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If the perimeter of the isosceles right triangle shown is [#permalink]
1
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
Image

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
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Re: If the perimeter of the isosceles right triangle shown is [#permalink]
1
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


We know, sides of an Isosceles Right Triangle are: \(x\), \(x\), \(\sqrt{2}x\)

So, \(x + x + \sqrt{2}x = 1 + \sqrt{2}\)
\(2x + \sqrt{2}x = 1 + \sqrt{2}\)
\(\sqrt{2}x (1 + \sqrt{2}) = 1 + \sqrt{2}\)
\(\sqrt{2}x = 1\)
\(x = \frac{1}{\sqrt{2}}\)

Area = \(\frac{1}{2}x^2 = \frac{1}{2}(\frac{1}{\sqrt{2}})^2 = \frac{1}{4}\)

Hence, option A
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Re: If the perimeter of the isosceles right triangle shown is [#permalink]
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