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

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Math Expert
Joined: 02 Sep 2009
Posts: 49303
In the figure above, x^2 + y^2 =  [#permalink]

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26 Sep 2017, 23:31
00:00

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 =

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

Attachment:

2017-09-27_1016_002.png [ 10.27 KiB | Viewed 879 times ]

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Joined: 22 May 2016
Posts: 1978
In the figure above, x^2 + y^2 =  [#permalink]

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Updated on: 28 Sep 2017, 07:16
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

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:

$$x^2 + y^2 = c^2$$
$$x^2 + y^2 = 5^2$$
$$x^2 + y^2 = 25$$

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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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Joined: 26 Mar 2013
Posts: 1808
Re: In the figure above, x^2 + y^2 =  [#permalink]

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28 Sep 2017, 06:04
1
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

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.

$$x^2 + y^2 = 5$$

The highlighted part should equal to (hypotenuse)...it should be 25.
Senior SC Moderator
Joined: 22 May 2016
Posts: 1978
In the figure above, x^2 + y^2 =  [#permalink]

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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.

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Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835
Re: In the figure above, x^2 + y^2 =  [#permalink]

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03 Oct 2017, 10:06
Bunuel wrote:

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.

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