Question

in rectangle pqrs, the diagonals intersect each other at point t. if pr = 9 and pq = 7, what is the area of △pqt? round to the nearest tenth. square units

Explanation:

Step1: Find the length of QR using Pythagorean theorem

In rectangle PQRS, triangle PQR is a right triangle with \( PR = 9 \) (hypotenuse) and \( PQ = 7 \) (one leg). By Pythagorean theorem \( PR^{2}=PQ^{2}+QR^{2} \), so \( QR=\sqrt{PR^{2}-PQ^{2}}=\sqrt{9^{2}-7^{2}}=\sqrt{81 - 49}=\sqrt{32}\approx5.656 \)

Step2: Find the area of rectangle PQRS

The area of rectangle \( A = PQ\times QR=7\times\sqrt{32}\approx7\times5.656 = 39.592 \)

Step3: Find the area of triangle PQT

In a rectangle, the diagonals bisect each other, so the area of triangle PQT is \( \frac{1}{4} \) of the area of the rectangle. So area of \( \triangle PQT=\frac{1}{4}\times39.592\approx9.9 \)

Answer:

\( 9.9 \)