The width of a rectangle is three-fourths it's lenght.the diagonal is 10 inches.what are it's dimensions

The width of a rectangle is three-fourths its length.
The diagonal is 10 inches.
What are its dimensions?

Let x be the length of the rectangle.
Since the width of the rectangle is three-fourths its length,
the width is 3/4*x
The diagonal is 10 inches.
Using the Pythagorean theorem
x^2 + ((3/4)*x)^2 = 10^2
x^2 + (9/16)*x^2 = 100
((16 + 9)/16)*x^2 = 100
Multiplying by 16
25x^2 = 1600
Dividing by 25
x^2 = 64
x = 8
8 inches is the length of the rectangle
(3/4)*x = (3/4)*8 = 3*2 = 6
6 inches is the width of the rectangle

w = 3L/4

Pythagorean theorem says

a^2 + b^2 = c^2 for right triangles
in this case a = w, b = l and c = the diagonal or 10

so, w^2 + L^2 = 10^2
substituing 3L/4 for w

(3L/4)^2 + L^2 = 100
9/16*L^2 + L^2 = 100
convert 2nd term to 16ths
9/16*L^2 + 16/16*L^2 = 100
25/16L^2 = 100
take the sqrt of both sides
5/4*L = 10

multiply both sides by 4/5
L = 40/5 = 8

W = 3/4 * L = 3*8/4=6

W = .75L : W² = 0.5625L²
W² + L² = 100
0.5625L² + L² = 100
L² = 100/(1.5625) = 64
L = 8
W = .75*8 = 6

Dimensions of the rectangle are 6 x 8 inches

(3L/4)^2 + L^2=10^2=100 , the Penthegorian Theorm.
9L^2/16 +L^2=100
25L^2/16=100
25L^2=1600
L^2=64
L=8 inches
W=(3/4)*8=6 inches
God bless you.

(3/4 L) ^2 + L ^2 = 100

25/16 L^2 = 100
L^2 = 1600/25 = 64
L= 8"
W = 6"