Algebra ..... plse help me?

Question

A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 squar centimeters grater then the first. What are the demmensions of the original rectangle?

Answer 1

While I agree with the first answer that using spell check may be worth your while, I also found the answer to be 80cm x 20cm. Good Job Guys!

Answer 2

Always be thinking: What do I know and what are they asking for?

OK. The first rectangle is 4 times as long as it is wide. Call its width x and that makes its length be 4*x.

The 2'nd rectangle is 5 cm longer than the first. That means its length is 4*x+5 cm. The 2'nd rectangle is also 2 cm wider than the first so it must be x+2 cm.

The area of the 1'st rectangle is x*4x (width times length) and the area of the second rectangle is (x+2)*(4*x+5) (again, width times length) Now we know that the 2'nd rectangle is 270 sq. cm. bigger than the 1'st one. So it must be that

(x+2)*(4*x+5) - x*4*x = 270

multiply it out to get

[4*x^2 + 13*x + 10] - 4*x^2 = 270

subtract the 4*x^2 term to get

13*x + 10 = 270

and you get

x = 20

Now go back and substitute the x in the 1'st rectangle to get 20 cm by 80 cm for its size. Its area is then 1600 sq. cm. (width times length)

The 2'nd rectangle must then be 22 cm wide and 85 cm long and have an area of 1870 sq. cm

And..... Lo and behold.... 1870-1600 = 270 sq. cm.

Just keep thinking: What do I know? How can I write it (algebraicly) so that it makes sense and I can put it into an equation? What is it that the question asks and how can I compute it from what I have?

OK, Pal. A rectangle is one third as wide as it is long. Another rectangle is 1 cm wider and 5 cm longer than the 1'st rectangle. The 2'nd rectantgle is 21 sq. cm bigger than the 1'st one. How big is each rectangle?

Show me what ya got

(Hint: If the 1'st rectangle is one third as wide as it is long, then it is three times as long as it is wide. It'll make the math a *bunch* easier if you see it that way )


Doug

Answer 3

Well, you got different answers from others, so I'll try it myself. Suppose rectangle 1 has length a and width b. Then a = 4b.
Rectangle 2 has length c and width d. c = a + 5, and d = b + 2.
Also, cd = ab + 270. Find a and b.

cd = (a+5)(b + 2) = ab + 270
(4b + 5)(b + 2) = 4b^2 + 270
4b^2 + 13b + 10 = 4b^2 + 270
13b = 260 ==> b = 20 ==> a = 80

The original rectangle is 80 cm x 20 cm.

Answer 4

ok

first rectangle has dimensions:

w=4x, l=x

second rectanlge has dimensions:

w=4x+5, l=x+2

when you solve for the area in the first rectangle,

you would use A=l*w which gives:

A=4x*x=4x^2

now, since the second rectangle is 270 cm^2 larger, the area for the second rectangle would be:

A=(4x^2)+270 cm^2

since you have the equation for area with respect to the first rectangle, you can solve for x in the second rectangle.

the area dimensions for the second rectangle are w=4x+5, l=x+2:

A=(4x+5)(x+2)=(4x^2)+(270) cm^2

solving (expanding brackets) gives you:

4x^2+5x+8x+10 = 4x^2+270

adding like terms to get a zero value on the right hand side leaves you with:

13x-260=0, (4x^2 on both sides cancel out)

solving for x, x = 20 cm.

since length is 4x, width = 4(20) = 80
and width is x, so length = 20.

dimensions are 80cm by 20cm

hope this helps :D

Answer 5

assuming:
the width of the first rectangle = x
the length of the the first rectangle = 4X
so the area of the first rectangle= width x length of the first rectangle
area of the first rectangle, A = X x 4X
so A = 4X^2

the width of the second rectangle = X + 2
the length of the the first rectangle = 4X + 5
so the area of the second rectangle= width x length of the second rectangle
A + 270 = (4X + 5)(X + 2)
substituting A from above
4X^2 + 270 = 4X^2 + 8X +5X + 10
260 = 13X
X = 20 cm

substituting X to the information above:
the width of the first rectangle = 20 cm
the length of the the first rectangle = 80 cm

Answer 6

Let x be the short side of the first triangle.

Area of first: (4x)(x) = a
Area of second: (4x+5)(x+2) = a + 270

Multiply the stuff out:
4x^2 = a
4x^2 + 8x + 5x + 10 = 4x^2 + 13x + 10 = a + 270.
Now subtract the first equation from the second equation. That is, subtract the left side of the first from the left side of the second, and the right side of the first from the right side of the second. (It is valid to do it the other way, or to subtract the same thing from both sides, but that will not help us here.)
You get:
13x + 10 = 270
13x = 260
x = 20 cm
So the width is 20 cm and the length is 80 cm. Sorry about the error earlier. Good luck!

Answer 7

Area of rectangle is A=LW fitrst rectangle: A1=4w(w) 2nd rectangle:4w+5 (w+2)=A1+270 , 4w square+13w+10=4w square+270 ,13w+10=270 ,13w=260 , w=20 thus the dimmensions of the first rectangle are W=20,L=80

Answer 8

Let the width of the first rectangle be X
The length would be 4X
Area is L x W (X) (4X) = 4X^2

The second rectangle is W=X+2 , L= 4X + 5
Area = (X+2) (4X + 5) = 4X^2 +13X +10
Area of second is 270 sq cm greater than first.
Add 270 sq cm to first to make them equal.
4X^2 + 270 = 4X^2 + 13X +10
collect terms and solve, 13X+260
X = 260/13 =20
Check:: 20 x80 = 1600, 22x 85 =1870
1870-1600 = 270

Answer 9

Algebra? More like you need English help. Maybe I should plse help you find a dctinary.

a 270 squar centimeters grater would make a lot of cheese shavings, let me tell you.

I could probably solve that one for you, but I won't because I don't have math next semester.

Answer 10

Original width - 20 centimeters
Original length - 80 centimeters

Answer 11

Let x=width of the fisrt retangle
then
(5+4x)(2+x)=270+4x^2
10+5x+8x+4x^2=270+4x^2
x=20
The dimensions are 80cmx20cm