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

Using the equation of the semi-circle derived from the circle:

$x^2 + y^2 = 25$

Substituting x = -4:

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

This simplifies to:

$16 + k^2 = 25$

Subtracting 16 from both sides:

$k^2 = 9$

Taking the square root:

$k = 3$

Thus, the value of k is 3.

To show that triangle ABP is right-angled at P, we can use the fact that segment AB is the diameter of the circle.

1. The coordinates of points A and B are A(-5, 0) and B(5, 0). The length of AB is:

   $AB = |5 - (-5)| = 10$

2. By the property of circles, any triangle inscribed in a circle where one side is the diameter is right-angled at the point opposite the diameter. Hence, since AB is the diameter, triangle ABP is right-angled at P.

To find the area of the region inside the semi-circle but outside triangle ABP, we can calculate:

1. The area of the semi-circle:

   A_{semi-circle} = rac{1}{2} imes rac{1}{2} imes ext{π} imes r^2 = rac{1}{2} imes ext{π} imes (5^2) = rac{25 ext{π}}{2}

   Using ext{π} ≈ 3.14:

   A_{semi-circle} ≈ rac{25 imes 3.14}{2} ≈ 39.27

2. The area of triangle ABP:

   The base AB = 10, height from point P to line AB = k = 3:

   A_{triangle} = rac{1}{2} imes AB imes h = rac{1}{2} imes 10 imes 3 = 15

3. The area of the region inside the semi-circle but outside triangle ABP:

   $A_{region} = A_{semi-circle} - A_{triangle} = 39.27 - 15 = 24.27$

Thus, the area of the region is approximately 24.27 square units.