Three congruent isosceles triangles DAO, AOB and OBC have AD=AO=OB=BC=10 and AB=DO=OC=12. These triangles are arranged to form trapezoid ABCD, as shown. Point P is on side AB so that OP is perpendicular to AB.
Point $X$ is the midpoint of AD and point Y is the midpoint of BC. When X and Y are joined, the trapezoid is divided into two smaller trapezoids. The ratio of the area of trapezoid ABYX to the area of trapezoid XYCD in simplified form is p:q. Find p+q.
+118608 I am not putting in every step. If you have any questions you can ask.
area O'AB : area O'XY : area O'DC
10^2 : 15^2 : 20^2
100 : 225 : 400
1 : 2.25 : 4
let OO'=2h
Area of O'AB = 0.5*12h= 6h
it follows that
area of O'XY = 2.25*6h = 13.5h and then that area of AXYB = 13.5h-6h = 7.5h
it also follows that
area of O'DC = 4*6h = 24h and then that area of XDCY = 24h-13.5h = 10.5h
so
area of AXYB : area of AXYB
7.5 : 10.5
5 : 7
Anyone who is interested can finish it from there 
