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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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Untitled.png [ 11.64 KiB | Viewed 14047 times ]
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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Attachment:

untitled.PNG [ 15.91 KiB | Viewed 12630 times ]

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
1
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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08 Jul 2017, 00:56
2
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;

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

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

$$4\sqrt{2}$$

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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08 Jul 2017, 05:26
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 $$\sqrt{2}$$ = $$4\sqrt{2}$$.

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

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

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22 Sep 2017, 06:09
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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22 Sep 2017, 06:19
santro789 wrote:
Can we not consider AD = 4 as the height of the triangle BCD?

The altitude (height) of a triangle is the perpendicular from the base to the opposite vertex. Does AD satisfy this for triangle BCD?
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Re: In the figure above, what is the area of triangular region B  [#permalink]

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16 Apr 2018, 09:11
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
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Re: In the figure above, what is the area of triangular region B  [#permalink]

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21 Aug 2018, 08:33
As BAD is 45:45:90 triangle so BD=4*1.44
So area of BCD is = 1*4*4*1.44/2
Area od BCD=8*1.44=4underroot2

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Re: In the figure above, what is the area of triangular region B &nbs [#permalink] 21 Aug 2018, 08:33
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