The length of a rectangle is 6 more than twice its width. Its area is 480 sq.cm. Find its dimension???

Question

I need a answer asap

Answer 1

Let length and width be x and y respectively.

Given, x = 2y +6   ------------(1)

And Area A = 480 sq m.  ---------(2)

Also we know area of a rectangle is

A = xy  ---------------(3)

Putting the values of (1) and (2) in (3) we get:

480 = (2y+6)y

or 480 = 2y^2 + 6y

or 2y^2 + 6y -480 = 0

so y = -17.06 or y=14.06 m  ( using quadratic formula y = {-b +/- sqrt(D)}/2a where D = b^2 -4ac

and a =2, b=6, c=-480

Neglecting negative value of y, since length cannot be negative.

y=14.06 m (approx)

Putting the value of y in (1) we get:

x = 34.12 m (approx)

So the length is 34.12 and width is 14.06 m.

 

Answer 2

If the width is w then length is 2w+6 (cm). The area is w(2w+6)=480.

That is, 2w²+6w-480=0, w²+3w-240=0. 

Using the formula, w=(-3±√(9+960))/2=(-3±√969)/2=(-3±31.1288)/2.

We need the positive square root, so w=14.0644cm approx. The length is 2w+6=34.1288cm. 

It’s unusual for questions of this type to have an irrational solution, so there may be an error in the question.

For example, if length=2w+16 instead of 2w+6, then the area is w(2w+16)=480.

That is, 2w²+16w-480=0, w²+8w-240=0=(w+20)(w-12). The width is therefore 12cm and the length=2w+16=40cm.