The legs of a right-angled triangle ABC are 8 and 10 cm from the vertex of the right angle C, the median CE and the bisector CD are drawn. Find the area of the triangle CED?
We find the area of the triangle ABC:
S ABC = 1/2 * AC * BC = 1/2 * 10 * 8 = 40 (cm²).
The CE median gives us two equal triangles:
S ACE = S BCE = S ABC / 2 = 20 (cm²).
S CED = S BCE – S BCD.
Consider triangles ACD and BCD, they have equal heights, respectively, the ratio of the areas of these triangles is equal to the ratio of the sides to which the heights are dropped.
By the property of the angle bisector, we have:
AD / BD = AC / BC = 10/8.
S ACD / S BCD = 10/8
S ABC = 10 + 8 = 18 parts = 40 (cm²).
S BCD = 40/18 * 8 = 320/18 = 160/9 (cm²).
S CED = S BCE – S BCD = 20 – 160/9 = 180/9 – 160/9 = 20/9 (cm²).
The area of the CED triangle is 20/9 cm².
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