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M24-22

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Bunuel
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M24-22 [#permalink] New post   16 Sep 2014, 01:21
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In triangle ABC, point O lies on side AC and segment BO is perpendicular to AC. If \(\angle BAO = \angle BCA = 45\) degrees and \(BC = 1\), what is the area of triangle ABO?

A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. 1
D. \(\sqrt{2}\)
E. 2
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A
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Bunuel
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Re: M24-22 [#permalink] New post   09 May 2018, 02:35
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bpdulog wrote:
In triangle ABC, point O lies on side AC and segment BO is perpendicular to AC. If \(\angle BAO = \angle BCA = 45\) degrees and \(BC = 1\), what is the area of triangle ABO?


A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. 1
D. \(\sqrt{2}\)
E. 2

Triangle ABC is right-angled and isosceles. Its area \(= 1*\frac{1}{2} = \frac{1}{2}\). The area of triangle ABO = half the area of triangle ABC \(= \frac{1}{4}\).

Answer: ADoes anyone have a picture explanation? I am not seeing the hint indicating that ABO triangle is 1/2 the area of ABC


Here it is:



Hope it helps.


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Triangle.png
Triangle.png [ 31.18 KiB | Viewed 13281 times ]
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Bunuel
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Re M24-22 [#permalink] New post   16 Sep 2014, 01:21
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Official Solution:

In triangle ABC, point O lies on side AC and segment BO is perpendicular to AC. If \(\angle BAO = \angle BCA = 45\) degrees and \(BC = 1\), what is the area of triangle ABO?

A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. 1
D. \(\sqrt{2}\)
E. 2

Triangle ABC is right-angled and isosceles. Its area \(= 1*\frac{1}{2} = \frac{1}{2}\). The area of triangle ABO = half the area of triangle ABC \(= \frac{1}{4}\).

Answer: A
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Re: M24-22 [#permalink] New post   18 Jun 2015, 10:30
Bunuel

Can you explain if the formula Side of A/Side of B = (Area of A/Area of B)^2 would be applicable to this case? IF not, why? And if yes, how?
PS: Your solution is much simpler, but just in case...
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Re: M24-22 [#permalink] New post   19 Jun 2015, 01:03
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michaelyb wrote:
Bunuel

Can you explain if the formula Side of A/Side of B = (Area of A/Area of B)^2 would be applicable to this case? IF not, why? And if yes, how?
PS: Your solution is much simpler, but just in case...


It should be \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).

Well, yes you can do this way too by realizing that both triangles BOC and ABC are 45-45-90 and thus similar. Then you can find the area of ABC (1/2), AC (\(\sqrt{2}\)) and then AC/BC = (area of ABC)^2/((area of AOC)^2):

\((\frac{\sqrt{2}}{1})^2 = \frac{1/2}{x}\) --> x = 1/4.

But, this is not a good way to solve.
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Re: M24-22 [#permalink] New post   19 Jun 2015, 04:48
Thanks, Bunuel!
Very helpful, as usual.
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Re: M24-22 [#permalink] New post   09 May 2018, 02:22
Does anyone have a picture explanation? I am not seeing the hint indicating that ABO triangle is 1/2 the area of ABC
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Re: M24-22 [#permalink] New post   31 Jul 2018, 12:44
In this case, wouldn't the value of AC be sqrt 2.

If so, it would then lead AO = (sqrt 2)/2 (half of the base of isosceles triangle)
Then BO = 1/2
Area of AOB = 1/2*1/2*(sqrt 2)/2

If we follow this approach, where we are still considering triangle AOB as half, why is the solution different from 1/4?
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Re: M24-22 [#permalink] New post   31 Jul 2018, 20:42
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Vineela.mk wrote:
In this case, wouldn't the value of AC be sqrt 2.

If so, it would then lead AO = (sqrt 2)/2 (half of the base of isosceles triangle)
Then BO = 1/2
Area of AOB = 1/2*1/2*(sqrt 2)/2

If we follow this approach, where we are still considering triangle AOB as half, why is the solution different from 1/4?




Yes, \(AC = \sqrt{2}\) and \(AO=\frac{\sqrt{2}}{2}\) but BO is not 1/2. \(BO = AO = \frac{\sqrt{2}}{2}\) because BAO is also isosceles. If you proceed from here you'll get the same answer as in the solution.
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Re: M24-22 [#permalink] New post   10 May 2019, 06:37
Bunuel I arranged the triangle such that AB is the height and BC is the base. Then I calculated the area as 1*1*/2=1/2. Please tell me what I am doing wrong, as I thought I could arrange the triangle to choose my height and base to calculate the area.
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Re: M24-22 [#permalink] New post   10 May 2019, 06:59
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crazi4ib wrote:
Bunuel I arranged the triangle such that AB is the height and BC is the base. Then I calculated the area as 1*1*/2=1/2. Please tell me what I am doing wrong, as I thought I could arrange the triangle to choose my height and base to calculate the area.


I think you calculated the area of ABC, while we need the area of ABO. Have you checked the diagram here: https://gmatclub.com/forum/m24-184387.html#p2104867
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Re: M24-22 [#permalink] New post   12 May 2019, 00:44
Bunuel wrote:
crazi4ib wrote:
Bunuel I arranged the triangle such that AB is the height and BC is the base. Then I calculated the area as 1*1*/2=1/2. Please tell me what I am doing wrong, as I thought I could arrange the triangle to choose my height and base to calculate the area.


I think you calculated the area of ABC, while we need the area of ABO. Have you checked the diagram here: https://gmatclub.com/forum/m24-184387.html#p2104867

Bunuel you're right. Misread the question again! Thank you!
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Re: M24-22 [#permalink] New post   26 Mar 2021, 11:39
It's rated under 700 level question.
Are we sure about that?
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Re: M24-22 [#permalink]
26 Mar 2021, 11:39
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