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

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Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

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

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26 Sep 2017, 23:31
00:00

Difficulty:

35% (medium)

Question Stats:

59% (00:32) correct 41% (01:04) wrong based on 45 sessions

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

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

[Reveal] Spoiler:
Attachment:

2017-09-27_1016_002.png [ 10.27 KiB | Viewed 642 times ]
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135189 [0], given: 12671

VP
Joined: 22 May 2016
Posts: 1105

Kudos [?]: 393 [0], given: 640

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

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27 Sep 2017, 18:06
Bunuel wrote:

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

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

[Reveal] Spoiler:
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:

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

[Reveal] Spoiler:
C

Last edited by genxer123 on 28 Sep 2017, 07:16, edited 1 time in total.

Kudos [?]: 393 [0], given: 640

VP
Joined: 26 Mar 2013
Posts: 1288

Kudos [?]: 302 [1], given: 166

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

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28 Sep 2017, 06:04
1
KUDOS
genxer123 wrote:
Bunuel wrote:

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

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

[Reveal] Spoiler:
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.

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

[Reveal] Spoiler:
A

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

Kudos [?]: 302 [1], given: 166

VP
Joined: 22 May 2016
Posts: 1105

Kudos [?]: 393 [0], given: 640

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
[Reveal] Spoiler:
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.

Kudos [?]: 393 [0], given: 640

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1771

Kudos [?]: 971 [0], given: 5

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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 971 [0], given: 5

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