An athletic track 14 m wide consists of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. Calculate the area of the shaded region. `(" Take" pi = (22)/(7) )`

Answer

To find the area of the shaded region of the athletic track, we will break down the problem into manageable steps. The track consists of two straight sections and two semi-circular ends.

### Step 1: Calculate the area of the straight sections
The track has two straight sections, each measuring 120 m in length and 14 m in width.

**Area of one straight section:**
\[ \text{Area} = \text{Length} \times \text{Width} = 120 \, \text{m} \times 14 \, \text{m} = 1680 \, \text{m}^2 \]

**Total area of both straight sections:**
\[ \text{Total Area of Straight Sections} = 2 \times 1680 \, \text{m}^2 = 3360 \, \text{m}^2 \]

### Step 2: Calculate the area of the semi-circular ends
The inner radius of the semi-circular ends is given as 35 m. The width of the track is 14 m, so the outer radius will be:
\[ \text{Outer Radius} = \text{Inner Radius} + \text{Width} = 35 \, \text{m} + 14 \, \text{m} = 49 \, \text{m} \]

**Area of one outer semi-circle:**
\[ \text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (49)^2 \]
Calculating \( (49)^2 \):
\[ (49)^2 = 2401 \]
So,
\[ \text{Area} = \frac{1}{2} \times \frac{22}{7} \times 2401 = \frac{22 \times 2401}{14} = \frac{52822}{14} = 3773 \, \text{m}^2 \]

**Area of one inner semi-circle:**
\[ \text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (35)^2 \]
Calculating \( (35)^2 \):
\[ (35)^2 = 1225 \]
So,
\[ \text{Area} = \frac{1}{2} \times \frac{22}{7} \times 1225 = \frac{22 \times 1225}{14} = \frac{26950}{14} = 1925 \, \text{m}^2 \]

**Total area of both semi-circular ends:**
\[ \text{Total Area of Semi-Circular Ends} = 2 \times (3773 - 1925) = 2 \times 1848 = 3696 \, \text{m}^2 \]

### Step 3: Calculate the total area of the shaded region
Now, we combine the areas of the straight sections and the semi-circular ends to find the total area of the shaded region.

\[ \text{Total Area of Shaded Region} = \text{Total Area of Straight Sections} + \text{Total Area of Semi-Circular Ends} \]
\[ \text{Total Area of Shaded Region} = 3360 \, \text{m}^2 + 3696 \, \text{m}^2 = 7056 \, \text{m}^2 \]

Thus, the area of the shaded region is **7056 m²**.