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# What is the area of the triangle above?

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What is the area of the triangle above? [#permalink]

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17 Aug 2017, 23:42
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Difficulty:

25% (medium)

Question Stats:

92% (00:24) correct 8% (00:23) wrong based on 47 sessions

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What is the area of the triangle above?

(A) 6 + √5
(B) 3 + √5
(C) (3 + √5)/2
(D) 2
(E) 1

[Reveal] Spoiler:
Attachment:

2017-08-18_1027.png [ 3.19 KiB | Viewed 535 times ]
[Reveal] Spoiler: OA

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What is the area of the triangle above? [#permalink]

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18 Aug 2017, 01:49

Lets name the vertices A,B and C, s.t AC = √5, AB = 1, and BC = 2
Because $$AC^2 = AB^2 + BC^2$$, the triangle is a right angled at B.

Therefore the area = $$\frac{1}{2}$$*Base*Height = $$\frac{1}{2}*2*1 = 1$$(Option E)
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Last edited by Bunuel on 20 Aug 2017, 03:18, edited 1 time in total.
Edited the typo.

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Re: What is the area of the triangle above? [#permalink]

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18 Aug 2017, 04:56
Bunuel wrote:

What is the area of the triangle above?

(A) 6 + √5
(B) 3 + √5
(C) (3 + √5)/2
(D) 2
(E) 1

[Reveal] Spoiler:
Attachment:
2017-08-18_1027.png

1. We must check whether this triangle is a right triangle.
2. Since ($$\sqrt{5}$$)^2 is equal to 1^2 + 2^2, we can confirm that this is a right triangle.
3. So the area : $$\frac{1}{2}$$ * base * height = $$\frac{1}{2}$$ * 1 * 2 = 1

E.
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Re: What is the area of the triangle above? [#permalink]

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19 Aug 2017, 02:39
pushpitkc wrote:

Lets name the vertices A,B and C, s.t AC = √5, AB = 1, and BC = 2
Because $$AC^2 = AB^2 + BC^2$$, the triangle is a right angled at B.

Therefore the area = $$\frac{1}{2}$$*Base*Height = $$\frac{1}{2}*2*1 = 2$$(Option E)

There is typo as highlighted

Kudos [?]: 302 [0], given: 166

Re: What is the area of the triangle above?   [#permalink] 19 Aug 2017, 02:39
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# What is the area of the triangle above?

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