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answer

by Sadowski » Thu May 24, 2007 6:16 am
Since we're given the 45 deg angle, we know that each side of the quadrilateral makes a 45-45-90 triangle.

We also know that the hypotenuse of the 45-45-90 is 5*sqrt(2)

The proportions for a 45-45-90 are 1-1-sqrt(2), so the height of the quadrilateral is 5.

Finally, multiply the base with the height = [5*sqrt(2)] * [5] = 25*sqrt(2)

Hope that helps.
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by f2001290 » Thu May 24, 2007 6:26 am
Hi sadwoski

Can i assume this Quadrilateral to be a Rhombus ?

I couldn't get your 1st statement that each side of quad make 45 - 45 -90
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by jayhawk2001 » Thu May 24, 2007 1:27 pm
Good one Sadowski.

Just to elaborate, the area of a rhombus (yes, quadrilateral with
4 equal sides) is =

1/2* product of length of diagonals
(or)
product of side * perpendicular distance between two sides (height)

In this case, if we drop a line from B perpendicular to AD, we will
get a 45-45-90 triangle with 5*sqrt-2 as the hypotenuse.

So, we know height = 5.

Area hence follows as side * height = 5*sqrt-2 * 5 = 25 * sqrt-2

For more details, visit your friendly wikipedia link --

https://en.wikipedia.org/wiki/Rhombus#Area
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by Cybermusings » Fri May 25, 2007 5:11 am
Area of a parallelogram = Base * height

Angle BAD = 45

Drop a perpendicular from vertex B on AD

Now the triangle becomes a 45-45-90 triangle with sides in the ratio 1:1:sqr rt.2

Hypotenuese of the Triangle = BA = 5* sqr rt. 2...Hence height = 5

Hence Area of the figure = 5 * sqr rt. 2 * 5 = 25 sqr rt. 2
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