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# In the figure, BD is perpendicular to AC. BA and BC have length a. Wha

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Math Expert
Joined: 02 Sep 2009
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In the figure, BD is perpendicular to AC. BA and BC have length a. Wha [#permalink]

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26 Oct 2017, 02:52
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In the figure, BD is perpendicular to AC. BA and BC have length a. What is the area of the triangle ABC?

A. $$2x\sqrt{a^2-x^2}$$

B. $$x\sqrt{a^2-x^2}$$

C. $$a\sqrt{a^2-x^2}$$

D. $$2a\sqrt{x^2-a^2}$$

E. $$x\sqrt{x^2-a^2}$$

[Reveal] Spoiler:
Attachment:

2017-10-26_1348.png [ 6.32 KiB | Viewed 851 times ]
[Reveal] Spoiler: OA

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In the figure, BD is perpendicular to AC. BA and BC have length a. Wha [#permalink]

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26 Oct 2017, 07:26
Bunuel wrote:

In the figure, BD is perpendicular to AC. BA and BC have length a. What is the area of the triangle ABC?

A. $$2x\sqrt{a^2-x^2}$$

B. $$x\sqrt{a^2-x^2}$$

C. $$a\sqrt{a^2-x^2}$$

D. $$2a\sqrt{x^2-a^2}$$

E. $$x\sqrt{x^2-a^2}$$

[Reveal] Spoiler:
Attachment:
2017-10-26_1348.png

In an isoceles triangle perpendicular from the vertex to the base bisects the base.

Hence $$AC=DC$$ and $$AC=2AD=2DC$$

so in triangle ADB

$$AD=\sqrt{a^2-x^2}$$

So $$AC=2AD=2\sqrt{a^2-x^2}$$

Hence Area of triangle ABC $$= \frac{1}{2}*x*2\sqrt{a^2-x^2}$$

or, $$x\sqrt{a^2-x^2}$$

Option B
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Re: In the figure, BD is perpendicular to AC. BA and BC have length a. Wha [#permalink]

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30 Oct 2017, 13:54
Bunuel wrote:

In the figure, BD is perpendicular to AC. BA and BC have length a. What is the area of the triangle ABC?

A. $$2x\sqrt{a^2-x^2}$$

B. $$x\sqrt{a^2-x^2}$$

C. $$a\sqrt{a^2-x^2}$$

D. $$2a\sqrt{x^2-a^2}$$

E. $$x\sqrt{x^2-a^2}$$

[Reveal] Spoiler:
Attachment:
2017-10-26_1348.png

We can let AD = DC = b.

b^2 + x^2 = a^2

b^2 = a^2 - x^2

b = √(a^2 - x^2)

Thus, DC = √(a^2 - x^2) and AC = 2√(a^2 - x^2).

So the area of ABC is:

½ * 2√(a^2 - x^2) * x

x√(a^2 - x^2)

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Re: In the figure, BD is perpendicular to AC. BA and BC have length a. Wha   [#permalink] 30 Oct 2017, 13:54
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# In the figure, BD is perpendicular to AC. BA and BC have length a. Wha

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