Find the area of the quadrilateral with vertices (0,3), (3,0), (0,9), and (9,0).

Guest Jun 8, 2020

#1**+1 **

Let A = (3,0) B = (0,3) C = (9,0) D = (0,9) O = (0,0)

This is a trapezoid with AB and CD being the parallel sides (the bases).

The formula for the area of a trapezoid: A = ½·h·(b_{1} + b_{2})

b_{1} = side AB b_{2} = side CD

Since triangle(OAB) is 45^{o}-45^{o}-90^{o} right triangle with OA = 3 and OB = 3,, AB = 3·sqrt(2).

Since triangle(OCD) is 45^{o}-45^{o}-90^{o} right triangle with OC = 9 and OD = 9,, CD = 9·sqrt(2).

Now to find the height -- draw line segment OXY perpendicular to both AB and CD, with

X the point of intersection with AB and Y the point of intersection with CD.

Since triangle(OAX) is 45^{o}-45^{o}-90^{o} right triangle with OA = 3, OX = 3/sqrt(2).

Since triangle(OCY) is 45^{o}-45^{o}-90^{o} right triangle with OC = 9, OY = 9/sqrt(2).

XY = OY - OX = 9/sqrt(2) - 3/sqrt(2) = 6/sqrt(2) or 3·sqrt(2).

This is the height of the trapezoid.

A = ½·h·(b_{1} + b_{2}) = ½·[ 3·sqrt(2) ]·[ 3·sqrt(2) + 9·sqrt(2) ]

= ½·[ 3·sqrt(2) ]·[ 12·sqrt(2) ]

= 36

geno3141 Jun 8, 2020