In the figure above, is the area of triangular region ABC

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

10 Jun 2012, 00:28

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70% (02:19) correct

29% (01:11) wrong

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OG13DS79v2.png [ 10.04 KiB | Viewed 2569 times ]

In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(1) (AC)^2=2(AD)^2

(2) ∆ABC is isosceles.

Source: OG13 DS79

[Reveal] Spoiler: OA

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Re: Triangular region ABC equal to triangular region DBA? [#permalink]

10 Jun 2012, 01:13

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I started with
Area = .5bh
Area(ABC) = .5*AC*CB
Area(DBA) = .5*AD*AB

So we need to know whether AC*CB = AD*AB

1) (AC)^2 = 2(AD)^2

This tells us the relationship between AC and AD but not enough to derive the full areas. We need to know about CB and AB.

Insufficient.

[strike]AD[/strike]
BCE

2) ∆ABC is isosceles.

This tells us that AC = CB, and furthermore, we then know AB, but nothing about AD.

Insufficient.

1+2) Here, we know the relationship between AC, AD, and CB explicitly. We can also determine AB.

Sufficient. Remember, we don't need to derive the area. Just determine whether or not we can determine if they will be equal.

:. Answer is C.

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Re: Triangular region ABC equal to triangular region DBA? [#permalink]

10 Jun 2012, 02:00

Thanks, vandygrad11!

After some more calculations I found the following solution, which is similar to yours:

Start with rephrasing the question:

Area of ABC = Area of DBA?
0.5 AC*CB = 0.5 = AD*AB ?
AC*CB = AD*AB?

Use Pythagorean theorem to simplify further:

AB^2=AC^2+CB^2
AB=\sqrt{AC^2+CB^2}

Therefore, the question reduces to:

AC*AB = AD*\sqrt{AC^2+CB^2}?

(1) AC = \sqrt{2}*AD

So we can get rid of AC. Still need AD, AB and CB though:

AC*AB = AD*\sqrt{AC^2+CB^2}?
\sqrt{2}*AD*AB = AD*\sqrt{AC^2+CB^2}?
\sqrt{2}*AB =\sqrt{2*AD^2+CB^2}?

We cannot simplify or reduce further. Insufficient.

(2) ABC = isosceles means AC = CB

So we can replace AC:

AC*AB = AD*\sqrt{AC^2+CB^2}?
CB*AB = AD*\sqrt{2*CB^2}?
CB*AB = AD*\sqrt{2}*CB?
AB = AD*\sqrt{2}?

We cannot simplify or reduce further. Insufficient.

(1) and (2)

(1) actually answers the question we are left with in (2):

AB = AD*\sqrt{2}?

Sufficient. Answer is C.

I am not sure if I could have done this in 2 mins.

Key take aways:

- Rephrase and simplify question as much as possible, given known formulas etc.
- To check C: use work you do for checking (1) also for checking (2) and vice versa.
- Don't get thrown off my complicated formulas :lol:

Hope it helps others. Please let me know if I made any mistake!

Thanks a lot!

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

23 Oct 2012, 11:36

does anyone have any other alternative method to solve?

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

23 Oct 2012, 22:39

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teal wrote:

does anyone have any other alternative method to solve?

You can solve sufficiency questions of geometry by drawing some diagrams too.

Attachment:

Ques3.jpg [ 4.81 KiB | Viewed 1680 times ]

We need to compare areas of ABC and DAB. Notice that given triangle ABC with a particular area, the length of AD is defined. If AD is very small, (shown by the dotted lines) the area of DAB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.

Now look at the statements:

(1) (AC)^2=2(AD)^2
AD = AC/\sqrt{2}
The area of ABC is decided by AC and BC, not just AC. We can vary the length of BC to see the relation between AC and AD is not enough to say whether the areas will be the same (see diagram). So insufficient.

Attachment:

Ques4.jpg [ 7.15 KiB | Viewed 1681 times ]

(2) ∆ABC is isosceles.
We have no idea about the length of AD so insufficient.

Using both, ratio of sides of ABC are 1:1:\sqrt{2} = AC:BC : AB
Area of ABC = 1/2*1*1 = 1/2

Area of DAB = 1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2

Areas of both the triangles is the same
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Re: In the figure above, is the area of triangular region ABC [#permalink]

24 Oct 2012, 02:04

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jfk wrote:

Attachment:

OG13DS79v2.png

In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(1) (AC)^2=2(AD)^2

(2) ∆ABC is isosceles.

Source: OG13 DS79

The area of a right triangle can be easily expressed as half the product of the two legs.
Area of triangle ABC is 0.5AC\cdot{BC} and that of the triangle DBA is 0.5AD\cdot{AB}.

The question in fact is "Is AC\cdot{BC} = AB\cdot{AD}?"

(1) From AC^2=2AD^2, we deduce that AC=\sqrt{2}AD, which, if we plug into AC\cdot{BC} = AB\cdot{AD}, we get \sqrt{2}AD\cdot{BC}=AB\cdot{AD}, from which \sqrt{2}BC=AB. This means that triangle ABC should necessarily be isosceles, which we don't know.
Not sufficient.

(2) Obviously not sufficient, we don't know anything about AD.

(1) and (2): If AC=BC=x, then AB=x\sqrt{2}, and AD=\frac{x}{\sqrt{2}}.
Then AC\cdot{BC}=AB\cdot{AD}=x^2.
Sufficient.

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

13 Mar 2013, 23:53

Nice and succinct explanation Eva...

Re: In the figure above, is the area of triangular region ABC [#permalink] 13 Mar 2013, 23:53

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In the figure above, is the area of triangular region ABC

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