In triangle PQR, we have <P = 30 degrees, <Q = 60 degrees, and <R=90 degrees. Point X is on line PR such that line QX bisects <PQR. If PQ = 12, then what is the area of triangle PQX$?
since PQR = 60°, PQX = 30°
So triangle PQX is an isosceles triangle with base angles of 30° and base=12
Thus, its altitude is 6/√3 and its area = 36/√3 = 12√3
To find the area of triangle PQX, we first need to find the length of line QX.
Since line QX bisects <PQR, we can use the Angle Bisector Theorem to find the ratio of the lengths of QX and XR. The ratio of QX to XR is equal to the ratio of the lengths of the sides opposite to Q and R, respectively.
In triangle PQR, the side opposite to Q is PR, so the ratio of QX to XR is equal to the ratio of the lengths of PQ to QR.
The given length of PQ is 12, and since the triangle is a right triangle, we can use trigonometry to find the length of QR.
Using the sine function, we have sin(Q) = opposite/hypotenuse, so sin(60) = QR/12.
Simplifying, we find √3/2 = QR/12.
Cross-multiplying, we get QR = 12√3/2 = 6√3.
Now, we can find the ratio of QX to XR. Since QX bisects <PQR, the ratio is 1:1. Therefore, QX = XR = 6√3.
To find the area of triangle PQX, we can use the formula for the area of a triangle: Area = 1/2 * base * height.
The base of the triangle is PQ, which is given as 12. The height of the triangle is QX, which we found to be 6√3.
Plugging in these values into the formula, we have:
Area = 1/2 * 12 * 6√3
= 6 * 6√3
= 36√3.
Therefore, the area of triangle PQX is 36√3.