In rectangle ABCD, AC = 20, BC = 12, DC = 3x – y. and AD = 2x + y. Solve for x and y. Show all work to receiv

Question

In rectangle ABCD, AC = 20, BC = 12, DC = 3x – y. and AD = 2x + y. Solve for x and y. Show all work to receive credit.

Answer 1

ְAC is a diagonal, thus
(i) 20^2 = (AC)^2 = (AB)^2 + (BC)^2

(ii) AB = CD = 3x - y
(iii) 2x+y = AD = BC = 12

Plug (iii),(ii) into (i)
20^2 = (3x - y)^2 + 12^2

400 = (3x - y)^2 + 12^2

400 = (3x - y)^2 + 144 // - 144

256 = (3x - y)^2

Because 3x - y is a side's length

3x - y = sqrt(256) = 16

We also have that 2x + y = 12

3x - y = 16
2x + y = 12

Add them
5x = 28
x = 5.6

Plug into 2x + y = 12
2*5.6 + y = 12

11.2 + y = 12 // - 11.2

y = 0.8

Answer 2

AC is a diagonal and all angles are astounding triangles so in accordance to pythagorean theorem AC squared equals AB squared plus BC squared. in case you do the math, AB is sixteen. AB=DC so 3x-y=sixteen. advert=BC so 2x+y=12. upload the two equations: 3x-y=sixteen + 2x+y=12 5x=28 x=28/5 y=4/5