question on right triangles

why you should *assume* 3-4-5 sides of right triangle when you are not said so and not to consider the similar triangles sharing two angles 90` at PSQ and RSQ and side QS ???
PS/QS=QS/SR or 16/QS=QS/9 or QS^2=16*9 or QS^2=(4*3)^2
to find the area of PQR you need QS (height) and PR (PS+SR) which are 12 and (16+9)=25
hence S(PQR)=1/2 *25*12=150

queenisabella wrote:why can't you assume that this is a 3:4:5 (or 9:16:25) triangle?

QS is a height drawn through the right angle of triangle PQR.

A height drawn through the right angle of a triangle forms THREE SIMILAR TRIANGLES.

In the figure above:
The angles of triangle PQS are x-y-90.
The angles of triangle QRS are x-y-90.
The angles of triangle PQR are x-y-90.
Since each triangle has the same combination of angles, all 3 triangles are similar.

Since triangle PQS is similar to triangle QRS, the ratio of the legs in each triangle must be the same.
In triangle PQS:
(side opposite x)/(side opposite y) = h/16.
In triangle QRS:
(side opposite x)/(side opposite y) = 9/h.
Since the ratios must be equal:
h/16 = 9/h
h² = 9*16
h = 3*4 = 12.

Area of PQR = (1/2)bh = (1/2)(12)(25) = 150.

The correct answer is D.

Your instincts are good: we should always be on the lookout for special triangles.
Each of the 3 triangles here is a multiple of a 3:4:5 triangle.
In triangle PQS, the sides are 12, 16, and 20.
In triangle QRS, the sides are 9, 12, and 15.
In triangle PQR, the sides are 15, 20 and 25.