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In the figure above, if JL and KL are parallel to x and y axis respect

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In the figure above, if JL and KL are parallel to x and y axis respect  [#permalink]

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New post 17 Dec 2017, 01:56
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A
B
C
D
E

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Question Stats:

89% (01:07) correct 11% (01:17) wrong based on 52 sessions

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Re: In the figure above, if JL and KL are parallel to x and y axis respect  [#permalink]

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New post 17 Dec 2017, 05:29
Bunuel wrote:
Image
In the figure above, if JL and KL are parallel to x and y axis respectively, what is the area of ∆ JKL?

(A) 4.5
(B) 5
(C) 7.5
(D) 8
(E) 15

Attachment:
2017-12-17_1255.png


For area of JKL we need base and height of triangle

Base of the triangle = JL = 5 units (Change in X co-ordinate from -4 to +1)
Height of the triangle = KL = 3 units (Change in Y co-ordinate from -1 to +2)


Area = (1/2)*Base*Height = (1/2)*5*3 = 7.5

Answer: option C
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Re: In the figure above, if JL and KL are parallel to x and y axis respect  [#permalink]

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New post 17 Dec 2017, 11:53
Bunuel wrote:
Image
In the figure above, if JL and KL are parallel to x and y axis respectively, what is the area of ∆ JKL?

(A) 4.5
(B) 5
(C) 7.5
(D) 8
(E) 15

Attachment:
2017-12-17_1255.png


Area of a triangle = \(\frac{base*height}{2}\)

Distance between two points = \(\sqrt{(x2-x1)^2+(y2-y1)^2}\)

Since JL and KL are parallel to x and y, this means that JL will always have the same y-coordinate and KL will always have the same x-coordinate, so the coordinates of point L is (1,-1) and the right angle is JLK (means that line JL is the base and KL is the height).

\(Base = \sqrt{(-4-1)^2+(-1-(-1))^2}=\sqrt{5^2}=5\)

\(Height = \sqrt{(1-1)^2+(2-(-1))^2}=\sqrt{3^2}=3\)

\(Area = \frac{5*3}{2}=7.5\)

(C) is the answer.
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Re: In the figure above, if JL and KL are parallel to x and y axis respect   [#permalink] 17 Dec 2017, 11:53
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In the figure above, if JL and KL are parallel to x and y axis respect

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