Question

Geometry problem #1

Answer 1

let each side of sq = a
total area = 5a^2
thus area of triangle NDM = (5/2)a^2
let FA and CD meet at X ,,BA and ED at Y,,and MX =x,,let NY =y
now area of triangle NDM = area of YAXD + area of NYA + area of NXM
=> (3)a^2 = ay + ax
=>y+x = 3a
=>y^2 + x^2 + 2yx = 9a^2
we also see that triangles MAX and NAY are similar
=>x/a = a/y
=>xy = a^2
--> x^2 + y^2 = 7a^2
now (y+a)^2 + (x+a)^2 = MN^2
=>(y^2+x^2) + 2a^2 + 2a(3a) = MN^2
=>15a^2 = MN^2

MN = a*sqrt(15)

Answer 2

now area of triangle NDM = area of YAXD + area of NYA + area of NXM
=> (3)a^2 = ay + ax

=>(2.5)a^2 = ay + ax ?

Answer 3

well it is 2.5a^2 = a^2 + 1/2 ay + 1/2 ax
simplify a bit and you'll what i wrote..

Answer 4

I'm sorry... still need more time to catch up with your solution :|

Answer 5

Got it! Thanks !!!! :)