Question

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

Explanation:

Step1: Find side QR via Pythagoras

In right $\triangle PQR$, $PR^2 = PQ^2 + QR^2$.
$$QR = \sqrt{PR^2 - PQ^2} = \sqrt{9^2 - 7^2} = \sqrt{81 - 49} = \sqrt{32} = 4\sqrt{2} \approx 5.6569$$

Step2: Calculate area of $\triangle PQR$

Area = $\frac{1}{2} \times$ base $\times$ height.
$$\text{Area} = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 7 \times 4\sqrt{2} = 14\sqrt{2} \approx 19.7989$$

Step3: Round to nearest tenth

Round the calculated area to 1 decimal place.
$$19.7989 \approx 19.8$$

Answer:

19.8 square units