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In the figure above, MNO and OPM are isosceles right triangles. What

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Joined: 02 Sep 2009
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In the figure above, MNO and OPM are isosceles right triangles. What  [#permalink]

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17 Aug 2017, 23:34
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In the figure above, MNO and OPM are isosceles right triangles. What is the length of OP?

(A) 8
(B) 10
(C) 5√3
(D) 7√2
(E) 5√5

Attachment:

2017-08-18_1026.png [ 11.24 KiB | Viewed 953 times ]

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Joined: 02 Sep 2016
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GMAT 1: 660 Q50 V29
Re: In the figure above, MNO and OPM are isosceles right triangles. What  [#permalink]

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18 Aug 2017, 00:22
MNO is an isosceles right triangle.
It implies OM= Square root of (25+25) = 5 *(square root 2)
Now again OMP is isosceles right triangle. So
OP= Square root of (50+50)=10

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Joined: 22 May 2016
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In the figure above, MNO and OPM are isosceles right triangles. What  [#permalink]

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18 Aug 2017, 10:25
Bunuel wrote:

In the figure above, MNO and OPM are isosceles right triangles. What is the length of OP?

(A) 8
(B) 10
(C) 5√3
(D) 7√2
(E) 5√5

Attachment:
2017-08-18_1026.png

Both isosceles right triangles have legs of equal length. Hence both legs of small triangle have length 5. Find the length of the hypotenuse MO of the small triangle, which is the leg of the larger triangle. Then use that length to find OP, hypotenuse of large triangle.

1. Length of MO, where legs NO = MN = 5

Find length of hypotenuse MO with Pythagorean theorem,** or use property of isosceles right triangles (45-45-90), whose sides have ratio

$$1: 1: \sqrt{2}$$

Legs NO and MN = 5, so MO=$$5\sqrt{2}$$

2. Find OP

MO is one of the legs of large triangle OPM.
MO = $$5\sqrt{2}$$

By side ratio property, hypotenuse OP = $$5\sqrt{2} * \sqrt{2}$$

$$5 * 2 = 10$$

**
For MO (=x): $$5^2 + 5^2 = x^2$$
50 = $$x^2$$
x = $$5\sqrt{2}$$

For OP (=x):
Legs have equal length. MO =$$5\sqrt{2}$$ = MP

$$(5\sqrt{2})^2 + (5\sqrt{2})^2 = x^2$$

100 = $$x^2$$
x = 10 = OP
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In the figure above, MNO and OPM are isosceles right triangles. What   [#permalink] 18 Aug 2017, 10:25
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