1st what are the dimensions of the original rectangle
2006-11-07 08:51:51 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
w1=width of 1st rect.
L1=length of 1st rect
A1=area of 1st rect
w2=width of 2nd rect
L2=length of 2nd rect
A2=area of 2nd rect
w1=x
L1=4x
A1=length*width
=4*(square of x)
w2=x+2
L2=4x+5
A2=(x+2)(4x+5)
=x(4x+5)+2(4x+5)
=4*(square of x)+5x+8x+10
=4*(square of x)+13x+10
According to the given condition
A1+270=A2
4*(square of x)+270=4*(square of x)+13x+10
270=13x+10 //4*(square of x) gets cancelled
13x=260
x=20
therefore width of 1st rect=20cm
lenght of 1st rect=20*4=80cm
2006-11-07 09:07:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First rectangle: Let x=width, 4x=length
Second rectangle: 4x + 5=length, x + 2=width.
Dimensions of second multiplied together=270 sq cm.
Solve for x and substitute into first dimensions.
2006-11-07 09:00:35 · answer #2 · answered by Someone 4 · 0⤊ 0⤋
4x long as wide >> L = 4W
2nd rect. is 5cm longer and 2cm wide >> len = 4W+5, wid = W+2
area of 2nd is 270sqcm great than 1st
>> (4W+5)(W+2) = 270+4W*W
2006-11-07 09:00:55 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
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2016-12-28 15:29:35 · answer #4 · answered by schneir 3 · 0⤊ 0⤋
I'll help set it up, but leave the solution to you.
For the first rectangle:
x * ( 4x ) = A
For the second rectangle:
(x + 2) * (4x + 5) = A + 270
Now you have a two equations with two unknowns, and should be able to solve.
2006-11-07 09:04:12 · answer #5 · answered by James B 3 · 0⤊ 0⤋
If you meant the 2nd is 270 greater than the first, here are your equations: (a = L1, b = W1)
a/b = 4
(a+5)*(b+2) = 270 + ab
Solve these for a & b
2006-11-07 09:02:40 · answer #6 · answered by Steve 7 · 0⤊ 0⤋
1st rectangle:
L = 4x
B =x
2nd rectangle
L = 4x + 5
B = x +2
Area of rectangle
(5 +4x) * (2 + x) = 270 + (4x)*(x)
10+8x+5x +4x^2 = 270 + 4x^2
13x = 260
x = 20 cm
Length = 4x
= 80 cm
2006-11-07 11:51:11 · answer #7 · answered by Forgettable 5 · 1⤊ 0⤋