tonebeeze wrote:
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y?
A. \(y > \sqrt {2}\)
B. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
C. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)
D. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)
E. \(y < \frac {\sqrt {3}}{4}\)
Concept:
For any two positive, real numbers, it will always be true that the Arithmetic Mean of those two numbers will be greater than or equal to the Geometric Mean of those two numbers.
the A.M. will equal the G.M. when the two positive values are the SAME. Since we are told that X < Y, it must be the case that the A.M. exceeds the G.M. for these two positive values
We are given that: x < y < z
A.M. of (x ; y) > G.M. of (x ; y)
and since the hypotenuse is the longest side for any right triangle, the Legs of the right triangle must be: x and y
The Area = (1/2) (x) (y) = 1
Or
xy = 2
A.M. > G.M.
(x + y) / 2 > sqrt(x * y)
x + y > 2 * sqrt(xy)
We know that xy = 2 from the given information about the area of the right triangle.
x + y > 2 * sqrt(2)
And we can use the area formula again to substitute for the variable — x
x*y = 2
So: x = (2/y)
(2/y) + y > 2 * sqrt(2)
(2 + y^2) / y > 2 * sqrt(2)
—can multiply both side by positive Y and we get the quadratic:
(y)^2 — 2 * sqrt(y) + 2 > 0
—-factoring the quadratic we get:
(y - sqrt(2))^2 > 0
or
y - sqrt(2) > 0
y > sqrt(2)
*A*
Posted from my mobile device