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Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are

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Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 05 Jun 2015, 06:18
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Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are 7 inches apart and 6 and 8 inches long, respectively. What is the radius of the circle in inches?

(A) 4
(B) 5
(C) \(4\sqrt{2}\)
(D) \(5\sqrt{2}\)
(E) 7

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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 05 Jun 2015, 06:32
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Bunuel wrote:
Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are 7 inches apart and 6 and 8 inches long, respectively. What is the radius of the circle in inches?

(A) 4
(B) 5
(C) \(4\sqrt{2}\)
(D) \(5\sqrt{2}\)
(E) 7



Answer: Option
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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 08 Jun 2015, 05:31
2
2
Bunuel wrote:
Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are 7 inches apart and 6 and 8 inches long, respectively. What is the radius of the circle in inches?

(A) 4
(B) 5
(C) \(4\sqrt{2}\)
(D) \(5\sqrt{2}\)
(E) 7


MANHATTAN GMAT OFFICIAL SOLUTION:

Redraw the figure as closely to scale as possible (use the grid on your scrapboard!), labeling the known dimensions and the radius in question.
Image

In order for the trapezoid vertices to lie on the circle, the trapezoid must be symmetrical about the dotted line, which passes through the center of the circle. By drawing this vertical and the radii to points B and C we have created two right triangles, allowing us to use the Pythagorean Theorem.

In fact, we might play an educated hunch that the triangles are 3–4–5 common right triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3 and 4 side. The unknown vertical sides are thus 4 and 3, which sum to 7 as they must.

Algebraically, we can set up the following equations from the picture:
x^2 + 3^2 = r^2
y^2 + 4^2 = r^2
x + y = 7

Setting the two equations for r^2 equal:
x^2 + 3^2 = y^2 + 4^2
x^2 – y^2 = 4^2 – 3^2
(x + y)(x – y) = 7

Since (x + y)(x – y) = 7, (x – y) = 1.

Solve for x and y:
(x + y) = 7
(x – y) = 7
2x = 8
x = 4
y = 7 – x = 3

The radius of the circle is 5, because r^2 = 3^2 + 4^2 = 25.

The correct answer is B.

Attachment:
2015-06-08_1627.png
2015-06-08_1627.png [ 71 KiB | Viewed 11731 times ]

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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 02 Jan 2016, 06:31
Is this a general rule: if trapezoid is inscribed in the circle, then it MUST BE isosceles trapezoid? Something like the triangle inscribed in a "half circle" is always right triangle?
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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 02 Jan 2016, 07:04
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MrSobe17 wrote:
Is this a general rule: if trapezoid is inscribed in the circle, then it MUST BE isosceles trapezoid? Something like the triangle inscribed in a "half circle" is always right triangle?


Hi,

Yes any trapezium in a circle will be an Isosceles Trapezium...

Since two sides are parallel so the line joining the centre and Mid points of those two parallel line will be a straight line and will serve the purpose of line of symmetry.
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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 01 Aug 2016, 00:55
Alternate Solution (Approximating Method):

Area of Square inside Trapezoid = 6x6= 36

Area of remaining part of Trapezoid = Sum of Area of 2 remaining triangles inside the Trapezoid = 1/2 .b .h = 1/2 x (8-6) x7 = 1/2 x 2 x 7 = 7

Total Area of Trapezoid = 36 + 7 = 43

To find radius of the circle use the answer choices to estimate which of the respective answer choices leads to an approximately correct corresponding circle area:

A) 4. Area= π. 4^2 = Aprox. 48 (TOO SMALL)
B) 5. Area= π. 5^2 = Aprox. 75 (CORRECT)
C) 7. Area= π. 7^2 = Aprox. 150 (TOO LARGE)
D) 4√2. Area= π. (4√2) ^2 = Aprox. 96 (TOO LARGE)
E) 5√2. Area= π. (5√2) ^2 = Aprox. 150 (TOO LARGE)

Answer is (B)
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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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New post 12 Mar 2017, 06:35
In order for the trapezoid vertices to lie on the circle, the trapezoid must be symmetrical about the dotted line, which passes through the center of the circle. By drawing this vertical and the radii to points B and C we have created two right triangles, allowing us to use the Pythagorean Theorem. In fact, we might play an educated hunch that the triangles are 3–4–5 common right triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3 and 4 side. The unknown vertical sides are
thus 4 and 3, which sum to 7 as they must. Algebraically, we can set up the following equations from the picture:

x^2 + 3^2 = r^2
y^2 + 4^2 = r^2
x + y = 7
Setting the two equations for r2 equal:
x^2 + 3^2 = y^2 + 4^2
x^2 – y^2 = 4^2 – 3^2
(x + y)(x – y) = 7
Since (x + y)(x – y) = 7, (x – y) = 1.
Solve for x and y:
(x + y) = 7
(x – y) = 7
2x = 8
x = 4
y = 7 – x = 3
The radius of the circle is 5, because r^2 = 3^2 + 4^2 = 25.
The correct answer is B.
Attachments

pythagores.PNG
pythagores.PNG [ 72.42 KiB | Viewed 6777 times ]


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Re: Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are  [#permalink]

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