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

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

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New post 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
2017-10-26_1348.png [ 6.32 KiB | Viewed 498 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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New post 26 Oct 2017, 07:26
Bunuel wrote:
Image
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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New post 30 Oct 2017, 13:54
Bunuel wrote:
Image
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)

Answer: B
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Kudos [?]: 979 [0], given: 5

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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