curved surface area of a cone is the area of the cone's surface excluding its base.

  • The total surface area of a cone is the combined area of the curved surface and the base.

  • This article contains exercise notes, important questions, exemplar solutions, exercises and video links for class 9 chapter 11 maths exercise 11.1 - Surface Areas and Volumes, which you can download as PDFs.

  • There are eight questions in Maths Chapter 11 ex 11.1 class 9 solutions  which are fully solved by experts at Vedantu.


Formulas Used in Class 9 Chapter 11 Exercise 11.1

  • Curved Surface Area of a Cone:  $\pi rl$

  • Total Surface Area of a Cone: $\pi r\left ( r+l \right )$
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NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes Exercise 11.1
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Surface Area and Volume L-1 | Surface Area & Volume of Cuboid & Cube | CBSE 9 Maths Ch 13 | Term 2
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Access NCERT Solutions for Maths Class 9 Chapter 11 - Surface Areas and Volumes

Exercise 11.1

1. Diameter of the base of a cone is $\text{10}\text{.5 cm}$ and its slant height is $\text{10 cm}$. Find its curved surface area. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 10 cm}$

The diameter of the base of cone $\text{= 10}\text{.5 cm}$

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\frac{\text{10}\text{.5}}{\text{2}}\text{ cm = 5}\text{.25 cm}$

The curved surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ 5}\text{.25 }\!\!\times\!\!\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{22 }\!\!\times\!\!\text{ 0}\text{.75 }\!\!\times\!\!\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 165 c}{{\text{m}}^{\text{2}}}$

Therefore, the curved surface area of the cone is $\text{165 c}{{\text{m}}^{\text{2}}}$.


2. Find the total surface area of a cone, if its slant height is $\text{21 m}$and diameter of its base is $\text{24 m}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 21 m}$

The diameter of the base of cone $\text{= 24 m}$

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\frac{\text{24}}{\text{2}}\text{ m = 12 m}$

The total surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ r}\left( \text{l + r} \right)$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  12  }\!\!\times\!\!\text{  }\left( \text{21 + 12} \right) \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  12  }\!\!\times\!\!\text{  33} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1244}\text{.57 }{{\text{m}}^{\text{2}}}$

Therefore, the total surface area of the cone is $\text{1244}\text{.57 }{{\text{m}}^{\text{2}}}$.


3. Curved surface area of a cone is $\text{308 c}{{\text{m}}^{\text{2}}}$ and its slant height is $\text{14 cm}$. Find

i. Radius of the Base

Ans:

It is given that the slant height $\left( \text{l} \right)$ of the cone $\text{= 14 cm}$

The curved surface area of the cone $\text{= 308 c}{{\text{m}}^{\text{2}}}$

Let us assume the radius of base of cone be $\text{r}$.

We know that curved surface area of the cone $\text{=  }\!\!\pi\!\!\text{ rl}$

$\therefore \text{ }\!\!\pi\!\!\text{ rl = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  r  }\!\!\times\!\!\text{  14} \right)\text{ cm = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{308}}{\text{44}}\text{ cm}$

$\Rightarrow \text{r = 7 cm}$

Hence, the radius of the base is $\text{7 cm}$.


ii. Total Surface Area of the Cone. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

The total surface area of the cone is the sum of its curved surface area and the area of the base.

Total surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ rl +  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + 154} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 462 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the cone is $\text{462 c}{{\text{m}}^{\text{2}}}$.


4. A conical tent is $\text{10 m}$ high and the radius of its base is $\text{24 m}$. Find

i. slant height of the tent

Ans:


slant height of the tent


From the figure we can say that $\text{ABC}$ is a conical tent.

It is given that the height $\left( \text{h} \right)$ of conical tent $\text{= 10 m}$

The radius $\left( \text{r} \right)$ of conical tent $\text{= 24 m}$

Let us assume the slant height as $\text{l}$.

In $\text{ }\!\!\Delta\!\!\text{ ABD}$, we will use Pythagorean Theorem.

$\therefore \text{A}{{\text{B}}^{\text{2}}}\text{ = AD}{{\text{ }}^{\text{2}}}\text{ + B}{{\text{D}}^{\text{2}}}$

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\text{h}}^{\text{2}}}\text{ + }{{\text{r}}^{\text{2}}}$

\[\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\left( \text{10 m} \right)}^{\text{2}}}\text{ + }{{\left( \text{24 m} \right)}^{\text{2}}}\]

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = 676 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{l = 26 m}$

The slant height of the tent is $\text{26 m}$.


ii. cost of canvas required to make the tent, if cost of $\text{1 }{{\text{m}}^{\text{2}}}$ canvas is $\text{Rs}\text{. 70}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

The curved surface area of the tent, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  24  }\!\!\times\!\!\text{  26} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \frac{13728}{7} \right)\text{ }{{\text{m}}^{\text{2}}}$

It is given that the cost of $\text{1 }{{\text{m}}^{\text{2}}}$ of canvas $\text{= Rs}\text{. 70}$

So, the cost of $\frac{13728}{7}\text{ }{{\text{m}}^{\text{2}}}$ canvas $\text{= Rs}\text{. }\left( \frac{\text{13728}}{\text{7}}\text{  }\!\!\times\!\!\text{  70} \right)\text{ = Rs}\text{. 137280}$

Hence, the cost of canvas required to make the tent is $\text{Rs}\text{. 137280}$.


5. What length of tarpaulin $\text{3 m}$ wide will be required to make conical tent of height $\text{8 m}$ and base radius $\text{6 m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $\text{20 cm}$. $\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of tent $\text{= 6 m}$

The height $\left( \text{h} \right)$ of tent $\text{= 8 m}$

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{6}}^{\text{2}}}\text{ + }{{\text{8}}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{\text{100}} \right)\text{ m}$

$\Rightarrow \text{l = 10 m}$

The curved surface area of the tent, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\!\!\times\!\!\text{  6  }\!\!\times\!\!\text{  10} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

It is give the width of tarpaulin $\text{= 3 m}$

Let us assume the length of the tarpaulin sheet required be $\text{x}$.

It is given that there will be a wastage of $\text{20 cm}$.

So, the new length of the sheet $\text{=}\left( \text{x - 0}\text{.2} \right)\text{ m}$

We know that the area of the rectangular sheet required will be the same as the curved surface area of the tent.

$\therefore \left[ \left( \text{x - 0}\text{.2} \right)\text{  }\!\!\times\!\!\text{  3} \right]\text{ m = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{x - 0}\text{.2 m = 62}\text{.8 m}$

$\Rightarrow \text{x = 63 m}$

The length of tarpaulin sheet required is $\text{63 m}$.


6. The slant height and base diameter of a conical tomb are $\text{25 m}$ and $\text{14 m}$ respectively. Find the cost of white-washing its curved surface at the rate of $\text{Rs}\text{. 210}$ per $\text{100 }{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of tomb $\text{= 7 m}$

The slant height $\left( \text{l} \right)$ of tomb $\text{= 25 m}$

The curved surface area of the conical tomb, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{22}{7}\text{  }\!\!\times\!\!\text{  7  }\!\!\times\!\!\text{  25} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 }{{\text{m}}^{\text{2}}}$

It is given that the cost of white-washing $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 210}$

So, the cost of white-washing $550\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \frac{\text{210}}{\text{100}}\text{  }\!\!\times\!\!\text{  550} \right)\text{ = Rs}\text{. 1155}$

Hence, the cost of white-washing the curved surface area of a conical tomb is $\text{Rs}\text{. 1155}$.


7. A joker’s cap is in the form of a right circular cone of base radius $\text{7 cm}$ and the height $\text{24 cm}$. Find the area of the sheet required to make $\text{10}$ such caps. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of conical cap $\text{= 7 cm}$

The height $\left( \text{h} \right)$ of conical cap $\text{= 24 cm}$

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{7}}^{\text{2}}}\text{ + 2}{{\text{4}}^{\text{2}}}} \right)\text{ cm}$

$\Rightarrow \text{l = }\left( \sqrt{625} \right)\text{ cm}$

$\Rightarrow \text{l = 25 cm}$

The curved surface area of one conical cap, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{22}{7}\text{  }\!\!\times\!\!\text{  7  }\!\!\times\!\!\text{  25} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{10}$ conical caps $\text{= }\left( \text{550  }\!\!\times\!\!\text{  10} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 5500 c}{{\text{m}}^{\text{2}}}$

Therefore, the total area of the sheet required is $\text{5500 c}{{\text{m}}^{\text{2}}}$.


8. A bus stop is barricaded from the remaining part of the road, by using $\text{50}$ hollow cones made of recycled cardboard. Each cone has a base diameter of $\text{40 cm}$ and height $\text{1 m}$. If the outer side of each of the cones is to be painted and the cost of painting is $\text{Rs}\text{. 12}$ per ${{\text{m}}^{\text{2}}}$, what will be the cost of painting all these cones? 

$\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14 and take }\sqrt{\text{1}\text{.04}}\text{=1}\text{.02} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of cone $\text{= }\frac{\text{40}}{\text{2}}\text{ = 20 cm = 0}\text{.2 m}$

The height $\left( \text{h} \right)$ of cone $\text{= 1 m}$

So the slant height of the cone, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\left( \text{0}\text{.2} \right)}^{\text{2}}}\text{ + }{{\left( \text{1} \right)}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{1.04} \right)\text{ m}$

$\Rightarrow \text{l = 1}\text{.02 m}$

The curved surface area of one cone, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\!\!\times\!\!\text{  0}\text{.2  }\!\!\times\!\!\text{  1}\text{.02} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 0}\text{.64056 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{50}$ cones $\text{= }\left( \text{50  }\!\!\times\!\!\text{  0}\text{.64056} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 32}\text{.028 }{{\text{m}}^{\text{2}}}$

It is given that the cost of painting $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 12}$

So, the cost of painting $32.028\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{32}\text{.028  }\!\!\times\!\!\text{  12} \right)\text{ = Rs}\text{. 384}\text{.336}$

We can also write the cost approximately as $\text{Rs}\text{. 384}\text{.34}$.

Therefore, the cost of painting all the hollow cones is $\text{Rs}\text{. 384}\text{.34}$.


Conclusion

NCERT of class 9 maths surface area and volume exercise 11.1, provides a clear understanding of calculating the surface areas of geometric shapes, specifically right circular cones. Students should focus on grasping the concepts of curved surface area and total surface area to solve related problems effectively. Vedantu’s detailed solutions offer step-by-step explanations that simplify complex problems. Practicing these exercises enhances problem-solving skills and builds confidence, helping students perform well in their Class 9 Maths exams.


Class 9 Maths Chapter 11: Exercises Breakdown

Exercise

Number of Questions

Exercise 11.2

9 Questions & Solutions (4 Short Answers, 5 Long Answers)

Exercise 11.3

9 Questions & Solutions (9 Long Answers)

Exercise 11.4

10 Questions & Solutions (5 Short Answers, 5 Long Answers)


CBSE Class 9 Maths Chapter 11 Other Study Materials


Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes Exercise 11.1

1. What are the benefits of referring to NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Exercise 11.1?

NCERT Solutions are the best study materials to find answers to all the textbook exercise problems at one place. Using NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Exercise 11.1, students will be able to find out surface areas and volumes of a cone. If they have any doubts regarding any question, they can refer to the solutions provided by subject matter experts. E-learning platforms like Vedantu provide authentic and well-researched NCERT Solutions designed as per the latest guidelines and syllabus to help students in learning a chapter better. Students can refer to exercise-wise NCERT Solutions for Class 9 Maths Chapter 11 to score well in the exams.

2. How to calculate the surface area and volume of a cone?

The surface area and volume of a cone can be calculated with the help of the formulas given below:

  • Curved Surface Area of a Cone= πrl

  • Total Surface Area of a Cone=  πrl + πr2 = πr (l + r)

  • The volume of a Cone= 1/3 πr2h

Where r is the base radius, h is the height and l is the slant height of the cone.

3. Where can I avail exercise-wise NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes?

Vedantu provides well-prepared NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes for all the exercises. Chapter 11 of CBSE Class 9 Maths is an exhaustive chapter as it includes a total of nine exercises. Students can find solutions to all the exercises’ questions on Vedantu’s site provided by expert tutors. This is an important chapter from the examination point of view. Hence, students must solve each and every problem and refer to Vedantu’s site for the needed solutions. They can also register for master classes for better doubt clearance.

4. What are the salient features of Vedantu’s NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1?

NCERT Solutions by Vedantu are the most comprehensive study resource available online. Class 9 Mathematics NCERT Solutions for Chapter 11 Surface Areas and Volumes Exercise 11.1 as well as other exercises include step-by-step explanations of the problems. Students must practice all the questions of the exercise as it is really important from the exam point of view. Some of the salient features of this study material curated by Vedantu are:

  • It helps in self-assessment

  • Provides effective revision and practice

  • Prepared by subject experts

5. What are the topics covered in Class 9 Maths Chapter 11 Exercise 11.1?

Class 9 Maths Chapter 11 Exercise 11.1 covers the following important topics.

  • Diameter of a cone

  • Surface area of a cone

  • Radius of a base 

6. How could I learn Chapter 11 Exercise 11.1 of Class 9 Maths in the efficient and fastest way?

Fastest way doesn’t mean that you take shortcuts in solving the answers. It means the learning time. Suppose you are not having much time for preparation for exams, and you want to learn the concepts in a faster yet effective way, then in such a case you can visit Vedantu, which has all the concepts for you in a concise form. The PDF of NCERT solutions for Class 9 Maths Chapter 11 Exercise 11.1 carries proper explanations for the same.

7. Where can I download the latest NCERT Solutions for Chapter 11 Exercise 11.1 of Class 9 Maths?

It is easy to download the latest NCERT Solutions for Chapter 11 Exercise 11.1 of Class 9 Maths just by clicking on one specific link. How to approach that correct link is very important. Firstly, visit Vedantu website (vedantu.com). Then click on the page NCERT Class 9 Maths book, where you will be re-directed towards the download of the latest NCERT Solutions for Class 9 Maths. Select the relevant chapter and the exercise for the latest NCERT Solutions for Chapter 11, Exercise 11.1. 

8. Can I take the help of the Maths experts at Vedantu for Exercise 11.1 of Chapter 11 of Class 9 Maths?

Yes, you can take the help of Maths experts at Vedantu. This facility is available to everyone who wants to get the right guidance regarding their subjects. They will help you by clearing all your doubts. By getting help from the Maths experts, you can solve the questions properly. They are ready to help you out till you are satisfied and understand the concepts thoroughly. You can also download the PDF of NCERT solutions for Class 9 Maths free of cost. These solutions are available at free of cost on Vedantu’s website(vedantu.com) and mobile app.

9. How can I improve in Chapter 11 Exercise 11.1 of Class 9 Maths?

If you are not able to improve your Maths for Class 9 Chapter 11 Exercise 11.1, then don’t simply jump to solving exercises . First understand the concept and try solving example sums. Then proceed to solving the sums from 11.1. However, solve all the problems to improve yourself in the said exercise.

10. What common mistakes should I avoid while solving these problems in ex 11.1 class 9?

Common mistakes include using incorrect formulas, miscalculating measurements, and not properly understanding the relationship between different parts of a cone.

11. What strategies can help me solve these problems effectively in class 9 ex 11.1?

In class 9 ex 11.1 effective strategies include memorizing the key formulas, practicing regularly, and carefully following the steps outlined in the solutions to ensure accuracy.

12. What should I focus on while practicing class 9 11.1?

In class 9 11.1 focus on understanding the formulas for the curved and total surface area of a cone and practicing their application through various problems.

13. What are the key concepts to understand from class 9 maths exercise 11 exercise 11.?

The key concepts of class 9 maths exercise 11 exercise 11 include understanding the properties of a right circular cone, and learning how to calculate both the curved surface area and the total surface area using the appropriate formulas.