We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

In trapezoid ABCD with bases AB and CD, we have AB = 52, BC = 12, CD = 39, and DA = 5. What is the area of ABCD?

Mathgenius Aug 24, 2018

#1**+2 **

Since two of the sides must be parallel, assume that AB is parallel to CD

Drop a perpendicular from C to AB and let the intersection be the point E

Let EB = x

Similarly, drop a perpendicular from D to AB and let the intersection be the point F

Let FA = 52 - (39 + x) = 13 - x

So....the height of the triangles = height of the trapezoid

The height of the triangle on the right =√[12^2 - x^2] = CE

The height of the triangle on the left is √ [5^2 - (13 - x)^2 ] = DF

And, as stated, CE = DF

Equating the quantities under the roots, we have

12^2 - x^2 = 5^2 - (13 - x)^2

144 - x^2 = 25 - (169 - 26x + x^2)

144 - x^2 = 25 - 169 + 26x - x^2

144 = 25 - 169 +26x

144 = -144 + 26x

288 = 26x

x = 288/26 = 144/13

So...the height of the trapezoid = √[12^2 - (144/13)^2 ] = 60/13

And the area = (1/2)(height)(sum of the bases) =

(1/2)(60/13) ( 39 + 52) =

(1/2) (60/13) (91) =

(1/2)(60) (91/13) =

(1/2)(60)(7) =

(1/2)(420) =

210 units^2

CPhill Aug 24, 2018