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In △ ABC, ∠ BAC = 90 °, ab = 3, AC = 4, ad bisects ∠ BAC and intersects BC with D, then the length of BD is () A. 157B. 125C. 207D. 215
Let ∵ BC = AB2 + ac2 = 32 + 42 = 5, ∵ BAC = 90 °, ab = 3, AC = 4, ∵ BC = AB2 + ac2 = 32 + 42 = 5, ∵ ad bisects ∵ BAC, ∵ the distance from point d to AB and AC is equal, then s △ ABC = 12 × 3H + 12 × 4H = 12 × 5 × 125, the solution is h = 127, s △ abd = 12 × 3 × 127 = 12bd · 125
In △ ABC, ab = 13, AC = 15, high ad = 12, then the length of BC is ()
A. 14b. 14 or 4C. 8D. 4 or 8
There are two right triangles in this graph. By using the Pythagorean theorem, we can get: CD2 = 152-122 = 81, CD = 9. Similarly, we can get BD2 = 132-122 = 25, BD = 5, BC = 14. There is another way to draw this graph. That is, in this case, BC = 9-5 = 4, so we choose B
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