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 =
\(5\sqrt{2} * \sqrt{2}\) \(5 * 2 = 10\)
Answer B**
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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