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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
Attachment:
2017-08-18_1027.png [ 3.19 KiB | Viewed 2216 times ]
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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18 Aug 2017, 04:56
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Bunuel
What is the area of the triangle above?
(A) 6 + √5
(B) 3 + √5
(C) (3 + √5)/2
(D) 2
(E) 1
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.
