+44 In the diagram below, $\angle PQR = \angle PRQ = \angle STR = \angle TSR$, $RQ = 8$, and $SQ = 3$. Find $PQ$. [asy] unitsize(5 cm); pair A,B,C,D,E; A = (0, 0.9); B = (-0.4, 0); C = (0.4, 0); D = (-0.275, 0.16); E = (0.11, 0.65); draw(A--B); draw(A--C); draw(B--C); draw(B--E); draw(C--D); label("$P$",A,N); label("$Q$", B, S); label("$R$", C, S); label("$S$", D, S); label("$T$", E, W); [/asy]
A square is inscribed in a right triangle, as shown below. Find the area of the square. [asy] unitsize(1.5 cm); pair A, B, C, D, E, F, G; A = (0,0); C = (5,0); B = (3^2/5,3*4/5); D = extension(B, A + (0,-5), A, C); G = extension(B, C + (0,-5), A, C); E = extension(D, D + (0,1), A, B); F = extension(G, G + (0,1), B, C); draw(A--B--C--cycle); draw(D--E--F--G); draw(rightanglemark(A,B,C,5)); draw(shift((-0.2,0.1))*(A--B), Arrows(6)); draw(shift((0.1,0.2))*(B--C), Arrows(6)); label("$1$", (A + B)/2 + (-0.2,0.1), NW, red); label("$3$", (B + C)/2 + (0.1,0.2), NE, red); [/asy]
+474 Here is the answer to the second problem.
To find the area of the square inscribed in the right triangle, we can use the fact that the square is inscribed, meaning that its diagonal is also the hypotenuse of the right triangle.
Let the side length of the square be \( s \). Since the square is inscribed in the right triangle, its diagonal is equal to the hypotenuse of the right triangle.
By the Pythagorean theorem, the length of the hypotenuse of the right triangle is given by:
\[ \sqrt{1^2 + 3^2} = \sqrt{10} \]
Since the diagonal of the square is also the hypotenuse of the right triangle, its length is \( \sqrt{10} \).
Now, the diagonal of the square divides it into two congruent right triangles. Each right triangle has legs of length \( s \) and \( s \), and a hypotenuse of length \( \sqrt{10} \).
Using the Pythagorean theorem again, we have:
\[ s^2 + s^2 = (\sqrt{10})^2 \]
\[ 2s^2 = 10 \]
\[ s^2 = \frac{10}{2} \]
\[ s^2 = 5 \]
\[ s = \sqrt{5} \]
So, the side length of the square is \( \sqrt{5} \).
The area of the square is the square of its side length:
\[ \text{Area} = (\sqrt{5})^2 \]
\[ \text{Area} = 5 \]
Therefore, the area of the square inscribed in the right triangle is \( 5 \) square units.
