Please help me with this problem!?

Question

The length of a rectangle is 1 cm longer than its width. If the diagonal
of the rectangle is 5 cm, what are the dimensions (the length and width) of the
rectangle?

Answer 1

L = 1 + W
D = 5cm

using pythagoras theorem, D² = L² + W²

substitute L into equation,
5² = (1 + W)² + W²
25 = 1 + 2W + W² + W²
let 2W² + 2W - 24 = 0

simplify,
2(W² + W - 12) = 0

factorise,
W.......+4.......+4W
W.......-3........-3W
-------------------------
W²......-12......+W

W² + W - 12 = 0
(W + 4)(W - 3) = 0

W + 4 = 0
W = - 4 ===> (not possible for a dimension to be negative value)

W - 3 = 0
W = 3cm

L = 1 + W = 1 + 3 = 4cm

therefore, Length = 4cm, Width = 3cm

Answer 2

Let x be the width. Since the length is 1 cm longer than the width, length = x + 1

Using pythegorean theorum of a^2 + b^2 = c^2:

x^2 + (x + 1)^2 = 5^2
x^2 + x^2 + 2x + 1 = 25
2x^2 + 2x + 1 - 25 = 0
2x^2 + 2x - 24 = 0
2(x^2 + x - 12) = 0

Now factor:

2(x + 4)(x-3) = 0

x = -4 or x = 3

Since we can't have negative measurements, x must be 3.

So width = 3, and length = 4

To check, we sub them into our pythegorean theorum:

3^2 + 4^2 = 25
9 + 16 = 25
25 = 25

So it works.

Answer 3

This is a problem that requires applying the Pythagorean Theorem, a^2 + b^2 = c^2. The rectangle's sides are of length a and b, but you know that b = (a+1).

a^2 + (a+1)^2 = 5^2 = 25
a^2 + a^2 + 2a +1 = 25
2a^2 + 2a - 24 = 0
a^2 + a - 12 = 0

factor:
(a-3)(a+4) = 0

a = 3 or a = -4

Since a length can't be negative, you can discard that solution. Therefore, the sides are of length a = 3 and b = (a+1) = 4.

Answer 4

Call the length L, and the width W, and the diagonal D.

By the Pythagorean theorem, we know that
L^2 + W^2 = D^2
L^2 + W^2 = 5^2
L^2 + W^2 = 25

The problem also states that
L = W + 1

Combining these, we get
(W+1)^2 + W^2 = 25
2*W^2 + 2W + 1 = 25
2*W^2 + 2W = 24
W^2 + W = 12
W = 3

Plugging this in, we find that L = 4

So, the rectangle is 3 by 4 centimeters.

Answer 5

3 cm by 4 cm