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The area of an isoceles right triangle is...

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The area of an isoceles right triangle is...

by BTGmoderatorLU » Mon Nov 20, 2017 6:40 am
The area of an isoceles right triangle is $$18x^2+6x+\frac{1}{2}$$ Find the perimeter of the triangle.

$$\left(A\right)\ \ 36x^2+12x+2$$
$$\left(B\right)\ \ 9x^2+3x+\frac{1}{4}$$
$$\left(C\right)\ \left(2+2^{\frac{1}{2}}\right)\left(6x+1\right)$$
$$\left(D\right)\ \left(1+2^{\frac{1}{2}}\right)\left(6x+1\right)$$
$$\left(E\right)18x+3$$

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
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Answer

by EconomistGMATTutor » Tue Nov 21, 2017 9:04 am
Hello LUANDATO.

Let's take a look at you question.

We have an isoceles right triangle with area: $$A=\ 18x^2+6x+\frac{1}{2}=\frac{1}{2}\cdot\left(36x^2+12x+1\right)=\frac{1}{2}\cdot\left(6x+1\right)\cdot\left(6x+1\right).$$ It implies that the length of two cathetus are: 6x+1.

So, the lenght of the third side (hypotenuse) is: $$hip=\sqrt{\left(6x+1\right)^2+\left(6x+1\right)^2}=\sqrt{2\left(6x+1\right)^2}=\left(6x+1\right)\sqrt{2}=\left(6x+1\right)\cdot2^{\frac{1}{2}}.$$

Finally, the perimeter of the triangle is $$\left(6x+1\right)+\left(6x+1\right)+\left(6x+1\right)\cdot2^{\frac{1}{2}}=2\left(6x+1\right)+2^{\frac{1}{2}}\left(6x+1\right)=\left(2+2^{\frac{1}{2}}\right)\left(6x+1\right).$$

So, the correct answer is C.

I hope this explanation may help you.

I'm available if you'd like a follow up.

Regards.
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