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# On the xy-plane, what is the perimeter of a triangle with vertices at

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On the xy-plane, what is the perimeter of a triangle with vertices at  [#permalink]

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29 Mar 2020, 08:34
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25% (medium)

Question Stats:

88% (01:30) correct 12% (02:33) wrong based on 34 sessions

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On the xy-plane, what is the perimeter of a triangle with vertices at points A(-1,-3), B(3,2), and C(3,-3)?

(A) 12

(B) $$10 + 2\sqrt{3}$$

(C) $$7 + 5\sqrt{2}$$

(D) 15

(E) $$9 + \sqrt{41}$$

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Re: On the xy-plane, what is the perimeter of a triangle with vertices at  [#permalink]

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29 Mar 2020, 11:36

Solution

Given
In this question, we are given that
• The vertices of a triangle are A(-1,-3), B(3,2), and C(3,-3)

To find
We need to determine
• The perimeter of triangle ABC

Approach and Working out
Length of AB = Distance between (-1, -3) and (3, 2) = $$\sqrt{4^2 + 5^2} = \sqrt{41}$$
Length of BC = Distance between (3, 2) and (3, -3) = $$\sqrt{5^2} = 5$$
Length of AC = Distance between (-1, -3) and (3, -3) = $$\sqrt{4^2} = 4$$
• Hence, perimeter of ABC = $$4 + 5 + \sqrt{41} = 9 + \sqrt{41}$$

Thus, option E is the correct answer.

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Re: On the xy-plane, what is the perimeter of a triangle with vertices at  [#permalink]

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29 Mar 2020, 11:53
If you draw it out, you'll realize it's a right triangle with side lengths 4, 5, and an unknown hypotenuse. To find the hypotenuse, use the Pythagorean Theorem, a^2 + b^2 = c^2.

4^2 + 5^2 = c^2
41 = c^2
sqrt(41) = c

Therefore, the perimeter is 4 + 5 + sqrt(41), or 9 + sqrt(41). The correct answer is (E).
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Joined: 03 Aug 2019
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On the xy-plane, what is the perimeter of a triangle with vertices at  [#permalink]

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29 Mar 2020, 20:13
On plotting the points, it turns out to be a right angle triangle,
So on applying Pythagoras theorem,
with base = 4 [3-(-1)] and height = 5 [2-(-3)]

hypotenuse =$$\sqrt{4^2+ 5^2}$$ =$$\sqrt{{41}}$$

So perimeter = 4+5+$$\sqrt{{41}}$$
=9+$$\sqrt{{41}}$$

Option E.
On the xy-plane, what is the perimeter of a triangle with vertices at   [#permalink] 29 Mar 2020, 20:13