Triangle SPO has PO = 39 and PS = 6, so its area is 1/2 * 39 * 6 = 117.
To find the area of the triangle shown below, we need to know the lengths of two sides and the angle between them. We are given that RS = 18 and RP = 12. We are also given that PO bisects angle RPS. This means that angle RPO = angle SPO = a/2.
To find the length of PO, we can use the Law of Cosines on triangle SPO.
PO^2 = SO^2 + OP^2 - 2 * SO * OP * cos(a/2)
We know that SO = RS/2 = 9 and OP = RP/2 = 6. We also know that cos(a/2) = cos(180 - a/2) = -cos(a/2). Substituting these values into the Law of Cosines, we get:
PO^2 = 9^2 + 6^2 - 2 * 9 * 6 * (-cos(a/2))
PO^2 = 3^2 * 13^2
PO = 3 * 13
PO = 39
Now that we know the length of PO, we can find the area of triangle SPO using the formula for the area of a right triangle:
Area = 1/2 * base * height
Area = 1/2 * PO * PS
We know that PO = 39 and PS = RP/2 = 6. Substituting these values into the formula, we get:
Area = 1/2 * 39 * 6
Area = 117
Therefore, the area of the triangle is 117.