ITEM 3 Your father has a triangular piece of land that measures 8 cm long, and one of the sides is 6 cm . The angle between two sides is \( 30^{\circ} \). Your father wants to construct a circular fence around histriangular piece of land but does not know how to do it. Task As a student of S .1 , help your father to construct a circle through the vertices of the triangularpiece of land. ii. Work out the area of the circle (taking \( \pi=\frac{22}{7} \) ) iii. If your father gives you his bank A.T.M to withdraw for him the money to buy the construction materials and He gives the pin as 179 in decimal base, what do you think will bethe four-digit pin number in base five will you use to withdraw the money?

Sure, I'd be happy to help you with both parts of the task! --- ### **ii. Calculating the Area of the Circle** To construct a circle that passes through all three vertices of the triangular piece of land (commonly known as the **circumcircle**), we first need to determine the **circumradius (R)** of the triangle. Once we have the radius, we can easily calculate the area of the circle. #### **Given:** - **Sides of the Triangle:** - \( AB = 8 \) cm - \( AC = 6 \) cm - **Included Angle:** - \( \angle BAC = 30^\circ \) - **Value of \( \pi \):** - \( \pi = \frac{22}{7} \) #### **Step 1: Find the Third Side (\( BC \)) Using the Law of Cosines** The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles: \[ BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC) \] Plugging in the known values: \[ BC^2 = 8^2 + 6^2 - 2 \cdot 8 \cdot 6 \cdot \cos(30^\circ) \] \[ BC^2 = 64 + 36 - 96 \cdot \left( \frac{\sqrt{3}}{2} \right) \] \[ BC^2 = 100 - 48\sqrt{3} \] \[ BC \approx \sqrt{100 - 48 \times 1.732} \approx \sqrt{100 - 83.138} \approx \sqrt{16.862} \approx 4.107 \text{ cm} \] #### **Step 2: Calculate the Area of the Triangle (\( A \))** Using the formula for the area of a triangle with two sides and the included angle: \[ A = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC) \] \[ A = \frac{1}{2} \cdot 8 \cdot 6 \cdot \sin(30^\circ) \] \[ A = 24 \cdot 0.5 = 12 \text{ cm}^2 \] #### **Step 3: Determine the Circumradius (\( R \))** The formula for the circumradius of a triangle is: \[ R = \frac{a}{2 \sin(\angle A)} \] Where: - \( a \) is the side opposite angle \( A \) (which is \( BC \) in this case). \[ R = \frac{4.107}{2 \times 0.5} = 4.107 \text{ cm} \] #### **Step 4: Calculate the Area of the Circumcircle** Using the area formula for a circle: \[ \text{Area} = \pi R^2 \] \[ \text{Area} = \frac{22}{7} \times (4.107)^2 \approx \frac{22}{7} \times 16.86 \approx 53 \text{ cm}^2 \] **Thus, the area of the circumcircle is approximately **\( \mathbf{53 \, \text{cm}^2} \)**. --- ### **iii. Converting the PIN from Decimal to Base Five** Your father's PIN is given as **179** in the decimal (base ten) system. To convert this to a **four-digit base five** number, follow these steps: #### **Step 1: Understand Base Five** In base five, each digit represents a power of five: - The rightmost digit is \( 5^0 \) - Next is \( 5^1 \) - Then \( 5^2 \) - And so on. #### **Step 2: Convert 179 to Base Five** 1. **Find the highest power of 5 less than or equal to 179:** - \( 5^3 = 125 \) (since \( 5^4 = 625 \) is too large) 2. **Determine how many times 125 fits into 179:** - \( 179 \div 125 = 1 \) with a remainder of \( 54 \) 3. **Next, divide the remainder by \( 5^2 = 25 \):** - \( 54 \div 25 = 2 \) with a remainder of \( 4 \) 4. **Then, divide the new remainder by \( 5^1 = 5 \):** - \( 4 \div 5 = 0 \) with a remainder of \( 4 \) 5. **Finally, the last remainder is \( 4 \), which is \( 5^0 \):** Putting it all together: - \( 5^3 \): 1 - \( 5^2 \): 2 - \( 5^1 \): 0 - \( 5^0 \): 4 **So, 179 in decimal is \( \mathbf{1204_5} \) in base five.** **Therefore, the four-digit PIN in base five is **`1204`**.** --- I hope this helps your father with constructing the fence and managing the necessary transactions!