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# M24-22

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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16 Sep 2014, 00:21
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Difficulty:

55% (hard)

Question Stats:

64% (01:14) correct 36% (01:14) wrong based on 137 sessions

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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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16 Sep 2014, 00:21
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 $$= \frac{1}{4}$$.

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Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
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WE: Consulting (Consulting)

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18 Jun 2015, 09:30
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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19 Jun 2015, 00:03
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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19 Jun 2015, 03:48
Thanks, Bunuel!
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09 May 2018, 01:22
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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09 May 2018, 01:35
1
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.

Attachment:
Triangle.png

>> !!!

You do not have the required permissions to view the files attached to this post.

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31 Jul 2018, 11:44
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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31 Jul 2018, 19:42
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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M24-22 &nbs [#permalink] 31 Jul 2018, 19:42
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# M24-22

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