NCERT Solutions Maths Exercise 11.1 Class 9 Chapter 11 Surface Areas and Volumes PDF (2024)

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:

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

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.