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Similar triangles, third grade math exercise.
(1) If PQ//BC is needed, then (30-3x)/4x=30/20, and the solution is x= 10/3.

(2) When S △ BQC and S△ABC are based on CQ and CA respectively, and the heights are equal, CQ/CA= 1/3, CQ= 10.

Let the exercise time be X, x=CQ/3= 10/3, BP=20-4x=20/3=AB/3, S△BPQ? /? S△ABC=BP/AB= 1/3

(3) Let BA=BC, Angle C= Angle A, and let the movement time be X. If △APQ is similar to △CQB, then

In one example, BC/CQ=AQ/AP, 20/3x=(30-3x)/4x, x= 10/9, and AP=40/9.

On the other hand, BC/CQ=AP/AQ, 20/3x=4x/(30-3x), and the solution is x=5 or-10. If negative numbers are removed, x=5 and AP=20, that is, point P coincides with point B..