Posted by Preston on Monday, October 14, 2013 at 12:21am.
1. A = lw
w =4l-3
4l-3(l) = 297
4l^2 -3l = 297
4l^2 -3l - 297 =0
(4l+33)(l-9)= 0
l = 9
w =33
2. a^2 + b^2 = c^2
short leg = x
Long leg = x+1
x^2 + (x+1)^2 = 7^2
x^2 +x^2 + x + x + 1 = 49
2x^2 +2x + 1 = 49
3. x, x +2
x (x+2) = 575
x^2+ 2x = 575
x^2 + 2x -575 =0
(x -23)(x + 25) = 0
x =23
23,25 and -23, -25
1 )
W = 4 L - 3
A = L * W = 297
L * ( 4 L - 3 ) = 297
4 L ^ 2 - 13 L = 297 Subtract 297 to both sides
4 L ^ 2 - 3 L - 297 = 297 - 297
4 L ^ 2 - 3 L - 297 = 0
Solutions :
L = - 33 / 4
and
L = 9
The length can't be negative so L = 9 in
W = 4 L - 3
W = 4 * 9 - 3
W = 36 - 3
W = 33 in
2 )
b = the longer leg = x + 1
c = hypotenuse = sqrt [ x ^ 2 + b ^ 2 ] = 7
c = hypotenuse = sqrt [ x ^ 2 + ( x + 1 ) ^ 2 ] = 7
c = sqrt ( x ^ 2 + x ^ 2 + 2 x + 1 )
c = sqrt ( 2 x ^ 2 + 2 x + 1 ) = 7 Square both sides
2 x ^ 2 + 2 x + 1 = 49 Subtract 49 to both sides
2 x ^ 2 + 2 x + 1 - 49 = 49 - 49
2 x ^ 2 + 2 x - 48 = 0
2 ( x ^ 2 + x - 24 ) = 0 Divide both sides by 2
x ^ 2 + x - 24 = 0
3 )
x = 1st number
x + 2 = 2nd number
x ( x + 2 ) = 575
x ^ 2 + 2 x = 575 Subtract 575 to both sides
x ^ 2 + 2 x - 575 = 0
Solutions :
x = - 25
and
x = 23
Numbers must be negative so :
1st number
x = - 25
2nd number
x + 2 = - 25 + 2 = - 23
4 )
c = width of the frame
New width = 7 + 2 c
New length = 13 + 2 c
( 7 + 2 c ) ( 13 + 2 c ) = 160
13 * 7 + 13 * 2 c + 2c * 7 + 2 c * 2 c = 160
91 + 26 c + 14 c + 4 c ^ 2 = 160
4 c ^ 2 + 40 c + 91 = 160 Subtract 160 to both sides
4 c ^ 2 + 40 c + 91 - 160 = 160 - 160
4 c ^ 2 + 40 c - 69 = 0
Solutions :
c = - 23 / 2
and
c = 3 / 2
Width of the frame can't be negative so :
c = 3 / 2 in
( 7 + 2 * 3 / 2 ) ( 13 + 2 * 3 / 2 ) =
( 7 + 3 ) ( 13 + 3 ) = 10 * 16 = 160 in ^ 2
5 )
P = 2 W + 2 L = 2 ( W + L ) = 30
A = W * L = 54
W * L = 54 Divide both sides by W
L = 54 / W
W ^ 2 - 15 W + 54 = 0 Divide both sides by 2
W + 54 / W = 15 Multiply both sides by W
W * W + ( 54 / W ) * W = 15 * W
W ^ 2 + 54 = 15 W Subtract 15 w to both sides
W ^ 2 + 54 - 15 W = 15 W -15 W
W ^ 2 + 54 - 15 W = 15 W -15 W
Solutions :
W = 6 in
and
W = 9 in
L = 54 / W
When W = 6 in then L = 54 / 6 = 9 in
When W = 9 in then L = 54 / 9 = 6 in
Assuming the width is smaller than the length the dimensions are :
W = 6 in , L = 9 in
In Question 5 delete
W ^ 2 - 15 W + 54 = 0 Divide both sides by 2
The length of the base measures
65ft less than twice the width. The perimeter of this base is 770
ft. What are the dimensions of the base?