In triangle $ABC,$ $AB = 15,$ $BC = 9,$ and $AC = 12.$ Find the length of the shortest altitude in this triangle.




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]

 Apr 30, 2024

For problem 2:


Since ∠PQR=∠PRQ, triangle PQR is isosceles with PQ=PR. Let x represent the length of segment PQ (and segment PR).


We are also given that RQ=8 and SQ=3. Segments RQ and SQ are the legs of right triangle QRS. We can use the Pythagorean Theorem to solve for QR (which we already know is 8):








Since we now know that RS=55​, and both triangles PQR and TRS are right triangles with a right angle at R, these triangles are similar by the AA Similarity criterion (both have a right angle and share the same acute angle measure).


Therefore, corresponding side lengths in these similar triangles will be proportional. In particular:


PQ/RS=RQ/PR (We can substitute PR with x since they are equal)




Cross-multiplying gives:




Taking the square root of both sides (remembering there might be positive and negative square roots), we get:


x= 2*sqrt(11)


Therefore, the length of segment PQ is 2*sqrt(11) units.

 Apr 30, 2024

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