The circle c has centre (0, 0) and radius 5 units - Leaving Cert Mathematics - Question 2 - 2017

To find the value of k, we substitute the x-coordinate of point P, which is -4, into the equation of the semi-circle: $(-4)^2 + k^2 = 25$ Expanding gives: $16 + k^2 = 25$ Subtracting 16 from both sides results in: $k^2 = 9$ Taking the square root of both sides, we find: $k = 3$

For triangle ABP to be right-angled at P, we can check the slopes of the lines AB and AP:

1. Slope of AB: The points A(-5, 0) and B(5, 0) lie on the x-axis, hence the slope (m_AB) is: m_{AB} = rac{0 - 0}{5 - (-5)} = 0

2. Slope of AP: Point P has coordinates (-4, 3), so the slope (m_AP) is: m_{AP} = rac{3 - 0}{-4 - (-5)} = 3

3. Slope of BP: For Point B(5, 0) and P(-4, 3): m_{BP} = rac{3 - 0}{-4 - 5} = -rac{3}{9} = -rac{1}{3}

To check if triangle ABP is right-angled at P, we confirm: m_{AB} imes m_{BP} = 0 imes -rac{1}{3} = 0 Hence, the product of the slopes is 0, confirming that ABP is right-angled at P.

The area of the region can be calculated using the formula for the area of the semi-circle and the area of triangle ABP:

1. Area of the semi-circle: The formula for the area of a semi-circle is: A_{semi-circle} = rac{1}{2} imes rac{1}{2} imes ext{r}^2 imes ext{pi} Substituting r = 5, we have: A_{semi-circle} = rac{1}{2} imes rac{1}{2} imes 5^2 imes ext{pi} = rac{25 ext{pi}}{2} \ \ A_{semi-circle} \approx 39.27\

2. Area of triangle ABP: Given the vertices A(-5, 0), B(5, 0), P(-4, 3): Using the formula: A_{triangle} = rac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| Plugging in the values: A_{triangle} = rac{1}{2} \left| -5(0-3) + 5(3-0) + (-4)(0-0) \right| = rac{1}{2} \left| 15 + 15 \right| = 15 \

3. Final area:
   A_{region} = A_{semi-circle} - A_{triangle} \\ A_{region} \approx 39.27 - 15 = 24.27 \ Thus, the area of the region, correct to 2 decimal places, is: $24.27 \text{ square units}$