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In the figure above, \(AB\) is perpendicular to \(BC\) and

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In the figure above, \(AB\) is perpendicular to \(BC\) and

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In the figure above, \(AB\) is perpendicular to \(BC\), and \(BD\) is perpendicular to \(AC\). Which of the following relationships must be true?

A. \((BD)^2=(BC)^2+(AB)^2\)

B. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}+\frac{1}{(AB)^2}\)

C. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}-\frac{1}{(AB)^2}\)

D. \((BD)^2=(BC+AB)^2\)

E. \((BD)^2=(BC-AB)^2\)

OA B

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AAPL wrote:
Veritas Prep



In the figure above, \(AB\) is perpendicular to \(BC\), and \(BD\) is perpendicular to \(AC\). Which of the following relationships must be true?

A. \((BD)^2=(BC)^2+(AB)^2\)

B. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}+\frac{1}{(AB)^2}\)

C. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}-\frac{1}{(AB)^2}\)

D. \((BD)^2=(BC+AB)^2\)

E. \((BD)^2=(BC-AB)^2\)

OA B
Let ∆ABD and ∆CBD each be a 45-45-90 triangle with sides 1, 1 and √2:

For the figure above, only option B is true:
1/BD² = 1/BC² + 1/AB²
1/1² = 1/√2² + 1/√2²
1 = 1/2 + 1/2
1 = 1

The correct answer is B.

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Post
AAPL wrote:
Veritas Prep



In the figure above, \(AB\) is perpendicular to \(BC\), and \(BD\) is perpendicular to \(AC\). Which of the following relationships must be true?

A. \((BD)^2=(BC)^2+(AB)^2\)

B. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}+\frac{1}{(AB)^2}\)

C. \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}-\frac{1}{(AB)^2}\)

D. \((BD)^2=(BC+AB)^2\)

E. \((BD)^2=(BC-AB)^2\)

OA B
Let’s take AB = BC = 1. Then, by the Pythagorean theorem, AC = √(1^2 + 1^2) = √2. Using the area formula, first with AB as height and BC as base and then with BD as height and AC as base; we get that (AB)(BC) = (BD)(AC); therefore BD = 1/√2. Let’s substitute AB = BC = 1 and BD = 1/√2 into each answer choice:

A) Is (1/√2)^2 = 1^2 + 1^2? No

B) Is 1/(1/√2)^2 = (1/1^2) + (1/1^2)?

Is (√2)^2 = 1 + 1? Yes

C) Is 1/(1/√2)^2 = (1/1^2) - (1/1^2)?

Is (√2)^2 = 0? No

D) Is (1/√2)^2 = (1 + 1)^2? No

E) Is (1/√2)^2 = (1 - 1)^2? No

As we can see, the only possible choice is B.

Answer: B

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Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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