M24 Q22

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?

1/4
1/2 
1
radical2 
2

Please explain in details.
Thanks in advance

cipher · Answer

Refer the attached figure.,

Angle ABC = 90, BO is the prependicular to side AC and hence it will also be the prependicular bisector for the side AC ( as angles BAO = BCA = 45)

this also implies that angle ABO =CBO = 45 hence CO=BO=AO = X

hyp theorm  x^2 + x^2 = 1 
 x^2= \frac{1}{2}

x= \frac{1}{\sqrt{2}}

Area of trianle ABO = \frac{1}{2} * AO* BO

Area =  \frac{1}{2} * x^2 =  \frac{1}{2} *\frac{1}{2} = \frac{1}{4 }

ykaiim · Answer

1/4.

Since both the angles are 45 degrees, triangle ABC has to be a right angled triangled at B. So, AB = BC =1
Ar(ABC)=1/2x1x1 = 1/2

Therefore, Ar(ABO) = 1/2xAr(ABC) = 1/4.

gmat00 · Answer

Thanks guys :-D

M24 Q22

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