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# In the figure bellow, the rectangular at the corner measures 10cm x 20

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Manager
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In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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Updated on: 03 Mar 2015, 05:25
2
25
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Difficulty:

95% (hard)

Question Stats:

47% (02:26) correct 53% (02:41) wrong based on 170 sessions

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In the figure below, the rectangular at the corner measures 10cm x 20cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

A) 10
B) 40
C) 30
D) 50
E) None of the above

Attachment:
File comment: Pic

delete.png [ 3.49 KiB | Viewed 22730 times ]

Originally posted by ynaikavde on 02 Mar 2015, 14:49.
Last edited by Bunuel on 03 Mar 2015, 05:25, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 62637
In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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03 Mar 2015, 05:27
2
8
ynaikavde wrote:
In the figure below, the rectangular at the corner measures 10cm x 20cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

A) 10
B) 40
C) 30
D) 50
E) None of the above

Attachment:
The attachment delete.png is no longer available

Consider the diagram below:

(r - 10)^2 + (r - 20)^2 = r^2;
r = 10 or r = 50.
r cannot be 10, so it's 50.

Attachment:

Untitled.png [ 5.81 KiB | Viewed 21372 times ]

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Re: In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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20 Nov 2016, 06:11
1
Top Contributor
A) 10
B) 40
C) 30
D) 50
E) None of the above

Bunuel wrote:
(r - 10)² + (r - 20)² = r²

Notice that, once Bunuel creates the equation (r - 10)² + (r - 20)² = r², he doesn't try to solve it (a fair bit of work).
As he's doing that, he's looking for variations of some common Pythagorean triples. These are possible INTEGERS lengths that are possible in a right triangles.
The 4 most common Pythagorean triples are:
3-4-5
5-12-13
8-15-17
7-24-25
You'll see that each of these triples satisfies the Pythagorean Theorem (a² + b² = c²)
For example, 3² + 4² = 5² AND 5² + 12² = 13², etc)

So, when he plugs answer choice D (r = 50) into his equation (r - 10)² + (r - 20)² = r² he gets: (50 - 10)² + (50 - 20)² = 50²
Simplify to get: 40² + 30² = 50²
PERFECT, this is a variation of 3² + 4² = 5², but in this instance, each side is 10 times the length of the 3-4-5 right triangle.
As such, Bunuel automatically know that r = 50 is a solution to his equation.

RELATED VIDEO - the part about Pythagorean triples starts around 3:25

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Re: In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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20 Nov 2016, 10:04
1
ynaikavde wrote:
In the figure below, the rectangular at the corner measures 10cm x 20cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

A) 10
B) 40
C) 30
D) 50
E) None of the above

Attachment:
delete.png

we can easily eliminate A.
since the sides of the rectangular figure are 10x20, and since 2r = side of the big square, it must be true that 10 can't be the radius... otherwise, one side would be exactly half of the side of the big square...
method given by bunuel is a good approach...
r^2 = (r-10)^2 + (r-20)^2
r^2 = r^2 -20r +100 + r^2 -40r + 400
r^2 - 60r +500 = 0
(r-10)(r-50)=0
r=50
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In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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16 Jan 2018, 21:08
Bunuel wrote:
ynaikavde wrote:
In the figure below, the rectangular at the corner measures 10cm x 20cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

A) 10
B) 40
C) 30
D) 50
E) None of the above

Attachment:
delete.png

Consider the diagram below:

(r - 10)^2 + (r - 20)^2 = r^2;
r = 10 or r = 50.
r cannot be 10, so it's 50.

Need help.! Can we apply PA*PB=PT^2 if PB is the diameter of circle?
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Re: In the figure bellow, the rectangular at the corner measures 10cm x 20  [#permalink]

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10 Mar 2019, 11:37
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Re: In the figure bellow, the rectangular at the corner measures 10cm x 20   [#permalink] 10 Mar 2019, 11:37
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