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# Let ABC be a right triangle such that B is a right angle

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Let ABC be a right triangle such that B is a right angle. A circle with diameter of  BC meets side AC at D.  If the area of ABC is 150 and AC = 40 then what is BD?

Nov 13, 2020

#1
0 If we draw the diagram, we can see that BD is the height of triangle ABC. This gives us: A = 1/2(AC)*(BD).
Plugging in the numbers we get: 150 = 1/2*(25)*(BD).

We can solve this to get BD = 12.

Hope this helped!

Caffeine :)

Nov 13, 2020
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My mistake, I thought that AC was 25 for a moment (from a similar problem). Anyhow, Jugoslav's answer has a wonderful explaination.

Caffeine  Nov 14, 2020
#2
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Let ABC be a right triangle such that B is a right angle. A circle with a diameter of  BC meets side AC at D.  If the area of ABC is 150 and AC = 40 then what is BD?

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Area of Δ ABC:      A = 150 u2

The base of a triangle ABC:       AC = 40 units

BD = 2A / AC = 300 / 40          BD = 7.5 units Nov 13, 2020
edited by jugoslav  Nov 13, 2020