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In the figure above, what is the area of triangular region BCD ? : Problem Solving (PS)
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# In the figure above, what is the area of triangular region BCD ?

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

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

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

$$4\sqrt{2}$$

Hope its clear now.

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Re: In the figure above, what is the area of triangular region BCD ? [#permalink]
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 BCD ? [#permalink]
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 BCD ? [#permalink]
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 BCD ? [#permalink]
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 BCD ? [#permalink]
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 BCD ? [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:
Untitled.png
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

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