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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

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 \(= \frac{1}{4}\).

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...

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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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

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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