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# What is the area of the triangle in the figure above?

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Math Expert
Joined: 02 Sep 2009
Posts: 44600
What is the area of the triangle in the figure above? [#permalink]

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18 Dec 2017, 21:41
00:00

Difficulty:

5% (low)

Question Stats:

96% (00:15) correct 4% (00:13) wrong based on 50 sessions

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What is the area of the triangle in the figure above?

(A) 4.0
(B) 7.5
(C) 8.0
(D) 8.5
(E) 15.0

[Reveal] Spoiler:
Attachment:

2017-12-19_0836_002.png [ 8.18 KiB | Viewed 634 times ]
[Reveal] Spoiler: OA

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Joined: 08 Jul 2010
Posts: 2067
Location: India
GMAT: INSIGHT
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Re: What is the area of the triangle in the figure above? [#permalink]

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19 Dec 2017, 00:45
Bunuel wrote:

What is the area of the triangle in the figure above?

(A) 4.0
(B) 7.5
(C) 8.0
(D) 8.5
(E) 15.0

[Reveal] Spoiler:
Attachment:
2017-12-19_0836_002.png

Base of the triangle = Distance between (0,0) and (5,0) = 5 units

Height of the triangle = Distance between (0,0) and (0,3) = 3 units

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

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Joined: 24 Nov 2016
Posts: 148
Re: What is the area of the triangle in the figure above? [#permalink]

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19 Dec 2017, 10:45
Bunuel wrote:

What is the area of the triangle in the figure above?

(A) 4.0
(B) 7.5
(C) 8.0
(D) 8.5
(E) 15.0

[Reveal] Spoiler:
Attachment:
2017-12-19_0836_002.png

Area of a triangle: $$\frac{base*height}{2}$$

Distance between a point and the origin (0,0): $$\sqrt{x^2+y^2}$$

Base: $$\sqrt{x^2+y^2}=\sqrt{5^2+0^2}=5$$

Height: $$\sqrt{x^2+y^2}=\sqrt{0^2+3^2}=3$$

Area of a triangle: $$\frac{(5*3)}{2}=7.5$$

Intern
Joined: 13 Aug 2017
Posts: 1
Re: What is the area of the triangle in the figure above? [#permalink]

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19 Dec 2017, 10:54
If we calculate the distance between origin and point on y-axis will give us height

h= 3 units

distance between origin and x-axis point will give the base

b= 5 units

Area = [1/2] *base * height

A= 0.5*5*3 = 7.5 Units
Re: What is the area of the triangle in the figure above?   [#permalink] 19 Dec 2017, 10:54
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