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In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 01:17
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74% (02:02) correct 26% (01:39) wrong based on 147 sessions
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 03:02
Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif Let AD = x > AB = 3x Let radius = r> AC = diameter of the circle = 2r Now in right angled triangle ADC , right angled at D, \(AD^2+CD^2 = AC^2\) > \(x^2+(3x)^2 = (2r)^2\) > \(r^2 = \frac{5x^2}{2}\) Thus the ratio of rect area / circle area = Ratio =\(\frac{x*3x}{\frac{5\pi*x^2}{2}}\)> Ratio = \(\frac{6}{5\pi}\). E is the correct answer.



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In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 03:44
Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif Let, AB = 3 and AD = 1 i.e. BD = \(\sqrt{3^2 + 1^2}\) = \(\sqrt{10}\) = Diameter of Circle Area of Rectangle = AB x BD = 3 x 1 = 3 Area of Circle = (π/4)*Diameter^2 = (π/4)*10 = 5π/2 Area of Rectangle / Area of Circle = 3 / (5π/2) = 6/5πAnswer: Option E
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 11:40
Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif Given: AB = 3 * AD Let AD = x => AB = 3x. Let the diameter be d d^2 = x^2 + 9x^2 d = ( x * \(\sqrt{10}\) )/2 radius = d/2 = (x * \(\sqrt{10}\) )/4 Area of rectangle : area of circle x * 3x : π * (x/4\(\sqrt{10}\)) ^ 2 3x^2 : π * x^2 * (10/4) 12: 10 π => 6 : 5π Option E



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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 16:24
Lets assume AD = 2 so AB = 6 (3 times AD) By applying Pyth. Theory we know that AC ( which is the diameter of the circle) is  6^2 +2^2 = AC^2 AC = 2√10 Ratio = 2*6 / pi (2√10)^2 = 6/5pi is the answer > option E
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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22 Jul 2015, 18:57
ashokk138 wrote: Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif Given: AB = 3 * AD Let AD = x => AB = 3x. Let the diameter be d d^2 = x^2 + 9x^2 d = ( x * \(\sqrt{10}\) ) /2radius = d/2 = (x * \(\sqrt{10}\) )/4 Area of rectangle : area of circle x * 3x : π * (x/4\(\sqrt{10}\)) ^ 2 3x^2 : π * x^2 * (10/4) 12: 10 π => 6 : 5π Option E The highlighted part is a mistake which has led to several mistakes in the three steps in between.
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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26 Jul 2015, 11:03
Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif 800score Official Solution:Let AD = x. AB = 3x. Area of the rectangle is (AD) × (AB) = x × (3x) = 3x². Using the Pythagorean theorem, take AB and AD to get the diameter AC of the circle. x² + (3x)² = AC² x² + 9x² = AC² 10x² = AC² AC = x√10 and the radius is [x√10] / 2. The area of a circle is πr², so the area is π [(x√10) / 2]² = (10πx²)/4 = (5πx²)/2. The ratio is: 3x² : (5πx²)/2 = 3 : 5π/2 = 6 : 5π The correct answer is E.
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In the figure above, ABCD is a rectangle inscribed in a circle. If the
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28 Nov 2017, 14:51
Bunuel wrote: In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution.Attachment: rectanglecircle.gif Let's say AD has length 1 and AB has length 3 (this also means that side CD also has length 3) Since the blue triangle is a right triangle, we can use the Pythagorean Theorem to find the length of the 3rd side (the hypotenuse) If we let x = the length of the hypotenuse, we can write: 1² + 3² = x² Simplify: 1 + 9 = x² So, x² = 10, which means x = √10 At this point, we can see that the diameter of the circle is √10, which means the radius is (√10)/2What is the ratio of the area of the rectangle to the area of the circle?Area of rectangle = (length)(width) = ( 1)( 3) = 3 Area of circle = π(radius)² = π( √10)/2)² = π(10/4) = π(5/2) = 5π/2 So, the RATIO = 3/(5π/2) = (3)(2/5π) = 6/5π = 6 : 5π Answer: E Cheers, Brent
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the
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18 Dec 2018, 05:17
For my understanding: Is the statement "figure not drawn to scale" the obvious indicator that we can not assume the properties of a 906030 triangle? That was what I tried to do. So the trick here is to recognize that? I thought that: Since the diagonal of the rectangle passes the midpoint, I could conclude that it functions as hypothenuse which yields us the information that we will definitely have 90 degrees in the rectangle. Furthermore, I based my following assumptions (906030) triangle on the fact that the line that passes the midpoint would work as some sort of bisector to the 90 degree angles of the rectangle. (Is this where it all went wrong?) Is the only thing I can trust in this prompt the fact that we have a rectangle inscribed in a circle? All the previously mentioned conclusions can't be drawn with certainty, is that correct? I'm a little confused here...
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the &nbs
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