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# help

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

Jun 8, 2020

#1
+21955
+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·(b1 + b2)

b1 = side AB     b2 = side CD

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

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

Since triangle(OCY) is 45o-45o-90o 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·(b1 + b2​) = ½·[ 3·sqrt(2) ]·[ 3·sqrt(2) + 9·sqrt(2) ]

= ½·[ 3·sqrt(2) ]·[ 12·sqrt(2) ]

= 36

Jun 8, 2020
#2
+1352
+1

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

A(0,3)      B(3,0)      C(9,0)      D(0,9)

BC = 6          angle DCB = 45º

Area of the quadrilateral (ABCD) = (92 - 32) / 2

Jun 8, 2020