如图,在三角形ABC中,角C=90度,D是BC上一点,AB=17,AD=10,BD=9,求AC的长

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AC^2=AD^2-DC^2AB^2=AC^2+BC^2AB^2=AD^2-DC^2+BC^2AB^2=AD^2-(BC-BD)^2+BC^217^2=10^2-(BC-9)^2+BC^2解得BC=15AC^2=17^2-15^2AC=8