Question

Find the value of x.

AD = 2x + 4
DB = 6x - 20

http://image.tutorvista.com/content/feed/tvcs/diagonalrectangle45.JPG

Answer 1

nope it was just that...

Answer 2

\[(6x-20)(6x-20)=(2x-4)(2x-4)+AB^2\]\[36x^2-120x-120x+400=4x^2-8x-8x+16+AB^2\]\[(36x^2-4x^2)+(-240x+16x)+(400-16)=AB^2\]
\[32x^2-224x+384=AB^2\]
\[x^2-7x+12=AB^2\]
\[(x-3)(x-4)=AB^2\]

Answer 3

that's the only thing given sorry o-o

Answer 4

@Anime_Basics i made a mistake
i wrote (2x-4) instead of writing (2x+4)
(2x+4)(2x+4)+AB^2=36x^2-240x+400
4x^2+8x+8x+16+AB^2=36x^2-240x+400
32x^2-256x+384=AB^2
x^2-8x+12=AB^2
sqrt(x^2-8x+1)=AB

Answer 5

(x-2)(x-6) and square root 6 and 2