CD is the height of a right-angled triangle ABC angle C 90 SADC: SCDB = 25: 36. What are the AC and CB legs?

Let us prove that triangles ACD and CDB are similar.

Since CD is the height of triangle ABC, triangles ACD and CDB are rectangular.

Let the angle ABC = X0, then the angle BCD = (90 – X) 0.

Angle ACD = (90 – BCD) = (90 – (90 – X) = X0.

Then the angle ACD = BD = X0, and the right-angled triangles ACD and CDB are similar in acute angle.

For similar triangles, the ratio of their areas is equal to the square of the similarity coefficient of the triangles. Then Sads / Ssdv = K ^ 2 = 25/36.

K = √ (25/36) = 5/6.

Then AC / CB = 5/6.

Answer: The legs of a triangle are 5/6.

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