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In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with the center O and the diameter AB. AC is 4 inches shorter than AB. The ratio of the angle COB’s measure to the measure of angle OBC is 2:1. What is the length of OB?
A. 2√2
B. 2√3
C. 4
D. 4 + 2√2
E. 4 + 2√3
Kudos for a correct solution.
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ENGRTOMBA2018
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Bunuel
\(\angle{COB}=2\angle{OBC}\) ----> In triangle OBC, \(\angle{obc}+\angle{ocb}+\angle{cob} = 180\) ---> x+x+2x = 180 ---> x=45 and \(\angle{cob} = {90}^\circ\)
Similarly, in Triangle AOC, \(\angle{OCA} = \angle{OAC} = {45}^\circ\)
Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R
We can proceed forward in 2 ways:
1. Trigonometric way (not recommended for GMAT)
In triangle ACO, AO / AC =cos (45)--->\(\frac{R}{2R-4} = \frac{1}{\sqrt{2}}\) ---> \(R = 4+2\sqrt{2}\), D is the correct answer.
2. Simpler and straighforward:
After you calculate AC=2R-4, in right triangle AOC (right angled at \(\angle{AOC}\)), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:
(2R-4)^2 = R^2+R^2 --->\(R = 4+2\sqrt{2}\), D is the correct answer.
Alternately, for this method after you calculate AC = 2R-4, in a 45-45-90 degree right triangle , the sides are in the ratio \(1:1:\sqrt{2}----> R\sqrt{2} = 2R-4 ---> R =4+2\sqrt{2}\)
Sorry if this sounds like a stupid question although I'm not very clear on how you managed to get Ang ocb = Ang Obc => x
from the question it states that COB and OBC are 2:1, couldn't OCB be equal to a value other than OBC?
from the question it states that COB and OBC are 2:1, couldn't OCB be equal to a value other than OBC?
Engr2012
ENGRTOMBA2018
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kelvind13
No question is a "stupid question".
Look below for answers to your doubts.
In triangle OCB, you have been given that \(\angle {cob} = 2* \angle {obc}\). So now assume \(\angle {obc} = x ---> \angle{cob} = 2x\)
now as OC and OB are radii of the circle, thus OC = OB ---> In triangle OCB as OC = OB, angles opposite to equal sides are always equal (remember this, this is true for ALL triangles!)---> \(\angle {OCB} = \angle{OBC} = x\)
Thus in triangle OCB, we get x+x+2x = 180...etc
Bunuel
Given : COB = 2*OBC
But angle OBC = angle OCB [Angles opposite to the equal side of Triangle BOC in which OB = OC = Radius]
Also Angle OBC + Angle BOC + Angle OCB = 180
therefore if angle OBC = a
then 2a+a+a = 180
i.e. a = 45 and angle COB = 90
i.e Triangle OBC is an Isosceles Right angle Triangle
Let Side OA = OB = Radius = r
i.e. AB = 2r
i.e. AC = 2r-4
Now, Triangle OAC also is an Isosceles Right angle Triangle in which Hypotenuse AC = r√2 = 2r-4 (given)
2r - r√2 = 4
i.e. r = 4/(2-√2) = 2(2+√2) = 4+2√2
Answer: Option D
I strictly urge GMAT test takers to stay away from Trigonometrical relations if you are not already good at them
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kelvind13
Relatively easier explanation would be
Line OC = Line OB [Both are Radius]
So angle OCB = angle OBC [Angles opposite to equal sides are equal in any triangle ref: Isosceles triangle]
I hope it helps!
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ENGRTOMBA2018
Joined: 20 Mar 2014
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GMATinsight, that is what I have mentioned in my post above
GMATinsight
Hi,
another straight way to find that the angle CAO =45..
an angle at the circumference and the angle at the center on the same arc are in the ratio 1:2...
therefore angle CAO:angle COB =1:2...
so angle CAO and CBO are equal and 45..
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Bunuel
Hi,
an angle at the circumference and the angle at the center on the same arc are in the ratio 1:2...
therefore angle CAO:angle COB =1:2...
so angle CAO and CBO are equal and 45, as the third angle is 90..
now two sides are x each and the hyp/dia =x+4..
therefore in the right angle triangle, x\(\sqrt{2}\)=x+4...
therefore \(x\sqrt{2}-x=4\)...
\(x=4(\sqrt{2}+1)\)..
radius =\(\frac{(x+4)}{2}\)...
\(4+2\sqrt{2}\)..
ans D
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Hi Engr2012,
\(\angle{COB}=2\angle{OBC}\) ----> In triangle OBC, \(\angle{obc}+\angle{ocb}+\angle{cob} = 180\) ---> x+x+2x = 180 ---> x=45 and \(\angle{cob} = {90}^\circ\)
Similarly, in Triangle AOC, \(\angle{OCA} = \angle{OAC} = {45}^\circ\)
Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R
We can proceed forward in 2 ways:
1. Trigonometric way (not recommended for GMAT)
In triangle ACO, AO / AC =cos (45)--->\(\frac{R}{2R-4} = \frac{1}{\sqrt{2}}\) ---> \(R = 4+4\sqrt{2}\), D is the correct answer.
2. Simpler and straighforward:
After you calculate AC=2R-4, in right triangle AOC (right angled at \(\angle{AOC}\)), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:
(2R-4)^2 = R^2+R^2 --->\(R =4+4\sqrt{2}\), D is the correct answer.
Alternately, for this method after you calculate AC = 2R-4, in a 45-45-90 degree right triangle , the sides are in the ratio \(1:1:\sqrt{2}----> R\sqrt{2} = 2R-4 ---> R =4+4\sqrt{2}\)[/quote]
Hi Engr2012,
Thanks for all alternatives. However it seems you small type mistakes. The correct answer is \(R = 4+2\sqrt{2}\) .
\(\angle{COB}=2\angle{OBC}\) ----> In triangle OBC, \(\angle{obc}+\angle{ocb}+\angle{cob} = 180\) ---> x+x+2x = 180 ---> x=45 and \(\angle{cob} = {90}^\circ\)
Similarly, in Triangle AOC, \(\angle{OCA} = \angle{OAC} = {45}^\circ\)
Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R
We can proceed forward in 2 ways:
1. Trigonometric way (not recommended for GMAT)
In triangle ACO, AO / AC =cos (45)--->\(\frac{R}{2R-4} = \frac{1}{\sqrt{2}}\) ---> \(R = 4+4\sqrt{2}\), D is the correct answer.
2. Simpler and straighforward:
After you calculate AC=2R-4, in right triangle AOC (right angled at \(\angle{AOC}\)), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:
(2R-4)^2 = R^2+R^2 --->\(R =4+4\sqrt{2}\), D is the correct answer.
Alternately, for this method after you calculate AC = 2R-4, in a 45-45-90 degree right triangle , the sides are in the ratio \(1:1:\sqrt{2}----> R\sqrt{2} = 2R-4 ---> R =4+4\sqrt{2}\)[/quote]
Hi Engr2012,
Thanks for all alternatives. However it seems you small type mistakes. The correct answer is \(R = 4+2\sqrt{2}\) .
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,315
Concentration: Finance, Strategy
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Thanks Mo2men, I have corrected the typos.
Bunuel
800score Official Solution:
Let us use the fact that the ratio of the angle COB’s measure to the measure of the angle OBC is 2 to 1. Denote the measure of the angle OBC as x degrees. Then, the measure of the angle COB is 2x degrees. OB = OC, as the both are radii. Therefore the triangle BOC is isosceles and angle OCB = angle OBC = x degrees. Since the sum of all the angle measures in any triangle is 180 degrees, we can write the equation:
x + x + 2x = 180
4x = 180
x = 45
So, the measure of the angle BOC is 2x = 90 degrees. The angles AOC and COB are contiguous so angle AOC = 180° – 90° = 90°. Also AO = OC as radii. Therefore, triangle AOC is an isosceles right triangle (90°-45°-45°). The sides of such triangle have 1 : 1 : √2 ratio.
The question statement defines AC = AB – 4. Since AB is a diameter and AO is a radius AC = 2AO – 4. When we plug it in AO : AC = 1 : √2 we get AO : (2AO – 4) = 1 : √2 .
√2AO = 2AO – 4
4 = 2AO – √2AO
4 = (2 – √2)AO
AO = 4/(2 – √2)
AO = 4(2 + √2) / ((2 – √2)(2 + √2))
AO = (8 + 4√2) / 2
AO = 4 + 2√2
The correct answer is D.
If you don't remember the property of a 90°-45°-45° triangle, there is an alternative ending of the solution. Let us use the fact that AC is 4 inches shorter than AB. Denote radii by y inches. So, we can deduce three things:
1. AO = OB = OC = y
2. AB = 2y
3. AC = 2y – 4
Then we use the Pythagorean Theorem for triangle AOC and get the following equation:
(2y – 4)² = y² + y²
4y² – 16y + 16 = 2y²
2y² – 16y + 16 = 0
y² – 8y + 8 = 0
y = 4 + 2√2 and y = 4 – 2√2 are two solutions to this equation. But, the length of AC (2y – 4) must be a positive number, so y = 4 + 2√2 is the only option.
The correct answer is D.
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Engr2012
\(\angle{COB}=2\angle{OBC}\) ----> In triangle OBC, \(\angle{obc}+\angle{ocb}+\angle{cob} = 180\) ---> x+x+2x = 180 ---> x=45 and \(\angle{cob} = {90}^\circ\)
Similarly, in Triangle AOC, \(\angle{OCA} = \angle{OAC} = {45}^\circ\)
Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R
We can proceed forward in 2 ways:
1. Trigonometric way (not recommended for GMAT)
In triangle ACO, AO / AC =cos (45)--->\(\frac{R}{2R-4} = \frac{1}{\sqrt{2}}\) ---> \(R = 4+2\sqrt{2}\), D is the correct answer.
2. Simpler and straighforward:
After you calculate AC=2R-4, in right triangle AOC (right angled at \(\angle{AOC}\)), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:
(2R-4)^2 = R^2+R^2 --->\(R = 4+2\sqrt{2}\), D is the correct answer.
Can you show how the above formula is solved to get 4+2 *root of 2 ??
ENGRTOMBA2018
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LaxAvenger
Sure, look below for the explanation:
\((2R-4)^2 = R^2+R^2\) ----> \([2(R-2)]^2 = 2R^2\)
\(4(R-2)^2 = 2R^2\) ---> \(2(R-2)^2 = R^2\) ---> \(2(R^2+4-4R) = R^2\) (applied the relation \((a+b)^2 = a^2+b^2+2ab\))
Rearranging the terms you get,
\(R^2 - 8R + 8 =0\) ----> For a quadratic equation,\(ax^2+bx+c = 0\) , the roots of the equations are : \((x1,x2) = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\). Thus for the given equation, we get,
\(R = \frac{8\pm \sqrt{64-32}}{2}\) ----> \(R = 4 \pm 2\sqrt{2}\) . Now as the length of AC ( = 2R – 4) must be a positive number, so R = 4 + 2√2 is the only option.
Hope this helps.
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