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In the figure above, x^2 + y^2 =

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In the figure above, x^2 + y^2 =  [#permalink]

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New post 26 Sep 2017, 23:31
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

59% (00:39) correct 41% (01:05) wrong based on 57 sessions

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In the figure above, x^2 + y^2 =  [#permalink]

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New post Updated on: 28 Sep 2017, 07:16
Bunuel wrote:
Image
In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

Attachment:
2017-09-27_1016_002.png

All four figures are right triangles.

\(x^2 + y^2\) = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.

Short leg, SL, of inside triangle:
\((\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2\)
\((4 + 5) = 9 = (SL)^2\)
\(SL = 3\)

Long leg, LL, of inside triangle:
\((\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2\)
\((6 + 10) = 16 = (LL)^2\)
\(LL = 4\)

The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5. Let hypotenuse 5 = c:

\(x^2 + y^2 = c^2\)
\(x^2 + y^2 = 5^2\)
\(x^2 + y^2 = 25\)


Answer
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that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"


Originally posted by generis on 27 Sep 2017, 18:06.
Last edited by generis on 28 Sep 2017, 07:16, edited 1 time in total.
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Re: In the figure above, x^2 + y^2 =  [#permalink]

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New post 28 Sep 2017, 06:04
1
genxer123 wrote:
Bunuel wrote:
Image
In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

Attachment:
2017-09-27_1016_002.png

All four figures are right triangles.

\(x^2 + y^2\) = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.

Short leg, SL, of inside triangle:
\((\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2\)
\((4 + 5) = 9 = (SL)^2\)
\(SL = 3\)

Long leg, LL, of inside triangle:
\((\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2\)
\((6 + 10) = 16 = (LL)^2\)
\(LL = 4\)

The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.

\(x^2 + y^2 = 5\)

Answer


The highlighted part should equal to (hypotenuse)...it should be 25.
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In the figure above, x^2 + y^2 =  [#permalink]

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New post 28 Sep 2017, 07:14
Mo2men wrote:
genxer123 wrote:
Bunuel wrote:
In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png

. . .
The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.

\(x^2 + y^2 = 5\)


The highlighted part should equal to (hypotenuse)...it should be 25.

Mo2men , nice catch. I was thinking of 5 as a fixed value derived from other fixed values, such that 5 = \(c^2\). I missed a step.

5 = \((\sqrt{c^2}) = (\sqrt{5^2}) = \sqrt{25}\).

I'll edit.

Thanks, and kudos.
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that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

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Re: In the figure above, x^2 + y^2 =  [#permalink]

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New post 03 Oct 2017, 10:06
Bunuel wrote:
Image
In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625


We see that x^2 + y^2 is equal to the square of the hypotenuse of a right triangle whose sides are not shown. Let’s first determine the base of this triangle, which we can call b:

(√4)^2 + (√5)^2 = b^2

4 + 5 = b^2

9 = b^2

b = 3

Let’s now determine the height of the triangle, which we can call h:

(√6)^2 + (√10)^2 = h^2

6 + 10 = h^2

16 = h^2

h = 4

Thus, x^2 + y^2 = 3^2 + 4^2 = 25.

Answer: C
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Re: In the figure above, x^2 + y^2 = &nbs [#permalink] 03 Oct 2017, 10:06
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