Question 320.
Find AG in the figure as under.
Answer 320.
As the answer is independent of the radius of the circle, the best choice of the radius will be such that DC passes through the center of the circle. As one vertical line is at a distance 4 from A and the other at a distance 9 from A, the center must be at a distance (1/2) * (4 + 9) = 6.5 from A
=> radius of the circle can be taken as 6.5.
Refer to the figure drawn.
Δs AOC and BOD are isosceles and congruent.
[OA = OB = radius and OC = OD = radius and vertically opposite angles O are equal.]
=> x
= DF
= √[(6.5)^2 - (2.5)^2]
= 6.
Though my above answer was selected as the best, it involves an unproved assumption of the answer being independent of the radius of the circle.
One of my contacts, Rakesh Dubey, who is very good at solving challenging problems of geometry besides those of other topics of maths, provided a superior solution in which he has solved the problem without any assumption as above. He had posted his solution in comments section after the question was closed for answering. His solution is as under.
Refer to the following figure.
Join A and B to C and D.
In right triangle ADB, it can be proved with simple geometry that
DF^2 = AF * FB ... ... ... ... ... ( 1 )
In right Δs CAG and BDF,
angle ACG = angle DBF
=> Δs CAG and BDF are similar
=> AC / BD = AG / DF = x / DF
=> x = (AC * DF) / BD
=> x^2 = (AC^2 * DF^2) / BD^2 ... ( 2 )
In right Δs CAE and BAC,
angle A is common
=> Δs CAE and BAC are similar
=> AC/BA = AE/AC
=> AC^2 = AE * BA ... ... ... ... ( 3 )
In right Δs BDF and BAD,
angle B is common
=> Δs BDF and BAD are similar
=> BD^2 = AB * BF ... ... ... ... ( 4 )
Plugging DF^2, AC^2 and BD^2 from ( 1 ), ( 3) and ( 4 ) in eqn. ( 2 ),
x^2 = [AE * BA * AF * FB) / (AB * BF)
=> x^2 = AE * AF = 4 * 9 = 36
=> x = AG = 6.
Link to YA!
Find AG in the figure as under.
As the answer is independent of the radius of the circle, the best choice of the radius will be such that DC passes through the center of the circle. As one vertical line is at a distance 4 from A and the other at a distance 9 from A, the center must be at a distance (1/2) * (4 + 9) = 6.5 from A
=> radius of the circle can be taken as 6.5.
Refer to the figure drawn.
[OA = OB = radius and OC = OD = radius and vertically opposite angles O are equal.]
=> x
= DF
= √[(6.5)^2 - (2.5)^2]
= 6.
Though my above answer was selected as the best, it involves an unproved assumption of the answer being independent of the radius of the circle.
One of my contacts, Rakesh Dubey, who is very good at solving challenging problems of geometry besides those of other topics of maths, provided a superior solution in which he has solved the problem without any assumption as above. He had posted his solution in comments section after the question was closed for answering. His solution is as under.
Refer to the following figure.
In right triangle ADB, it can be proved with simple geometry that
DF^2 = AF * FB ... ... ... ... ... ( 1 )
In right Δs CAG and BDF,
angle ACG = angle DBF
=> Δs CAG and BDF are similar
=> AC / BD = AG / DF = x / DF
=> x = (AC * DF) / BD
=> x^2 = (AC^2 * DF^2) / BD^2 ... ( 2 )
In right Δs CAE and BAC,
angle A is common
=> Δs CAE and BAC are similar
=> AC/BA = AE/AC
=> AC^2 = AE * BA ... ... ... ... ( 3 )
In right Δs BDF and BAD,
angle B is common
=> Δs BDF and BAD are similar
=> BD^2 = AB * BF ... ... ... ... ( 4 )
Plugging DF^2, AC^2 and BD^2 from ( 1 ), ( 3) and ( 4 ) in eqn. ( 2 ),
x^2 = [AE * BA * AF * FB) / (AB * BF)
=> x^2 = AE * AF = 4 * 9 = 36
=> x = AG = 6.
Link to YA!
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