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In the figure above, what is the area of triangular region B
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05 Jun 2020, 08:31
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) \(16\sqrt{2}\)
Re: In the figure above, what is the area of triangular region B
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05 Jun 2020, 08:31
05 Jun 2020, 08:31
Expert Reply
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Re: In the figure above, what is the area of triangular region B
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05 Jun 2020, 09:41
05 Jun 2020, 09:41
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Carcass wrote:
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
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:
Now focus on the blue right triangle...
Area of triangle = (base)(height)/2
So, area = (4√2)(4)/2
= (16√2)/2
= 8√2
Answer: C
Cheers,
Brent
