Anonymous

# Can some1 help me with this math problem please??

1. the length of a rectangle is 4 times it width. if the area of the rectangle is 20 m (squared), find its length and width.

2. If the length of a rectangle is 4 more than its width and the area of the rectangle is 6in. (squared), find its length and width

Relevance

1) L = 4W

A = LW = 20

Substitute L = 4W to get 20 = 4W x W

20 = 4W^2, divide both sides by 4 to get

5 = W^2, take square root of both sides to get W

W = √5

20 = L x W

20 = L x √5, divide both sides by √5 to get L

L = 20/√5 or, when rationalized, 4√5

L = 4√5 and W = √5

----

2) L = W + 4

A = 6 = L x W, substitute L = W + 4 to get

6 = (W + 4) x W

6 = W^2 + 4W, substract 6 from both sides to get

0 = W^2 + 4W - 6

This is not easily factored, so you'd have to use the quadratic equation to solve for W. Once you have W, just plug that in to L = W +4 to get the L value. Good luck!

• It's best to draw this out, which I can't do here. So draw yourself a rectangle. Your length is L (I would normally use a cursive lower case L to represent this because just using a lower case l can be confusing) and the width is w.

What do you know about the length? We are told that it is 4 times the width. That means L=4*w which is the same thing as 4w, right?

What do you know about the area of a rectangle?

A=L * w or Lw

Now if A=Lw, and we know that L=4w for this rectangle, that means that, for this rectangle:

A=4w*w = 4w^2

What else do we know about this rectangle? We know that its area is 20m^2. That means A=20m^2, or for simplicity, 20. Let's substitute that into the equation:

20=4w^2

Basic algebra: divide out the 4

5=w^2

How do we solve w^2? To find w, we want to use the square root (sqrt). We have to do that to each side:

sqrt5=sqrt (w^2)

sqrt5=w

Now, since that 20 was actually m^2, the square root of the is m. So the w= sqrt5 meters

As for the length, we know that the length is 4 times the width, which means that the length is 4(sqrt5) m.

We then use our calculator to check if it works:

A=4(sqrt5) * (sqrt 5)

=20

2. The length of a rectangle is 4 more than its width.

What does that mean, exactly? Let's think about that, as I'm guessing that's where you are struggling:

Let's say the width is this long -----. The length is 4 more than that ---------. So, that means the length is w+4, right?

If A=L * w, and L = 2+4, then A = (w+4) * w.

Hopefully you know how to work out the rest. Give it a shot and then post your answer and we can see if it's correct.

• 1. Set Width = x and length = 4x

Since the area of a rectangle is length times width set up your equation like 4x times x = 20 and solve

2. Set width=x then length is x+4

Set up the equation like before x times x+4 = 6 and solve

• 1. since the length is 4 times the width, L = 4w

Area = Lw

Your area you gave is 20

20 = (4w)(w)

20 = 4w^2

Divide both sides by 4

5 = w^2

Take the square root of both sides

square root 5 = w

Now you can put w = square root 5 into your L= 4w formula

L = square root 5(2)

L = 2 square root 5

2. You have the length is 4 times its width, so L = 4w

The area you have is 6

Area = Lw

6 = (4w)(w)

6 = 4w^2

Divide both sides by 4

3/2 = w^2

Take the square root of both sides

w = square root 3/2

Now but w = square root 3/2 into L = 4w

L = 4(square root 3/2)

• for a: do the following sollution:

lenght is (x x4)m.

width is (x)m.

area is 21 m(squared)

u put the numbers in the formulae of area= lengh x width.

u will get x=2.5.

then u place that in lengh: (x+4) which will be (2.5x4)=10.

and do the same for width which is x= 2

then i think the answer will be lengh is 10 and 2. i guess! just tried.

for two:

lenght is (x +4)m.

width is (x)m.

u put the numbers in the formulae of area= lengh x width.

jsut work it out and do what i had done first by placing the value in lengh and in width! get it~??

• u hav 2 develop some equations inorder 2 solve this

u can sub. 4 lenght as x and width as y

and remember area is lenght times breadth

sorry i dont hav enough time 2 do this 4 u