find length of rectangle given diagonal and area

Rectangle has area=168 m^2 and diagonal of 25. Find length

This is how tried to attempt the problem

Area= L X W

168 = L x W ..........(1)

L^2 + W^2 =25^2 ............(2)

From (1) L = 168/W...........(3)

Substitute (3) into (2)

(168/W)^2 +W^2 = 625

28224/W^2 + W^2 = 625

The problem gets complicated as I proceed

Is this aproach correct if it is,

Is there a convinient method

Re: find length of rectangle given diagonal and area

Quote:

Originally Posted by

**Bonganitedd** Rectangle has area=168 m^2 and diagonal of 25. Find length

This is how tried to attempt the problem

Area= L X W

168 = L x W ..........(1)

L^2 + W^2 =25^2 ............(2)

From (1) L = 168/W...........(3)

Substitute (3) into (2)

(168/W)^2 +W^2 = 625

28224/W^2 + W^2 = 625

The problem gets complicated as I proceed

Is this aproach correct if it is,

Is there a convinient method

Sketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.

Re: find length of rectangle given diagonal and area

Its same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2

The area, LW =168

Re: find length of rectangle given diagonal and area

Quote:

Originally Posted by

**Bonganitedd** Rectangle has area=168 m^2 and diagonal of 25. Find length

This is how tried to attempt the problem

Area= L X W

168 = L x W ..........(1)

L^2 + W^2 =25^2 ............(2)

From (1) L = 168/W...........(3)

Substitute (3) into (2)

(168/W)^2 +W^2 = 625

28224/W^2 + W^2 = 625

The problem gets complicated as I proceed

Is this aproach correct if it is,

Is there a convinient method

Have a look at this webpage.

Re: find length of rectangle given diagonal and area

Hello, Bonganitedd!

Quote:

Rectangle has area=168 m^2 and diagonal of 25. Find the length.

This is how tried to attempt the problem

$\displaystyle \text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$

$\displaystyle L^2 + W^2 \:=\:25^2\;\;[2]$

$\displaystyle \text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$

Is this approach correct? . Yes

If it is, is there a convinient method?

We have: .$\displaystyle \frac{28,\!224}{W^2} + W^2 \:=\:625$

Multiply by $\displaystyle W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0 $

Factor: .$\displaystyle (W^2 - 49)(W^2 - 576) \:=\:0$

$\displaystyle \begin{array}{ccccccccc}W^2-49 \:=\:0 & \Rightarrow & W^2 \:=\:49 & \Rightarrow & W \:=\:7 \\ W^2 - 576 \:=\:0 & \Rightarrow & W^2 \:=\:576 & \Rightarrow & W \:=\:24 \end{array}$

$\displaystyle \text{Assuming }L > W\text{, we have: }\:W\,=\,7,\;\boxed{L \,=\,24}$

Re: find length of rectangle given diagonal and area

Once you get to "[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the "fourth degree polynomial" bothers you, you can let $\displaystyle x= W^2$ so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.

Re: find length of rectangle given diagonal and area

Not taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF.