NCERT Class 9 Maths Solutions: Surface Areas and Volumes

Question:

The problem involves calculating the surface area of a cone and relating it to the area of a rectangular tarpaulin. The key concepts are the formula for the curved surface area of a cone and the area of a rectangle. We’ll need to find the slant height of the cone to calculate its curved surface area, which will then tell us the required length of the tarpaulin.


The tarpaulin is used to make the conical tent. The shape of the tent is a cone. The tarpaulin will form the curved surface area of the cone.
We are given the height (h) of the conical tent as 8 m and the base radius (r) as 6 m.
To find the curved surface area of a cone, we need the slant height (l). The relationship between height, radius, and slant height is given by the Pythagorean theorem: l^2 = r^2 + h^2.
Calculate the slant height (l).
l = sqrt(r^2 + h^2)
l = sqrt(6^2 + 8^2)
l = sqrt(36 + 64)
l = sqrt(100)
l = 10 m

The curved surface area (CSA) of a cone is given by the formula: CSA = πrl.
Calculate the curved surface area of the tent.
CSA = 3.14 * 6 m * 10 m
CSA = 188.4 m^2

The tarpaulin is a rectangle with a given width of 3 m. Let the required length of the tarpaulin be L. The area of the tarpaulin is width * length = 3 * L.
This area of the tarpaulin must be equal to the curved surface area of the cone.
Area of tarpaulin = CSA of cone
3 * L = 188.4 m^2
L = 188.4 m^2 / 3 m
L = 62.8 m

However, the question states that extra length is required for stitching margins and wastage. This extra length is given as 20 cm.
We need to convert 20 cm to meters: 20 cm = 0.20 m.
The total length of tarpaulin required will be the calculated length plus the extra length.
Total Length = L + Extra Length
Total Length = 62.8 m + 0.20 m
Total Length = 63.0 m

Therefore, a length of 63.0 m of tarpaulin 3 m wide will be required.

Question:

The problem involves the concept of volume conservation when melting and recasting objects. The total volume of the smaller spheres will be equal to the volume of the larger sphere. We will also need the formulas for the surface area and volume of a sphere.

Formula for the volume of a sphere: $V = (4/3) \pi r^3$
Formula for the surface area of a sphere: $A = 4 \pi r^2$


Part 1: Radius r’ of the new sphere

Let the radius of each of the 27 solid iron spheres be ‘r’ and their surface area be ‘S’.
The volume of one such sphere is $V_{small} = (4/3) \pi r^3$.
The total volume of 27 such spheres is $V_{total} = 27 \times V_{small} = 27 \times (4/3) \pi r^3$.

These 27 spheres are melted to form a new sphere with radius ‘r” and surface area S’.
The volume of the new sphere is $V_{new} = (4/3) \pi (r’)^3$.

According to the principle of conservation of volume, the total volume of the smaller spheres is equal to the volume of the new sphere:
$V_{total} = V_{new}$
$27 \times (4/3) \pi r^3 = (4/3) \pi (r’)^3$

We can cancel out the common term $(4/3) \pi$ from both sides of the equation:
$27 r^3 = (r’)^3$

To find r’, we take the cube root of both sides:
$\sqrt[3]{27 r^3} = \sqrt[3]{(r’)^3}$
$3r = r’$

So, the radius of the new sphere is $r’ = 3r$.

Part 2: Ratio of S and S’

The surface area of one of the smaller spheres is $S = 4 \pi r^2$.
The surface area of the new sphere is $S’ = 4 \pi (r’)^2$.

We found that $r’ = 3r$. Substitute this into the formula for S’:
$S’ = 4 \pi (3r)^2$
$S’ = 4 \pi (9r^2)$
$S’ = 36 \pi r^2$

Now we need to find the ratio of S and S’:
Ratio = $S / S’$
Ratio = $(4 \pi r^2) / (36 \pi r^2)$

Cancel out the common terms $4 \pi r^2$:
Ratio = $1 / 9$

Therefore, the ratio of S and S’ is 1:9.

Question:

The surface area of a sphere is given by the formula $4\pi r^2$, where $r$ is the radius of the sphere. The ratio of two quantities is found by dividing one quantity by the other.


Let $r_1$ be the initial radius of the spherical balloon and $r_2$ be the final radius of the spherical balloon.
Given:
Initial radius, $r_1 = 7$ cm
Final radius, $r_2 = 14$ cm

The surface area of a sphere is given by the formula $A = 4\pi r^2$.

Let $A_1$ be the initial surface area of the balloon.
$A_1 = 4\pi r_1^2$
$A_1 = 4\pi (7)^2$
$A_1 = 4\pi (49)$
$A_1 = 196\pi$ cm$^2$

Let $A_2$ be the final surface area of the balloon.
$A_2 = 4\pi r_2^2$
$A_2 = 4\pi (14)^2$
$A_2 = 4\pi (196)$
$A_2 = 784\pi$ cm$^2$

We need to find the ratio of the surface areas of the balloon in the two cases, which is $A_1 : A_2$ or $\frac{A_1}{A_2}$.

Ratio = $\frac{A_1}{A_2} = \frac{196\pi}{784\pi}$

We can cancel out $\pi$ from the numerator and the denominator:
Ratio = $\frac{196}{784}$

Now, we simplify the fraction. We can see that $784 = 4 \times 196$.
Ratio = $\frac{196}{4 \times 196}$

Cancel out 196 from the numerator and the denominator:
Ratio = $\frac{1}{4}$

Therefore, the ratio of the surface areas of the balloon in the two cases is $1:4$.

Alternatively, we can directly find the ratio of the surface areas using the formula:
Ratio = $\frac{4\pi r_1^2}{4\pi r_2^2}$
Ratio = $\frac{r_1^2}{r_2^2}$
Ratio = $\left(\frac{r_1}{r_2}\right)^2$

Substitute the given values of $r_1$ and $r_2$:
Ratio = $\left(\frac{7}{14}\right)^2$
Ratio = $\left(\frac{1}{2}\right)^2$
Ratio = $\frac{1^2}{2^2}$
Ratio = $\frac{1}{4}$

The ratio of the surface areas of the balloon in the two cases is $1:4$.

Question:

The surface area of a sphere is given by the formula $A = 4 \pi r^2$, where $r$ is the radius of the sphere. The diameter ($d$) and radius ($r$) of a sphere are related by $d = 2r$ or $r = d/2$. The ratio of surface areas of two spheres will be the square of the ratio of their radii or diameters.


Let the diameter of the moon be $d_m$ and the diameter of the earth be $d_e$.
According to the question, the diameter of the moon is approximately one-fourth of the diameter of the earth.
So, $d_m = \frac{1}{4} d_e$.

Let the radius of the moon be $r_m$ and the radius of the earth be $r_e$.
We know that the radius is half of the diameter, so:
$r_m = \frac{d_m}{2}$
$r_e = \frac{d_e}{2}$

Now let’s find the ratio of their radii:
$\frac{r_m}{r_e} = \frac{d_m/2}{d_e/2} = \frac{d_m}{d_e}$
Since $d_m = \frac{1}{4} d_e$, we have:
$\frac{r_m}{r_e} = \frac{\frac{1}{4} d_e}{d_e} = \frac{1}{4}$

The surface area of the moon ($A_m$) is $A_m = 4 \pi r_m^2$.
The surface area of the earth ($A_e$) is $A_e = 4 \pi r_e^2$.

We need to find the ratio of their surface areas, which is $\frac{A_m}{A_e}$.
$\frac{A_m}{A_e} = \frac{4 \pi r_m^2}{4 \pi r_e^2}$
Cancel out $4 \pi$ from the numerator and denominator:
$\frac{A_m}{A_e} = \frac{r_m^2}{r_e^2} = \left(\frac{r_m}{r_e}\right)^2$

Substitute the ratio of the radii we found:
$\frac{A_m}{A_e} = \left(\frac{1}{4}\right)^2 = \frac{1^2}{4^2} = \frac{1}{16}$

Therefore, the ratio of the surface area of the moon to the surface area of the earth is 1:16.

The final answer is $\boxed{1:16}$.

Question:

When a right triangle is revolved about one of its sides, it forms a cone. The side of revolution becomes the height of the cone, and the other perpendicular side becomes the radius of the base of the cone. The volume of a cone is given by the formula V = (1/3) * pi * r^2 * h, where r is the radius of the base and h is the height.


Part 1: Revolved about the side 12 cm
The right triangle ABC has sides 5 cm, 12 cm, and 13 cm. Since it’s a right triangle, the sides 5 cm and 12 cm are the perpendicular sides, and 13 cm is the hypotenuse.
When revolved about the side 12 cm:
The height of the cone (h) will be 12 cm.
The radius of the base of the cone (r) will be 5 cm.
The volume of the solid (cone) obtained is V1 = (1/3) * pi * r^2 * h = (1/3) * pi * (5 cm)^2 * (12 cm) = (1/3) * pi * 25 cm^2 * 12 cm = 100 * pi cm^3.

Part 2: Revolved about the side 5 cm
When the same triangle is revolved about the side 5 cm:
The height of the cone (h) will be 5 cm.
The radius of the base of the cone (r) will be 12 cm.
The volume of the solid (cone) obtained is V2 = (1/3) * pi * r^2 * h = (1/3) * pi * (12 cm)^2 * (5 cm) = (1/3) * pi * 144 cm^2 * 5 cm = 240 * pi cm^3.

Part 3: Ratio of the volumes
The ratio of the volumes of the two solids obtained is V1 / V2.
Ratio = (100 * pi cm^3) / (240 * pi cm^3) = 100 / 240 = 10 / 24 = 5 / 12.
Alternatively, the ratio can be written as 5:12.

Question:

When a right-angled triangle is revolved about one of its sides, it forms a cone. The volume of a cone is given by the formula V = (1/3) * pi * r^2 * h, where ‘r’ is the radius of the base and ‘h’ is the height of the cone.


The right triangle ABC has sides 5 cm, 12 cm, and 13 cm. Since it’s a right triangle, the sides 5 cm and 12 cm are the legs, and 13 cm is the hypotenuse (as 5^2 + 12^2 = 25 + 144 = 169 = 13^2).
The triangle is revolved about the side 12 cm.
When revolved about the side 12 cm, this side becomes the height (h) of the cone.
The other leg, 5 cm, becomes the radius (r) of the base of the cone.
The hypotenuse, 13 cm, becomes the slant height of the cone, but this is not needed to calculate the volume.

Given:
Height of the cone, h = 12 cm
Radius of the base of the cone, r = 5 cm

The formula for the volume of a cone is:
V = (1/3) * pi * r^2 * h

Substitute the given values into the formula:
V = (1/3) * pi * (5 cm)^2 * (12 cm)
V = (1/3) * pi * (25 cm^2) * (12 cm)
V = (1/3) * pi * 300 cm^3
V = 100 * pi cm^3

The volume of the solid obtained is 100π cm³.

Question:

The total surface area of a hemisphere is the sum of the area of its curved surface and the area of its circular base. The formula for the curved surface area of a hemisphere is 2πr², and the formula for the area of a circle is πr². Therefore, the total surface area of a hemisphere is 2πr² + πr² = 3πr².


The question asks for the total surface area of a hemisphere with a given radius.
The formula for the total surface area of a hemisphere is given by:
Total Surface Area = Curved Surface Area + Area of the circular base
Curved Surface Area of a hemisphere = 2πr²
Area of the circular base of a hemisphere = πr²
So, Total Surface Area of a hemisphere = 2πr² + πr² = 3πr²

Given radius (r) = 10 cm
Given π = 3.14

Substitute the values into the formula:
Total Surface Area = 3 * 3.14 * (10 cm)²
Total Surface Area = 3 * 3.14 * 100 cm²
Total Surface Area = 3 * 314 cm²
Total Surface Area = 942 cm²

Therefore, the total surface area of the hemisphere is 942 cm².

Question:

The question requires the application of the formula for the volume of a cone. The formula relates the volume (V) of a cone to its base radius (r) and height (h) as V = (1/3)πr²h. To solve the problem, we need to rearrange this formula to find the radius when the volume and height are known.


The problem provides the height of a cone (h = 15 cm) and its volume (V = 1570 cm³). We are also given the value of pi (π = 3.14). The formula for the volume of a cone is V = (1/3)πr²h.
To find the radius (r), we can rearrange the formula:
r² = (3 * V) / (π * h)
Now, substitute the given values into the formula:
r² = (3 * 1570 cm³) / (3.14 * 15 cm)
r² = 4710 cm³ / 47.1 cm
r² = 100 cm²
To find the radius, take the square root of both sides:
r = √100 cm²
r = 10 cm
Therefore, the radius of the base of the cone is 10 cm.

Question:

The volume of a right circular cone is given by the formula V = (1/3)πr²h, where V is the volume, r is the radius of the base, and h is the height. The diameter of a circle is twice its radius (d = 2r).


We are given the volume (V) of a right circular cone and its height (h). We need to find the diameter of its base.
1. Recall the formula for the volume of a right circular cone: V = (1/3)πr²h.
2. Substitute the given values into the formula: 48π = (1/3)πr²(9).
3. Simplify the equation: 48π = 3πr².
4. Divide both sides by 3π to solve for r²: r² = 48π / 3π = 16.
5. Take the square root of both sides to find the radius (r): r = √16 = 4 cm.
6. Calculate the diameter (d) using the relationship d = 2r: d = 2 * 4 cm = 8 cm.

Therefore, the diameter of the base of the cone is 8 cm.