Carcass wrote:
Attachment:
#greprepclub If the perimeter of the isosceles right triangle shown is.jpg
If the perimeter of the isosceles right triangle shown is \(1+\sqrt{2}\), what is the area of the triangular region ?
A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. \(1\)
D. \(\frac{\sqrt{2}}{4}
\)
E. \(\frac{1+\sqrt{2}}{4}\)
Kudos for the right answer and explanation
The sides of an isosceles right triangle (aka a 45-45-90 special right triangle) are in the ratio \(1 : 1 : \sqrt{x}\)
So let's assign the following lengths to the right triangle
Since the perimeter is \(1+\sqrt{2}\), we can write: \(x + x + x\sqrt{2} = 1+\sqrt{2}\)
Simplify the left side: \(2x + x\sqrt{2} = 1+\sqrt{2}\)
Factor the left side: \((2 + \sqrt{2})x = 1+\sqrt{2}\)
Solve for: \(x =\frac{1+\sqrt{2}}{2 + \sqrt{2}}\)
Now that we know the value of x, we can find the area of the triangle.
Area of triangle \(= \frac{1}{2}(base)(height)\)
\(= \frac{1}{2}(\frac{1+\sqrt{2}}{2 + \sqrt{2}})(\frac{1+\sqrt{2}}{2 + \sqrt{2}})\)
\(= \frac{1}{2}(\frac{1+2\sqrt{2} + 2}{4 + 4\sqrt{2} + 2})\)
\(= \frac{1}{2}(\frac{3+2\sqrt{2}}{6 + 4\sqrt{2}})\)
\(= \frac{1}{2}(\frac{3+2\sqrt{2}}{2(3+2\sqrt{2})})\)
\(= \frac{1}{2}(\frac{1}{2})\)
\(=\frac{1}{4}\)
Answer: A