Triangle area

Question

T6016.png If the perimeter of ΔACD is 3+3\sqrt{3} , what is the perimeter of equilateral triangle ΔABC? A. 6 B. 9 C. 6\sqrt{3} D. 9+3\sqrt{3} E. 6+6\sqrt{3}

lawschoolsearcher · Answer

ΔACD  = 3 + 3\sqrt{3}

The perimeter of a 30-60-90 triangle is 
 x + 2x + x\sqrt{3} = 3x + \sqrt{3}

We know that 
 3x > x\sqrt{3}  using the THIRD SIDE RULE
Since  3x + x\sqrt{3} = 3 + 3\sqrt{3} 
and  3 < 3\sqrt{3}  so
 x\sqrt{3}=3 
 x=3/\sqrt{3} 
 x=3/\sqrt{3} 
 x=\sqrt{3}

ΔABC  = 3 x (2x) = 6x = 6\sqrt{3}

So C.

Spidy001 · Answer

ACD 30-60-90 triangle

then we have perimeter = x +2x+(\sqrt{3})x = 3+ 3*\sqrt{3}

=> x = \sqrt{3}

permiter of triangle ABC = 6x = 6*\sqrt{3}

Answer is C.

subhashghosh · Answer

x - Length of one side

x + x/2 + root(3)x/2 = 3 + 3root(3)

3x/2 + root(3)x/2 = 3 + 3root(3)

x/2(3 + root(3)) = 3(1 + root(3))

x/2 root(3)( 1 + root(3)) = 3(1 + root(3))

x = 2root(3)

3x = 6root(3)

Answer - C

fluke · Answer

Let the side of the triangle be "S".

AC=S 
Height of an equilateral triangle:  AD=\frac{\sqrt{3}}{2}S 
 CD=\frac{S}{2}

Perimeter of the triangle  	riangle{ACD}=AC+AD+CD=S+\frac{\sqrt{3}}{2}S+\frac{S}{2}

S+\frac{\sqrt{3}}{2}S+\frac{S}{2}=3+3\sqrt{3}

S(1+\frac{\sqrt{3}}{2}+\frac{1}{2})=3+3\sqrt{3}

S(\frac{3+\sqrt{3}}{2})=3+3\sqrt{3}

\frac{S}{2}=\frac{3+3\sqrt{3}}{3+\sqrt{3}}

S=2\sqrt{3}

Perimeter of  	riangle{ABC}=3S=6\sqrt{3}

Ans: "C"

Triangle area

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards