In the diagram below, we have BE = 7, AB = 5, and [CBDA]=15. What is [CDA]?
![[asy] size(125);pair C = (0,0);pair D = (4,0);pair EE = (2.5,0);pair A = (2.5,4);pair B = (2.5,2.5);draw(A--C--D--cycle);draw(C--B--D--cycle);draw(A--EE, dashed);draw(rightanglemark(B,EE,D,10));dot(Label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F7%2F3%2Fa%2F73afc7d94b4e634328ee398166742343981f7b91.png)
+9479 Also! 
Let's call side CD, which is the base of triangle CDA, " b " .
Let's call the area of CDA " a " .
(1/2)(b)(7 + 5) = a
(1/2)(b)(12) = a
6b = a
b = a/6
(1/2)(b)(7) = a - 15
(7/2)(b) = a - 15
(7/2)b + 15 = a
Plug in a/6 for b .
(7/2)(a/6) + 15 = a
(7/12)a + 15 = a Multiply through by 12 .
7a + 180 = 12a
180 = 5a
36 = a
+349 Is CPhill's ghost composing an answer right now? Because apparently, CPhill can compose answers while not being online... LOL
+129852 Parallel to CD, draw line EF through B
We have four triangles EBA, FBA, EBC and FBD....and their combined area is 15
And because triangle EAF is similar to triangle CAD.....then EB = 5/12 CE and BF = 5/12 ED ....and the height of EBA and FBA = BA = 5, while the height of EBC and FBD = EB = 7
So we have
Area EBA + Area of FBA + Area of EBC + Area of FBD = 15
(1/2)(5/12)CE*BA + (1/2)(5/12)ED *BA + (1/2)(5/12)CE *EB + (1/2)(5/12)ED *EB = 15
(1/2)(5/12)CE*5 + (1/2)(5/12)ED *5 + (1/2)(5/12)CE *7 + (1/2)(5/12)ED *7 = 15
[ (1/2)(5/12) CE * (5 + 7) ] + [ (1/2)(5/12) ED * (5 + 7) ] = 15
[ (1/2)(5/12) CE * (12) ] + [ (1/2)(5/12) ED * (12) ] = 15
[ (1/2)(5/12)* 12 [ CE + ED] = 15
(60/24) [ CD] = 15
CD = 15 * (24/60) = (1/4) * 24 = 6
So.....area CDA = (1/2)CD * EA = (1/2) * 6 * 12 = 36 units^2
+9479 Also!
Let's call side CD, which is the base of triangle CDA, " b " .
Let's call the area of CDA " a " .
(1/2)(b)(7 + 5) = a
(1/2)(b)(12) = a
6b = a
b = a/6
(1/2)(b)(7) = a - 15
(7/2)(b) = a - 15
(7/2)b + 15 = a
Plug in a/6 for b .
(7/2)(a/6) + 15 = a
(7/12)a + 15 = a Multiply through by 12 .
7a + 180 = 12a
180 = 5a
36 = a
