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# In the XY-plane, the vertices of a triangle have coordinates (0, 0), (

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Joined: 02 Sep 2009
Posts: 53066
In the XY-plane, the vertices of a triangle have coordinates (0, 0), (  [#permalink]

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16 Oct 2018, 03:55
00:00

Difficulty:

15% (low)

Question Stats:

95% (01:20) correct 5% (01:10) wrong based on 31 sessions

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In the XY-plane, the vertices of a triangle have coordinates (0, 0), (5, 5) and (10, 0). What is the perimeter of the triangle?

(A) 12

(B) 13

(C) 5 + 10√2

(D) 10 + 5√2

(E) 10 + 10√2

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Re: In the XY-plane, the vertices of a triangle have coordinates (0, 0), (  [#permalink]

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16 Oct 2018, 04:01
Bunuel wrote:
In the XY-plane, the vertices of a triangle have coordinates (0, 0), (5, 5) and (10, 0). What is the perimeter of the triangle?

(A) 12

(B) 13

(C) 5 + 10√2

(D) 10 + 5√2

(E) 10 + 10√2

Perimeter of the triangle = Sum of the lengths of three sides

$$Side_1$$ = Distance between (0, 0), (5, 5) $$= √(5-0)^2+(5-0)^2 = √50 = 5√2$$

$$Side_2$$ = Distance between (0, 0), (10, 0) $$= √(0-0)^2+(10-0)^2 = √100 = 10$$

$$Side_3$$ = Distance between (10, 0), (5, 5) $$= √(5-0)^2+(5-10)^2 = √50 = 5√2$$

i.e. Perimeter $$= 5√2 + 5√2 + 10 = 10+10√2$$

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Re: In the XY-plane, the vertices of a triangle have coordinates (0, 0), (  [#permalink]

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16 Oct 2018, 04:03
Let A(0,0),B(5,5), and C(10,0) be the vertices of the triangle ABC.

Perimeter of triangle=AB+BC+CA

AB=5√2
BC=5√2
CA=10
(Measure of sides are calculated using distance formula)

So, perimeter=10+10√2

Ans. (E)

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Re: In the XY-plane, the vertices of a triangle have coordinates (0, 0), (   [#permalink] 16 Oct 2018, 04:03
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