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Geometry Problem HELPPPPPPPPPP

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by arifaisal » Thu Nov 10, 2011 11:54 pm
Q1-Each side of a rhombus has length equal to 1 and one of its angles is 45 degree. What is the area of the rhombus?
(A)1
(B)1/square root 2
(C)1/2
(D)square root 2
(E) none

pls give full explanation
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by neelgandham » Fri Nov 11, 2011 12:24 am
Length of the side = 1
One of the Interior angles = 45

Area of a Rhombus is a^2 * sin(X), where a = side of the Rhombus and X is one of the interior angles. So, area of the Rhombus = 1*1*Sin 45 = 1/square root(2)

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by shankar.ashwin » Fri Nov 11, 2011 12:30 am
Area of rhombus = base * height.

base = 1

when you drop a perpendicular , you get a 45-45-90 triangle with hypotenuse as 1.

so 2x^2 = 1 (or) height = 1/Sqrt(2)

Area = 1* 1/Sqrt(2) B
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by sanju09 » Fri Nov 11, 2011 2:26 am
arifaisal wrote:Q1-Each side of a rhombus has length equal to 1 and one of its angles is 45 degree. What is the area of the rhombus?
(A)1
(B)1/square root 2
(C)1/2
(D)square root 2
(E) none

pls give full explanation

In a rhombus, if one of its sides is taken as its base and its distance from the opposite parallel side is taken as its height, then the area of this rhombus may also be given by base × height.

In the given problem, base = 1 and height is one leg of the special triangle 45º-45º-90º whose hypotenuse is 1, and hence the height would be 1/√2, using properties. The area of rhombus thus comes out to be

[spoiler]1 × 1/√2 = 1/√2.



B[/spoiler]
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