Question

The diagonal of a rectangle is 3 times as long as the width. If the length is 4 cm, what is the width of the rectangle?

Answer 1

`w= sqrt2` cm

Explanation:

Let `d` represent the diagonal
Let `l` represent the length
Let `w` represent the width

`d= 3w`
`l=4`

Since the angles in a rectangle are `90^@` each:
`l^2+w^2= d^2`

Substitute:
`4^2+w^2= (3w)^2`

`16+w^2= 9w^2`

Set the expression equal to 0:
`8w^2-16=0`

Factor with GCF:
`8(w^2-2)=0`

`w^2-2=0`
`w= +-sqrt2`
Since distance can't be negative:
`w= sqrt2` cm

Solving for `d`:
`d= 3*sqrt2`
`d=3sqrt2` cm

Answer 2

`"width "=sqrt2`

Explanation:

`"the diagonal 'splits' the rectangle into 2 right triangles"`

`"let width "=w" then diagonal "=3w`

`"solve the triangle with legs w and 4 and hypotenuse 3w"`

`"using "color(blue)"Pythagoras' theorem"`

`rArr(3w)^2=w^2+4^2`

`rArr9w^2=w^2+16`

`"subtract "w^2" from both sides"`

`rArr8w^2=16`

`"divide both sides by 8"`

`rArrw^2=2`

`color(blue)"take the square root of both sides"`

`sqrt(w^2)=+-sqrt2larrcolor(blue)"note plus or minus"`

`rArrw=+-sqrt2`

`w>0rArrw=sqrt2`