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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]

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New post 03 Feb 2014, 01:13
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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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) \(16\sqrt{2}\)

Problem Solving
Question: 71
Category: Geometry Triangles; Area
Page: 71
Difficulty: 550


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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 03 Feb 2014, 01:13
SOLUTION

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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) \(16\sqrt{2}\)

Using Pythagorean theorem \(BD=\sqrt{4^2+4^2}=4\sqrt{2}\).

The area of triangle BCD = \(\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}\).

Answer: C.
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 03 Feb 2014, 03:30
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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) \(16\sqrt{2}\)



Sol: In the given question Triangle ABD is isosele with sides AB=AD=4 and therefore BD=4\(\sqrt{2}\)

Area of triangle BCD is 1/2*4*4\(\sqrt{2}\)

600 level is okay
Ans is C
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 03 Feb 2014, 08:05
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its C.


BD will be 4square root 2. (that will be base) * 4/2=8 square root 2.
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 03 Feb 2014, 11:16
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Ans C

First we calculate BD by Hypotenuse Theorem:
BD=\sqrt{4^2+4^2}=4\sqrt{2}

Area BCD=1/2 * 4\sqrt{2} * 4= 8\sqrt{2}
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In the figure above, what is the area of triangular region BCD ? [#permalink]

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New post 08 Jul 2017, 01:56
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parkerd wrote:
This question has already been answered, but I just needed clarification on how \sqrt{4^2 + 4^2} = 4\sqrt{2}?


(A) 4√2
(B) 8
(C) 8√2
(D) 16
(E) 16√2


\(\sqrt{4^2 + 4^2}\)

Taking \(4^2\) common, we get;

\(\sqrt{4^2(1+1)}\)

\(\sqrt{4^2(2)}\) ------------ (\(\sqrt{4^2} = 4\))

\(4\sqrt{2}\)

Answer (A)...

Hope its clear now.

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In the figure above, what is the area of triangular region B [#permalink]

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New post 08 Jul 2017, 06:26
Bunuel 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)\(16\sqrt{2}\)

The area of right triangle BCD is \(\frac{1}{2}b*h\), where b = BC = 4, and h is BD because height must be perpendicular to base. Angle DBC of 90° indicates exactly that.To find length of BD, the hypotenuse of triangle ABD ...

Triangle ABD is an isosceles right triangle having sides in the ratio \(x : x : x\sqrt{2}\).

So to find BD, simply multiply 4 by \(\sqrt{2}\) = \(4\sqrt{2}\).

\((\frac{1}{2})\)\((4)\)\((4\sqrt{2})\)
= \(8\sqrt{2}\)

Answer C
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 22 Sep 2017, 07:09
Can we not consider AD = 4 as the height of the triangle BCD?
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 22 Sep 2017, 07:19
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Re: In the figure above, what is the area of triangular region B [#permalink]

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New post 16 Apr 2018, 10:11
Here BDsquare = 4square +4square
(BD)2 = 16+16
BDSquare = 32
BD= Root32
Area of Triangle BCD =( Base*Height )/2
=(4 * Root 32 )/2 =
= root 32 can be written as root 16*root 2 and by solving further
= 2*4*root2 - 8 root 2 = C
Re: In the figure above, what is the area of triangular region B   [#permalink] 16 Apr 2018, 10:11
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