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505-555 (Easy)|   Geometry|      
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If the lines shown in the picture touch the circle with center O only 29 Oct 2019, 02:13
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77% (01:06) correct 23% (01:21) wrong based on 114 sessions

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If the lines shown in the picture touch the circle with center O only at points A and B, what is the area of the triangle ABC?

(1) Triangle ADC has area of 25.
(2) Triangle BDC has area of 25.


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If the lines shown in the picture touch the circle with center O only 29 Oct 2019, 02:43
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Bunuel

If the lines shown in the picture touch the circle with center O only at points A and B, what is the area of the triangle ABC?

(1) Triangle ADC has area of 25.
(2) Triangle BDC has area of 25.


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If the lines shown in the picture touch the circle with center O only at points A and B, what is the area of the triangle ABC?
Since the figure is symmetrical w.r.t to line CDO, angle ADC = angle BDC = 90
Area ADC = CD * DA
Area BDC = CD * DB
Since DA = DB; and DA + DB = AB; DA=DB= 1/2*AB; Area ADC = Area BDC = 1/2 * Area ABC

(1) Triangle ADC has area of 25.
Area of triangle ADC = 1/2 * Area of triangle ABC = 25
Area of triangle ABC = 25*2 = 50
SUFFICIENT

(2) Triangle BDC has area of 25.
Area of triangle BDC = 1/2 * Area of triangle ABC = 25
Area of triangle ABC = 25*2 = 50
SUFFICIENT

IMO D
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If the lines shown in the picture touch the circle with center O only 09 Dec 2019, 03:04
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Hi! how D creates two right angle. Can anyone explain. I did not understand above explanation as it said ...figure is symmetrical to CDO..
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If the lines shown in the picture touch the circle with center O only 12 Dec 2019, 05:17
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mafuz
Hi! how D creates two right angle. Can anyone explain. I did not understand above explanation as it said ...figure is symmetrical to CDO..
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mafuz

This can be said by applying the properties of a triangle
Consider triangle ACO and BCO
Here AO=OB(BOTH ARE RADIUS OF CIRCLE WITH CENTRE O); Angle CAO and Angle CBO are 90 degree; CA=CB(both are tangent)

So we can say by property SAS that triangle CAO is congruent to Triangle CBO

Now, consider triangle ADO and Triangle BDO
Here, AO=BO(Radius); angle AOD = angle BOD(using congruent condition of traingle ACO and BCO); OD=OD(Same side)
So, traingle ADO and BDO are congruent

Now since ADC is a chord, and angle ADO and BDO are same; we can say they are 90 degree

Hope this helps!
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If the lines shown in the picture touch the circle with center O only 12 Dec 2019, 07:45
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Thanks a lot aviral
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If the lines shown in the picture touch the circle with center O only 02 Sep 2021, 22:49
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