M24-22

Question

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

Bunuel · Answer

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

michaelyb · Answer

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...

Bunuel · Answer

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.

michaelyb · Answer

Thanks, Bunuel!
Very helpful, as usual.

bpdulog · Answer

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

Bunuel · Answer

Here it is: https://gmatclub.com/forum/download/file.php?id=41209 Hope it helps. Triangle.png

Vineela.mk · Answer

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?

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=41209

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.

crazi4ib · Answer

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.

Bunuel · Answer

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

crazi4ib · Answer

Bunuel  you're right. Misread the question again! Thank you!

believerinthy · Answer

It's rated under 700 level question.
Are we sure about that?

M24-22

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