In a parallelogram PQRS, diagonal RP is extended to point T, such that RP = PT. Find the percentage by which the area of the triangle TSQ exceeds that of parallelogram PQRS, if PQ = 16, QR = 12 and PQR = 36°.40%50%60%Cannot be determined

Given that PQRS is a parallelogram, we know that opposite sides are equal and opposite angles are equal.

Step 1: Calculate the area of the parallelogram PQRS.

The area of a parallelogram is given by the formula: base * height. However, we don't have the height. We do have the length of the base (PQ = 16) and one angle (PQR = 36°). We can find the height by using the sine of the angle: height = QR * sin(PQR) = 12 * sin(36°).

Step 2: Calculate the area of triangle TSQ.

Since RP = PT, triangle PTS is an isosceles triangle. This means that angle PTS = angle PST = 36°. Therefore, angle TSQ = 180° - 2*36° = 108°.

The area of a triangle is given by the formula: 1/2 * base * height. In this case, the base is TS = 2RP = 2PT = 2*16 = 32. The height is SQ = QR = 12. However, we need to adjust this because the height is not perpendicular to the base. We can do this by using the sine of the angle: height = SQ * sin(TSQ) = 12 * sin(108°).

Step 3: Calculate the percentage by which the area of triangle TSQ exceeds that of parallelogram PQRS.

First, subtract the area of the parallelogram from the area of the triangle to find the difference. Then, divide this difference by the area of the parallelogram and multiply by 100 to get the percentage.

Without actual calculations, we cannot determine the exact percentage. Therefore, the answer is "Cannot be determined".