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 into the equation of the semi-circle. Given that P is on the semi-circle with equation: $x^2 + y^2 = 25$, we substitute x = -4:

$(-4)^2 + k^2 = 25$

This simplifies to: $16 + k^2 = 25$ $k^2 = 25 - 16$ $k^2 = 9$ $k = 3$

Thus, the value of k is 3.

To show that triangle ABP is right-angled at P, we need to find the slopes of lines AP and BP, and check if they are negative reciprocals.

1. Calculate slopes:

   • For line AP:
     • A(-5, 0) and P(-4, k)
     • The slope, m_AP = ( \frac{k - 0}{-4 - (-5)} = \frac{3 - 0}{-4 + 5} = \frac{3}{1} = 3 )
   • For line BP:
     • B(5, 0) and P(-4, k)
     • The slope, m_BP = ( \frac{k - 0}{-4 - 5} = \frac{3 - 0}{-4 - 5} = \frac{3}{-9} = -\frac{1}{3} )

2. Check for right-angle:

   • Since ( m_{AP} \cdot m_{BP} = 3 \cdot -\frac{1}{3} = -1 ), therefore triangle ABP is right-angled at P.

To find the area of the region inside the semi-circle but outside triangle ABP, we follow these steps:

1. Calculate area of the semi-circle: The area of a semi-circle is given by: $A_{semi-circle} = \frac{1}{2} \pi r^2$ where r is the radius. Therefore, $A_{semi-circle} = \frac{1}{2} \pi (5)^2 = \frac{25\pi}{2}$

2. Calculate area of triangle ABP: Triangle ABP has a base AB and height corresponding to point P. The base AB = 10 units (from A(-5,0) to B(5,0)), and height = k = 3. Area of triangle is: $A_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 10 \times 3 = 15$

3. Total area inside the semi-circle but outside triangle ABP: $A = A_{semi-circle} - A_{triangle}$ $A = \frac{25\pi}{2} - 15$ Evaluating:

   • Using ( \pi \approx 3.14:) $A \approx \frac{25 \times 3.14}{2} - 15 = 39.27 - 15 = 24.27 \text{ square units (approx)}$

Thus, the required area is 24.27 square units.