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A. 20 B. \(10\sqrt{3}\) C. \(20\sqrt{3}/3\) D. 10 E. \(10\sqrt{3}/3\)

This question CAN be solved without trigonometry. In fact trigonometry is NOT tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

Look at the figure below:

Attachment:

Triangle.png [ 18.53 KiB | Viewed 10529 times ]

Notice that triangle ABE is 30°-60°-90° right triangle (E=90° and A=30°), thus its sides are in the ratio \(1 : \sqrt{3}: 2\) (\(\sqrt{3}\) corresponds with AE and 2 corresponds with AB).

Now, since AE=2x+x=3x, then \(AE:AB=\sqrt{3}: 2\) --> \(3x:10=\sqrt{3}: 2\) --> \(x=\frac{5\sqrt{3}}{3}\) --> \(AC=4x=\frac{20\sqrt{3}}{3}\).

Re: BC=BD=DC=AD. If AB= 10, what is the length of AC? [#permalink]

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18 Mar 2013, 14:38

1

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A step shorter than Bunuel's method

Triangle BCD is equilateral. Triangle ADB is isosceles (with angle ADB =120).

Therefore, ABC is a right angle triangle, with angle BAC=30; ACB=60 & ABC=90.

We know that a right angle triangle with 30-60-90 combination has sides of ratio \(1x:\sqrt{3}x:2x\), with \(\sqrt{3}x\) corresponding to AB; \(1x\) corresponding to BC; and \(2x\) corresponding to AC.

Since we know AB = 10, so AC = \(\frac{20}{\sqrt{3}}\) which can also be written as \(20\sqrt{3}/3\)

A. 20 B. \(10\sqrt{3}\) C. \(20\sqrt{3}/3\) D. 10 E. \(10\sqrt{3}/3\)

You can do it using just the Pythagorean theorem too. Notice that you have an equilateral triangle BCD. If its side is a, its altitude BE (shown by Bunuel in the diagram) will be \(\sqrt{3}a/2\). Also the base AE will be a + a/2 (Since altitude of equilateral triangle bisects the base)

Re: BC=BD=DC=AD. If AB= 10, what is the length of AC? [#permalink]

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19 Mar 2013, 09:13

From BC=BD=DC, we know angle DBC= 60 and Angle BDA=120. And AD=AB so angle ABD=30. Therefor triangle ABC is right angle trangle with angle B=90 and Angle A=30.

I understood your approach but I solved this problem with another approach which is right but I am getting the different result. Can you tell me what is wrong with this approach.

angle BDC,BCD and DBC are all 60 degrees as it is equilateral triangle. Now in the triangle BAD angle BDA will be 120 because of straight line. so the rest of the two angles will be 30 each which makes the angle ABC a rectangle. So the triangle ABC is rectangle and if you solve it with this approach you will get the answer 10multiply by root3/3

Can you please tell me whats wrong with this. Thanks

You are right that triangle ABC is right angled at B. Also, AC = 2*BC, AB = 10

Re: BC=BD=DC=AD. If AB= 10, what is the length of AC? [#permalink]

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25 Nov 2013, 07:45

emmak wrote:

Attachment:

1.jpg

Note: figure not drawn to scale

BC=BD=DC=AD. If AB= 10, what is the length of AC?

A. 20 B. \(10\sqrt{3}\) C. \(20\sqrt{3}/3\) D. 10 E. \(10\sqrt{3}/3\)

Another way; since BD=BC=DC, triangle BDC equilateral triangle hence angle BCA = 60 degrees. it mens that other angles angle ABC + angle BAC = 180-60 = 120. Which means that AB = < AC. Which means AC >=10. Let BC = X ; Therefore AC= 2X. Using pythagoreus theorem AC^2 = AB^2 + BC^2 = 100 +x 4x^2 = 100 + x^2 3x^2 = 100 x = 10[square_root]3/3 AC = 2X = 20 [square_root]3/3

Re: BC=BD=DC=AD. If AB= 10, what is the length of AC? [#permalink]

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12 Dec 2013, 13:12

Note: figure not drawn to scale

BC=BD=DC=AD. If AB= 10, what is the length of AC?

If BC=BD=DC then we know triangle DBC is an equilateral triangle. Furthermore, we know that ADC all lie on a line together which means angle ADB = 180-60 = 120. Because AD = DB we know that this triangle is isosceles and that the two other angle measures in this triangle are 30 each. Looking at both triangles together, we see that ABC is a 30:60:90 triangle. Knowing this, and one side length (the length opposite 60) we can solve for BC. Because BC = DC = AD we can find the length of AC (which is AD+DC)

The ratio of the sides in a 30:60:90 is (x/2) : (√3/2 x) : x √3/2 x = 10 x = 20/√3 In a 30:60:90, the hypotenuse is twice the length of the shortest side. The shortest side is equal to x/2 or (20/√3)/2. Because the hypotenuse is twice that length, it is simply equal to 20/√3

Finally, to cancel out the root on the bottom multiply by (√3/√3) = 20√3/3

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BC=BD=DC=AD. If AB= 10, what is the length of AC? [#permalink]

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22 Apr 2016, 03:01

I would prefer nave method without a need to dissect the diagram. But you gain important insights about equilateral triangles from Karishma's solution.

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BC=BD=DC=AD. If AB= 10, what is the length of AC?
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