Point A is on side XZ of triangle XYZ such that XA = XY. Given that YXZ = 90˚, YZX = 30˚, and ZA = 6-sqrt12, find the area of triangle XYZ.
We begin by drawing a diagram of triangle XYZ and marking the given information:
Since XA = XY and YXZ is a right angle, we know that angle AXY is also a right angle. Thus, triangle AXY is a right triangle with legs of equal length XY and hypotenuse XA. Therefore, XA = XY = sqrt(2) * XY.
We can also find the length of side YZ using the Law of Sines:
sin(YZX) / YZ = sin(YXZ) / XY sin(30˚) / YZ = sin(90˚) / XY YZ = 2 * XY
Now, we can find the length of side XZ using the Pythagorean theorem:
XZ^2 = YZ^2 + XY^2 XZ^2 = (2 * XY)^2 + XY^2 XZ^2 = 5 * XY^2 XZ = XY * sqrt(5)
Finally, we can find the area of triangle XYZ using the formula:
area = 1/2 * base * height
where the base is XZ and the height is the distance from point Y to line XZ. To find the height, we drop a perpendicular from Y to XZ and label the point of intersection H:
Since triangle YZH is a 30-60-90 triangle, we know that YH = YZ / 2 = XY and HZ = YZ * sqrt(3) / 2 = sqrt(3) * XY. Thus, the height of triangle XYZ is YH = XY.
Therefore, the area of triangle XYZ is:
area = 1/2 * XZ * YH area = 1/2 * (XY * sqrt(5)) * XY area = 1/2 * (6 - sqrt(12)) * sqrt(2) * sqrt(5) area = (15 - 5 * sqrt(3)) square units
So the area of triangle XYZ is (15 - 5 * sqrt(3)) square units.
