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In the figure above, x = y. What is the ratio area of (triangle ABC)/

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In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png
2018-02-01_1104.png [ 14.24 KiB | Viewed 600 times ]
[Reveal] Spoiler: OA

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In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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New post 01 Feb 2018, 01:48
Bunuel wrote:
Image
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png


Area of triangle ADC = \(\frac{1}{2}*AD*x\)

Area of triangle ABC = Area of triangle ADB - Area of triangle ADC
Area of triangle ABC = \(\frac{1}{2}*AD*(x+y) - \frac{1}{2}*AD*x = \frac{1}{2}*AD*(2x-x)\) (as x=y) = Area of triangle ADC

Therefore, Ratio of area of (triangle ABC)/area of (triangle ADC) = 1(Option E)
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Re: In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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New post 01 Feb 2018, 03:17
Bunuel wrote:
Image
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png


Area of the triangle = (1/2)*Base*Height

For triangle ABC, Area of the triangle = (1/2)*y*AD

For triangle ADC, Area of the triangle = (1/2)*x*AD

Now since Area of ABC/Area of ADC = (1/2)*y*AD/(1/2)*x*AD = y/x

But since, x=y hence

Now since Area of ABC/Area of ADC = y/y = 1

Answer: option E
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Re: In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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New post 01 Feb 2018, 04:38
Bunuel wrote:
Image
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


Altitude of Triangle ABC is AH so area of triangle ABC is 1/2 x Y x AH and Area of Triangle ADC is 1/2 x X x AH. Now the ratio of 2 triangle will cancel everything out so answer is E

[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png
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In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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New post 01 Feb 2018, 09:27
Bunuel wrote:
Image
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png

To avoid having to keep track of side names, make a 3-4-5 right triangle, so that AD = 3, BD = 4, and AB = 5

x = y. BD = 4. So x = 2, y = 2

Area of ∆ ABC = \(\frac{b*h}{2}=\frac{2*3}{2}=3\)

Area of ∆ ADC: \(\frac{2*3}{2}=3\)

Ratio of areas?
\(\frac{3}{3}=1\)

Answer E
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Re: In the figure above, x = y. What is the ratio area of (triangle ABC)/ [#permalink]

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New post 02 Feb 2018, 11:16
Bunuel wrote:
Image
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?

(A) 1/2*(x + y)

(B) x + y

(C) xy

(D) 1/2

(E) 1


[Reveal] Spoiler:
Attachment:
2018-02-01_1104.png


Since x = y, we see that the base of triangle ABC is equal to that of triangle ADC. We also see that the height of triangle ABC is equal to that of triangle ADC.

Thus, the ratio area of (triangle ABC)/(triangle ADC) = 1.

Answer: E
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Re: In the figure above, x = y. What is the ratio area of (triangle ABC)/   [#permalink] 02 Feb 2018, 11:16
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