SOLUTION: the length of a painting is 5 centimeters more than its width. it is matted in a frame whose length is 8 more than twice the width of the painting and width is five less than twice
Question 1067683: the length of a painting is 5 centimeters more than its width. it is matted in a frame whose length is 8 more than twice the width of the painting and width is five less than twice the width of the painting. the frame alone had an area of 650 square cm find the dimensions of the painting and the frame Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! the length of a painting is 5 centimeters more than its width.
let L = the length of the painting
let w = the width
then
L = (w+5)
:
it is matted in a frame whose length is 8 more than twice the width of the painting and width is five less than twice the width of the painting.
2w+8 = length of frame
and
2w-5 = the width of the frame
:
the frame alone had an area of 650 square cm
find the dimensions of the painting and the frame
:
Overall area - painting area = frame area
(2w+8)*(2w-5) - w(w+5) = 650
FOIL
4w^2 - 10w + 16w - 40 - w^2 - 5w = 650
Combine like terms to form a quadratic equation
4w^2 - w^2 + 6w - 5w - 40 - 650 = 0
3w^2 + w - 690 = 0
Use the quadratic formula to find w, but this will factor to:
(3s+46)(w-15) = 0
the positive solution is all we want here
w = 15 cm is the width of the painting
then
L = 15 + 5 = 20 cm is the length of the painting
:
Find the dimensions of the frame
2(15) + 8 = 38 cm is the length of the frame
and
2(15) - 5 = 25 cm is the width
:
;
Confirm our solutions find the overall area and painting area
38*25 = 950
20*15 = 300
-------------subtract
frame A:650 sq/cm
