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

Question

https://gmatclub.com/forum/download/file.php?id=42498 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 Untitled.png

CounterSniper · Answer

angle ABO=angle ACO=90* (Property )
AB^2=10-1
AB=3
ar of quad = AB*BO=3

PKN · Answer

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)

RichardSaunders · Answer

PKN

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

PKN · Answer

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.

RichardSaunders · Answer

Very helpful. Thanks  PKN !

anilesh10 · Answer

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

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

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