length of a rectangle is 1 cm longer than its width. if the diagonal of the rectange is 4 cm, what are the dim

Question

what is the length and width.

Answer 1

this is a pythagorean theorem problem as the angles of a rectangle are right angles

x = width and x+1 = length 4 is the diagonal and the hypoteneuse of the triangle formed

x^2 + (x+1)^2 = 4^2

x^2 + x^2 + 2x + 1 = 16

2x^2 + 2s - 15 = 0

using the quadratic formula a=2 b=2 c=-15

x = -2 +- sqrt ( 4 - 1(2)(-15) / 4

x = -2 +- sqrt (34) / 4

since x must be positive then x = -2 + sqrt (34) /4

and x+1 = -1 + sqrt (34) /4

Answer 2

Say the width is x, then the length is x + 1. Using Pythagoras' theory, you can say that: x^2 + (x + 1)^2 = 16.
Re-arrange this to: 2x^2 + 2x - 15 = 0 and use the quadratic equation formula. This gives 2 answers for x. One is: 2.284 and the other is meaningless in this case (it is a negative value).
Therefore the width is 2.284cm and the length 3.284cm (to 3 dp)

Answer 3

Let x be the width, then x+1 is the length
x^2 + (x+1)^2 = 4^2 = 16
x^2 + x^2 + 2x + 1 = 16
2x^2 + 2x - 15 = 0
x = {-2 +- sqr[2^2 - 4(2)(-15)]}/2(2)
= { -2 +- sqr[4 + 120]}/4 = {-2 +- sqr[124]}/4
= {-2 +- 2sqr[31]}/4 = {-1 +- sqr[31]}/2

Answer 4

If the width is w , the length is w+1

for the diagonal with pythagorus theorem you have

d^2 = w^2+ (w+1)^2= w^2+w^2+2w+1
so d^2 = 16 = 2w^2+2w+1
2w^2+2w-15=0
you solve and find w=2.4 l=3.4