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In the figure above, what is the area of triangular region B

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In the figure above, what is the area of triangular region B [#permalink] New post   05 Jun 2020, 08:31
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In the figure above, what is the area of triangular region BCD ?

(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)
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Re: In the figure above, what is the area of triangular region B [#permalink] New post   05 Jun 2020, 08:31
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Re: In the figure above, what is the area of triangular region B [#permalink] New post   05 Jun 2020, 09:41
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Carcass wrote:
Image


In the figure above, what is the area of triangular region BCD ?

(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)

Let x = the length of the hypotenuse of the red right triangle
Image

Since we have a right triangle, we can apply the Pythagorean Theorem
We get: 4² + 4² = x²
Simplify: 16 + 16 = x²
Simplify: 32 = x²
So, x = √32 = 4√2

We get:
Image


Now focus on the blue right triangle...
Image

Area of triangle = (base)(height)/2
So, area = (4√2)(4)/2
= (16√2)/2
= 8√2

Answer: C

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Brent
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Re: In the figure above, what is the area of triangular region B [#permalink] New post   29 Oct 2024, 04:06
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Re: In the figure above, what is the area of triangular region B [#permalink]
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