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605-655 (Medium)|   Geometry|      
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metallicafan
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 10 Jul 2012, 08:57
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70% (02:28) correct 30% (02:10) wrong based on 146 sessions

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The perimeter of the triangle below is \(6\sqrt{3} + 6\). What is the length of the hypotenuse of the triangle?

(A) \(2\)
(B) \(4\)
(C) \(2\sqrt{3}\)
(D) \(3\sqrt{3}\)
(E) \(4\sqrt{3}\)

It seems easy, but it is not!, Is there a faster way to solve it than using the conjugate of the denominator (in this case, \((3 - \sqrt{3})\))?

Source: https://www.gmathacks.com
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E
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novemberain
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 10 Jul 2012, 09:14
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Sin 60 = KL/LM
(Sqrt 3)/2 * LM = KL ………………(1)

Sin 30 = KM /LM
½ * LM = KM………………………(2)
KL + KM + LM = 6*(Sqrt 3) + 6………..(3)

Solve (1),(2) and (3) to get LM
LM = 4* (sqrt3)
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 10 Jul 2012, 09:36
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metallicafan
The perimeter of the triangle below is \(6\sqrt{3} + 6\). What is the length of the hypotenuse of the triangle?
(A) \(2\)
(B) \(4\)
(C) \(2\sqrt{3}\)
(D) \(3\sqrt{3}\)
(E) \(4\sqrt{3}\)

It seems easy, but it is not!, Is there a faster way to solve it than using the conjugate of the denominator (in this case, \((3 - \sqrt{3})\))?

Source: https://www.gmathacks.com
Show more

Notice that triangle KLM is 30°-60°-90° right triangle. Now, in 30°-60°-90° right triangle the sides are always in the ratio \(1:\sqrt{3}:2\).

So, we have that \(x+\sqrt{3}x+2x=6\sqrt{3} + 6\) --> \(x=\frac{6\sqrt{3} + 6}{3+\sqrt{3}}=\frac{6(\sqrt{3} + 1)}{3+\sqrt{3}}\) --> multiply denominator and numerator by \(3-\sqrt{3}\): \(x=\frac{6(\sqrt{3} + 1)(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6(\sqrt{3} + 1)(3-\sqrt{3})}{9-3}=(\sqrt{3} + 1)(3-\sqrt{3})=2\sqrt{3}\).

Hypotenuse is \(2x=4\sqrt{3}\).

Answer: E.

Hope it's clear.
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 10 Jul 2012, 09:54
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Bunuel
metallicafan
The perimeter of the triangle below is \(6\sqrt{3} + 6\). What is the length of the hypotenuse of the triangle?
(A) \(2\)
(B) \(4\)
(C) \(2\sqrt{3}\)
(D) \(3\sqrt{3}\)
(E) \(4\sqrt{3}\)

It seems easy, but it is not!, Is there a faster way to solve it than using the conjugate of the denominator (in this case, \((3 - \sqrt{3})\))?

Source: https://www.gmathacks.com

Notice that triangle KLM is 30°-60°-90° right triangle. Now, in 30°-60°-90° right triangle the sides are always in the ratio \(1:\sqrt{3}:2\).

So, we have that \(x+\sqrt{3}x+2x=6\sqrt{3} + 6\) --> \(x=\frac{6\sqrt{3} + 6}{3+\sqrt{3}}=\frac{6(\sqrt{3} + 1)}{3+\sqrt{3}}\) --> multiply denominator and numerator by \(3-\sqrt{3}\): \(x=\frac{6(\sqrt{3} + 1)(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6(\sqrt{3} + 1)(3-\sqrt{3})}{9-3}=(\sqrt{3} + 1)(3-\sqrt{3})=2\sqrt{3}\).

Hypotenuse is \(2x=4\sqrt{3}\).

Answer: E.

Hope it's clear.
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Another way to do the calculations:

\(x=\frac{6(\sqrt{3}+1)}{3+\sqrt{3}}=\frac{6(\sqrt{3}+1)}{\sqrt{3}(\sqrt{3}+1)}=\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}\)
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 10 Jul 2012, 21:33
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metallicafan
The perimeter of the triangle below is \(6\sqrt{3} + 6\). What is the length of the hypotenuse of the triangle?
(A) \(2\)
(B) \(4\)
(C) \(2\sqrt{3}\)
(D) \(3\sqrt{3}\)
(E) \(4\sqrt{3}\)

It seems easy, but it is not!, Is there a faster way to solve it than using the conjugate of the denominator (in this case, \((3 - \sqrt{3})\))?

Source: https://www.gmathacks.com
Show more
Hi,

Let assume the triangle is equilateral,
then each side would be = \(\frac {6\sqrt{3} + 6}3\) =\(2\sqrt{3} + 2\)=2*1.7+2=5.4
but the hypotenuse would be greater than this side.
so, now on checking the options:
(A) \(2\)
(B) \(4\)
(C) \(2\sqrt{3}\) (=3.4)
(D) \(3\sqrt{3}\) (=5.1)
(E) \(4\sqrt{3}\) (>5.4)

Answer (E),

Regards,
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The perimeter of the triangle below is 6*3^1/2 + 6. What is 09 May 2023, 08:58
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