In the figure above, what is the area of triangular region B : GMAT Problem Solving (PS)
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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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03 Feb 2014, 00:13
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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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03 Feb 2014, 00:13
SOLUTION

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 = $$\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}$$.

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

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03 Feb 2014, 02:30
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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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03 Feb 2014, 07:05
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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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03 Feb 2014, 10:16
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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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Re: In the figure above, what is the area of triangular region B [#permalink]

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08 Feb 2014, 01:47
SOLUTION

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 = $$\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}$$.

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

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03 Oct 2015, 10:48
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Re: In the figure above, what is the area of triangular region B   [#permalink] 03 Oct 2015, 10:48
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