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In the figure above, line segments AB and AC are tangent to circle O.

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Bunuel
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In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   02 Aug 2018, 00:45
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In the figure above, line segments AB and AC are tangent to circle O. If the length of OB = 1 and the length of OA = √10, what is the area of quadrilateral ABOC? (Figure not drawn to scale.)


A. 3/2

B. √3

C. 2√2

D. 3

E. 2√3



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PKN
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   09 Aug 2018, 21:09
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RichardSaunders wrote:
PKN

How do we know that "Angle OBA=90=Angle OCA"?


Hi RichardSaunders,
By definition of tangent to a circle:-
Tangent is a line passing a circle and touching it at just one point.
The tangent line is always at the 90 degree angle (perpendicular) to the radius of a circle..
Link:- https://gmatclub.com/forum/math-circles-87957.html

Here, the lines AB and AC touch the given circle at 'B' and 'C' respectively. Hence, they are tangents. As per above definition, the radii OB and OC are perpendicular to the lines AB and AC respectively. So, \(\angle{OBA}=90=\angle{OCA}\)

Hope it helps.
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   02 Aug 2018, 04:02
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Bunuel wrote:

In the figure above, line segments AB and AC are tangent to circle O. If the length of OB = 1 and the length of OA = √10, what is the area of quadrilateral ABOC? (Figure not drawn to scale.)


A. 3/2

B. √3

C. 2√2

D. 3

E. 2√3



Show Spoiler
Attachment:
Untitled.png


Observations on the quadrilateral ABOC.
1) AB=AC (tangent property)
2) Angle OBA=90=Angle OCA
3)OB=OC=1 (radius of the circle)
Area of quadrilateral ABOC= Area of triangle OBA+ Area of triangle OCA
Area of triangle OBA=Area of triangle OCA (from (1) and (3), and OA being the common side)
In the right angled triangle OBA, we have AB=\(\sqrt{10-1}\)=3
Area of triangle OBA=\(\frac{1}{2}*AB*OB\)=\(\frac{3}{2}\)
Area of quadrilateral=2*Area of triangle OBA=\(2*\frac{3}{2}\)=3

Ans. (D)
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   09 Aug 2018, 17:28
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PKN

How do we know that "Angle OBA=90=Angle OCA"?
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   02 Aug 2018, 00:51
Bunuel wrote:

In the figure above, line segments AB and AC are tangent to circle O. If the length of OB = 1 and the length of OA = √10, what is the area of quadrilateral ABOC? (Figure not drawn to scale.)


A. 3/2

B. √3

C. 2√2

D. 3

E. 2√3



Show Spoiler
Attachment:
Untitled.png



angle ABO=angle ACO=90* (Property )
AB^2=10-1
AB=3
ar of quad = AB*BO=3
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RichardSaunders
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   11 Aug 2018, 16:36
Very helpful. Thanks PKN!
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink] New post   05 May 2020, 11:25
angle ABO is 90degree (tangent property)...and since AB=AC(tangent property) triangle ABO and triangle ACO is similar.
so by Pythagoras theorem we can find the value of AB which comes out to be 3.
area of parallelogram ABOC =2(1/2*3*1)=3
so ans is D
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Re: In the figure above, line segments AB and AC are tangent to circle O. [#permalink]
05 May 2020, 11:25
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