A right cylinder , a right cone and a hemishpere have the same heightr and sthe same base area .Find the ratio of their (i) volumes (ii) total surface areas.

Answer

To solve the problem of finding the ratio of volumes and total surface areas of a right cylinder, a right cone, and a hemisphere with the same height and base area, we can follow these steps:

### Step 1: Define the Variables
Let the radius of the base of the solids be \( R \) and the height of the solids be \( h \). Since they have the same base area and height, we can assume \( h = R \).

### Step 2: Calculate the Volume of Each Solid
1. **Volume of the Cylinder (V_c)**:
\[ V_c = \pi R^2 h = \pi R^2 R = \pi R^3 \]

2. **Volume of the Cone (V_co)**:
\[ V_{co} = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^2 R = \frac{1}{3} \pi R^3 \]

3. **Volume of the Hemisphere (V_h)**:
\[ V_h = \frac{2}{3} \pi R^3 \]

### Step 3: Find the Ratio of Their Volumes
Now, we can find the ratio of the volumes of the cylinder, cone, and hemisphere:
\[ \text{Ratio of volumes} = V_c : V_{co} : V_h = \pi R^3 : \frac{1}{3} \pi R^3 : \frac{2}{3} \pi R^3 \]

Cancelling out \( \pi R^3 \), we get:
\[ 1 : \frac{1}{3} : \frac{2}{3} \]

To express this in whole numbers, we can multiply by 3:
\[ 3 : 1 : 2 \]

### Step 4: Calculate the Total Surface Area of Each Solid
1. **Total Surface Area of the Cylinder (TSA_c)**:
\[ TSA_c = 2\pi R^2 + 2\pi R h = 2\pi R^2 + 2\pi R^2 = 4\pi R^2 \]

2. **Total Surface Area of the Cone (TSA_co)**:
The slant height \( L \) of the cone can be calculated as:
\[ L = \sqrt{R^2 + h^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]
Thus, the TSA of the cone is:
\[ TSA_{co} = \pi R^2 + \pi R L = \pi R^2 + \pi R (R\sqrt{2}) = \pi R^2 + \pi R^2\sqrt{2} = \pi R^2 (1 + \sqrt{2}) \]

3. **Total Surface Area of the Hemisphere (TSA_h)**:
\[ TSA_h = 2\pi R^2 + \pi R^2 = 3\pi R^2 \]

### Step 5: Find the Ratio of Their Total Surface Areas
Now, we can find the ratio of the total surface areas:
\[ \text{Ratio of TSA} = TSA_c : TSA_{co} : TSA_h = 4\pi R^2 : \pi R^2(1 + \sqrt{2}) : 3\pi R^2 \]

Cancelling out \( \pi R^2 \), we get:
\[ 4 : (1 + \sqrt{2}) : 3 \]

### Final Result
Thus, the final ratios are:
1. **Volumes**: \( 3 : 1 : 2 \)
2. **Total Surface Areas**: \( 4 : (1 + \sqrt{2}) : 3 \)