Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

64% (01:11) correct 36% (01:12) wrong based on 133 sessions

HideShow timer Statistics

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.
_________________

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.
_________________