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# M24 Q22

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M24 Q22 [#permalink]

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04 May 2010, 08:38
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
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Re: M24 Q22 [#permalink]

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05 May 2010, 11:08
gmat00 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?

1/4
1/2
1
radical2
2

Please explain in details.
Thanks in advance

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 = $$\frac{1}{2} * x^2$$= $$\frac{1}{2} *\frac{1}{2} = \frac{1}{4$$}
Attachments

untitled.JPG [ 7.9 KiB | Viewed 1524 times ]

Director
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Re: M24 Q22 [#permalink]

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05 May 2010, 11:18
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 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?

1/4
1/2
1
radical2
2

Please explain in details.
Thanks in advance

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Re: M24 Q22 [#permalink]

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06 May 2010, 13:29
Thanks guys
Re: M24 Q22   [#permalink] 06 May 2010, 13:29
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# M24 Q22

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