3. Application It takes 65,000 solar cells, each 1.25 in. by 2.75 in., to power the Helios Prototype, shown below. How much surface area, in square feet, must be covered with the cells? The cells on Helios are $$ 18 \% $$ efficient. Suppose they were only $$ 12 \% $$ efficient, like solar cells used in homes. How much more surface area would need to be covered to deliver the same amount of power?

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Let's break down the problem into two parts:

1. **Calculating the Required Surface Area with 18% Efficiency**
2. **Determining the Additional Surface Area Needed if Efficiency Drops to 12%**

---

### 1. **Calculating the Required Surface Area with 18% Efficiency**

**Given:**
- Number of solar cells: **65,000**
- Dimensions of each cell: **1.25 inches × 2.75 inches**

**Step-by-Step Calculation:**

1. **Calculate the Area of a Single Solar Cell:**
   $$
   \text{Area}_{\text{cell}} = 1.25 \, \text{in} \times 2.75 \, \text{in} = 3.4375 \, \text{square inches}
   $$

2. **Calculate the Total Area for All Cells:**
   $$
   \text{Total Area}_{\text{in}^2} = 65,000 \times 3.4375 \, \text{in}^2 = 223,437.5 \, \text{square inches}
   $$

3. **Convert Square Inches to Square Feet:**
   $$
   1 \, \text{square foot} = 144 \, \text{square inches}
   $$
   $$
   \text{Total Area}_{\text{ft}^2} = \frac{223,437.5}{144} \approx 1,552 \, \text{square feet}
   $$

**Result:**
$$
\boxed{1,\!552 \text{ square feet}}
$$

---

### 2. **Determining Additional Surface Area Needed if Efficiency Drops to 12%**

**Given:**
- Original Efficiency: **18%**
- New Efficiency: **12%**

**Concept:**
To deliver the same amount of power, the surface area needs to compensate for the drop in efficiency. The relationship between power ($$ P $$), efficiency ($$ \eta $$), and area ($$ A $$) is given by:
$$
P = \eta \times A
$$
To maintain the same power output when efficiency decreases, the area must increase proportionally:
$$
\eta_1 \times A_1 = \eta_2 \times A_2 \implies A_2 = \frac{\eta_1}{\eta_2} \times A_1 = \frac{18\%}{12\%} \times A_1 = 1.5 \times A_1
$$
This means the area needs to **increase by 50%**.

**Calculation:**
$$
\text{Additional Area} = 0.5 \times 1,552 \, \text{ft}^2 = 776 \, \text{square feet}
$$

**Result:**
$$
\boxed{776 \text{ square feet}}
$$

---

### **Summary**

- **Total Surface Area Needed at 18% Efficiency:** **1,552 square feet**
- **Additional Surface Area Needed if Efficiency Drops to 12%:** **776 square feet**
To power the Helios Prototype, you need to cover **1,552 square feet** with solar cells at 18% efficiency. If the efficiency drops to 12%, you would need an additional **776 square feet** to deliver the same amount of power.

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