+113 In right triangle ABC, leg AC is 2 cm and leg BC is 6 cm. A square is constructed of hypotenuse AB, the hypotenuse being one side of the square. What is the distance, in cm, from point C to the intersection point of the two diagonals of the square?
A) 4√3
B) 4√2
C) 6√2
D) 5√3
E) 5√2
+128408 Put A at (0,2), C at (0,0), and B at (6,0)
AB is √40 units in length
Construct two lines perpendicular to AB through A and B
The equation of the line through A is y = 3x + 2 .......the of the line through B is
y = 3x - 18
Constuct two circles with radiuses of √40 centered at A and B
The equation of the circle centered at A is x^2 + (y - 2)^2 = 40
The equation of the circle centered at B is (x - 6)^2 + y^2 = 40
And we can find the intersection of the circle centered at A and the perpendicular line to AB passing through A thusly :
x^2 + (3x + 2 - 2)^2 = 40
x^2 + 9x^2 = 40
10x^2 = 40
x = 2 and y = 3(2) + 2 = 8
So the intersection is at (2, 8)....label this point D
Likewise....we can find the intersection of the cirlce centered at B and the perpendicular line to AB passing through B thusly :
(x - 6)^2 + (3x - 18)^2 = 40
(x - 6)^2 + 9 ( x - 6)^2 = 40
10 (x - 6)^2 = 40
(x - 6)^2 = 4
x - 6 = 2
x = 8 and y = 3(8) - 18 = 6
So the intersection is (8, 6).....label this point E
And DB will be one diagonal of the square and AE will be the other diagonal
And the equation of DB is y = -2x + 12
And the eqation of AE is y = (1/2)x + 2
And we can find the inersection of the diagonals, thusly
-2x + 12 = (1/2)x + 2
10 = 2.5x
4 = x and y = -2(4) + 12 = 4
So the intersection of the diagonals is at (4,4)
And the distance between this point and C = √ [ 4^2 + 4^2] = √32 = 4√2
+36916 I wanted to see how Chris got the EXACT answer with a radical ....I got an equivalent numerical answer using the Law of Cosines (twice) See handwritten diagram:
