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Question

The length of a rctangle is 1 cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle

Answer 1

Let the width of the rectangle be w cm , then the length is (w+1) cm. The diagonal is 4 cm.
Therefore, by Pythagoras's Theorem,
w^2 + (w+1)^2 = 4^2
or w^2 + w^2 + 2w +1 = 16
or 2w^2 + 2w -15 =0
or w^2 +w -15/2 = 0
so (w +1/2)^2 -1/4 -15/2 =0
so (w+1/2)^2 = 31/4
so w = -1/2 + sq root (31/4) You ignore the second "possible" answer which is
w = -1/2- sq root (31/4) because this will give a negative value for the width.

so w = -0.5 + 2.783882181

therefore width = 2.283882181 cm. or 2.3 cm to one dec.pl.

If you were to draw the rectangle , you could not accurately use a value for the width with more than one decimal place.

But if you use w = 2.3 to w = 2.2838...
then the more decimal places you use for the value for the width and length, the more accurate the final answers.

Finally, the complete answer is
the width is about 2.3 cm and the length is about 3.3 cm.

As a check (2.3)^2 + (3.3)^2 should equal 4^2
and 5.29 + 10.89 = 16.18 i.e. approx. 16.00

Also if we use w = 2.284, we get 5.216656 + 10.784656 , which = 16.0013120 i.e. approx 16.00

Sorry this answer is so long-winded.

Answer 2

Let W be the width. Then the length is W + 1.

Also, notice that the diagonal of the rectangle makes a right triangle wih the length and the width, so 4^2 equals the length squared plus the width squared.

You should end up with a single expression in terms of W, which you can simplify and solve.

Answer 3

Width = x. Length is x + 1.Right ?
For a rectangle, (Diagonal)^2 = x^2 + (x+1)^2 Right?
16 = x^2 +x^2 +2x +1 or 2x^2+2x -15 =0. You can solve this quadratic.May be there is a printing error. If the diagonal is 5.
the sides are 3 & 4.

Answer 4

Let w be the width. Then, the length is w+1. Diagonal is 4.
By pythogorean theorem, d^2=lengh^2+width^2
16=w^2+(w+1)^2
16=w^2+w^2+2w+1
2w^2+2w-15=0
solve this equation for w. To find the length, add 1 to w. Note that w cannot be negative.

Answer 5

A squared + B squared = C squared.
A=Length B=Width C= Hypotenuse (diagonal)
A=width +1 B=width C= 4cm
there's the formulas, you do the math

(J Patrick gives you the whole scoop)

Answer 6

Obtain a quadratic equation, and solve it.

If one side has length ' x ', the other has length (x + 1).

Use Pythagoras's Theorem on two sides and the diagonal. Then

x^2 + (x + 1)^2 = 4^2, so that 2x^2 + 2x + 1 = 16, or

2x^2 + 2x - 15 = 0.

Then find the appropriate solution. You'll need to use the standard formula for the solution of a quadratic equation. (HINT: what sign MUST it have?!)

Live long and prosper.

Answer 7

draw a triangle length x+1 height, x diagonal 4
pythagoras said diagonal squared = length squared x height squared
4 squared = 16 = (x+1) squared x (x) squared
= 16 = (x+1) x (x+1) + (x)squared
= 16 = (x)squared +2x +1 + (x) squared
16 = (2x)squared +2x +1
15 = (2x)squared +2x
divide by 2 = 7.5 = (x)squared + x

bring everything to the one side
(x)squared + x -7.5 = 0

use the formula (-b +or - sq root of(b squared - 4ac)/ 2a

remember the answer must be positive

-1+ or -sq root of 1sq - 4(1)(7.5)/2(1)

-1 + or - sq root 31 allover2

x = -3.28388 or x = 2.28388

remember it has to be positive

height i = 2.28388 and length is (2.283882 +1) = 3.28388

check
2.28388squared +3.28388 squared = 4 squared

Answer 8

width and length = width + 1 cm

pythagorean

w^2 + (w+1)^2 = 16
w^2 + w^2+2w+1 = 16
2w^2 + 2w= 15
w^2+2 = 7 1/2
w^2 = 5 1/2
w = 2.3452078
L=3.3452078

something like that gotta run

Answer 9

You have a triangle that is x on one side x+1 on another and the hypot is 4. Is that enough?
if not
x^2 + (x+1)^2 = 4^2 solve for x
Is that enough?

Answer 10

all you do is draw 4cm line,lets say top right to bottom left. and make a rectangle with your line going from your top to bottom.it will look like two triangles