Question 388129: 3 squares with the first square's side length is 4 units, the next one to the right has side length of x units and the next square to the right has side length of 9 units. there is a straight line which's coordinates lies on the topleft vertices of each 3 squares.
=> what is the side length of the middle square?
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! 3 squares with the first square's side length is 4 units, the next one to the right has side length of x units and the next square to the right has side length of 9 units. there is a straight line which's coordinates lies on the topleft vertices of each 3 squares.
=> what is the side length of the middle square?
a vertex is one of the corners of one of the squares,
vertices is plural of vertex
set lower left corner of first square on origin (0,0)
top left corner first square: (0,4)
top left corner 2nd square: (4,x)
top left corner 3rd square: (4 + x,9)
slope = rise/run = (y2 - y1)/(x2 - x1)
slope between (0,4) and (4,x) --> (x - 4)/(4 - 0) = (x - 4)/4
slope between (4,x) and (4 + x,9) --> (9 - x)/(4 + x - 4) = (9 - x)/x
2 slopes need to be equal
(x - 4)/4 = (9 - x)/x
x(x - 4)/4 = (9 - x)
x(x - 4) = (9 - x)4
x^2 - 4x = 36 - 4x
x^2 - 4x + 4x - 36 = 0
x^2 - 36 = 0
(x + 6)(x - 6) = 0
x can not be -6
x = 6
side length is 6
check:
(x - 4)/4 = (6 - 4)/4 = 2/4 = 1/2
(9 - x)/x = (9 - 6)/6 = 3/6 = 1/2, same so yes
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