
It is currently Tue Apr 23, 2019 7:12 am

View unanswered posts  View active topics
Author 
Message 
questioner

Post subject: GMAT Geometry Posted: Wed Aug 04, 2010 9:38 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

The figure above shows a square inscribed within an equilateral triangle where the top of the square is parallel to the bottom of the triangle. If the side x of the square measures 12 inches, then what is the perimeter of the equilateral triangle (in inches)? A. 36 B. 36√3 C. 24√3 + 24 D. 24√3 + 36 E. 72
Use this link for a detailed flash graphical explanation:
(D) Triangle questions are common on the GMAT, but they are easy if you can break the question down into simple steps and understand the rules associated with the common triangle types. This question breaks down into a few steps:
1) Since the triangle is equilateral, all its sides are the same length. The perimeter will be 3 times the length of one of its sides.
2) The small triangle on top of the square is also equilateral because it is similar to the larger triangle.
3) The sides of the small equilateral triangle must all be 12.
4) Looking at the small triangle to the left of the square, we know that the lower left angle is 60 degrees because it shares the angle with the larger equilateral triangle. Because another angle is 90 degrees, this triangle must be a 306090 triangle.
5) The lengths of the sides of a 306090 triangle are always in the ratio of 1 : √3 : 2. If we divide the height of the triangle by √3, we will have the length of the short side. Multiplying the value of the short side by 2 will give us the length of the hypotenuse: Hypotenuse = (12 / √3) × 2= 24/√3 = (24√3)/3= 8√3.
6) The length of one side of the large triangle is 8√3 + 12.
7) The perimeter is 3 times this amount: 3 × (8√3 + 12) = 24√3 + 36 inches.
The correct answer is choice (D). 
Hi there,
Why is the Hypotenuse 24/√3 divided by 3? Wouldn't the length of the side simply be 24/√3 + 12?
Thanks.


Top 


Gennadiy

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Wed Aug 04, 2010 9:53 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

When we found that hypotenuse of a small triangle (shaded on the picture above) is 24/√3, we use a common trick to get rid of irrational number in the denominator:
24/√3 = (24 × √3) / (√3 × √3) = (24√3) / 3 = 8√3
So the the side of the largest triangle, (8√3 + 12), is the same value as you have calculated: (24/√3 + 12). The only difference is that we got rid of irrationality in denominator.


Top 


questioner

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Wed Dec 01, 2010 2:44 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Once you divide 12 by √3 what is the lenght you get and what is the hypotenuse?


Top 


Gennadiy

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Wed Dec 01, 2010 2:56 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

12 / √3 = (12 × √3) / (√3 × √3) = 12√3 / 3 = 4√3
4√3 is the length of the leg of the shaded triangle. The hypotenuse of the shaded right triangle is twice that length (the property of any 30⁰  60⁰  90⁰ tirangle), 8√3.


Top 


questioner

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Mon Dec 20, 2010 6:28 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

I disagree with your answer. Since H = 24/√3, the answer should therefore be 74/√3 + 36.


Top 


Gennadiy

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Mon Dec 20, 2010 6:38 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

You probably mean, the answer should be 72/√3 + 36. But it is the same as the answer choice D. 24√3 + 36.
72/√3 + 36 = (72 × √3) / (√3 × √3) + 36 = (72√3) / 3 + 36 = (72/3) × √3 + 36 = 24√3 + 36
We should rationalize fractions with square roots in denominators. It makes it easier to work with those.


Top 


questioner

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Fri Mar 11, 2011 10:07 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Is there an easier way to explain the perimeter for this problem?


Top 


Gennadiy

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Fri Mar 11, 2011 10:51 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

The given explanation is quite simple and the rest of simplest variations would be very similar to the given one (with minor changes).
However, let me try to make it even simpler for you:
1. We, basically deal with three facts:  an equilateral triangle (Its sides are all equal and angles equal 60⁰ each)  a right 90⁰60⁰30⁰ triangle (Its side, which is opposite to 30⁰ angle, equals 1/2 of hypotenuse. Its side, which is opposite to 60⁰ angle equals √3/2 of the hypotenuse.)  a square (its sides are equal, angles are 90⁰ each)
2. The solution:  All angles of the equilateral triangle are 60⁰. And all angles of the square are 90⁰.
 Therefore the both triangles at the base of the largest one are 90⁰60⁰30⁰. The side, which is opposite to the 60⁰ angle, is √3/2 of the hypotenuse. Therefore the hypotenuse is (2/√3) × 12 = 8√3. The side, which is opposite to the 30⁰, is 1/2 of the hypotenuse. It equals 8√3/2 = 4√3.
 Therefore the bottom side of the equilateral triangle is 4√3 + 12 + 4√3 = 12 + 8√3
 The perimeter of the equilateral triangle is 3 × (12 + 8√3) = 36 + 24√3, since its sides are equal.
That's it. 
How does this solution differ from the given one? The only difference is NOT dealing with the smaller equilateral triangle (which is formed by the square and the largest triangle). The hardest moment in this solution is the same: we have to deal with 906030 triangle in this question. 
Is it possible to solve this question without knowing the properties of 90⁰60⁰30⁰ triangle? Yes, however it takes longer. Therefore it's better for you to memorize the properties of 90⁰60⁰30⁰ and 90⁰45⁰45⁰ triangles.
ALTERNATIVE SOLUTION: 1. The triangle made by the top side of the square and sides of the largest triangle is similar to the largest one (because one angle is the same and the bases are parallel, so the angles at the bases are equal). Therefore the smaller triangle at the top is equilateral as well. All it's sides are 12 inches.
2. Two right triangles at the base are equal because one side of each equals 12 inches and the angles are equal: 90⁰, 60⁰, 30⁰.
3. Let's denote the side of the right triangle at the base of the largest one by x. Then the base side of the largest triangle is (x + 12 + x).
4. All sides of the largest equilateral triangle are equal. So they all are 12 + 2x. Therefore the hypotenuse of the right triangle is 12 + 2x – 12 = 2x.
5. If we apply the Pythagorean Theorem to the right triangle, we get the following equation: (2x)² = 12² + x² x² = 12²/3 x = 12/√3 = 4√3
6. One side of the largest equilateral triangle is 12 + 2x = 12 + 8√3. The perimeter is (12 + 8√3) × 3 = 36 + 24√3.


Top 


questioner

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Wed May 18, 2011 5:19 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Hi. I am having trouble with this question. I was just wondering why 24/√3 is again divided by 3 to equal 8/√3. Please help.


Top 


Gennadiy

Post subject: Re: math (test 1, question 1): geometry, triangles Posted: Wed May 18, 2011 5:24 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498


Top 


Who is online 
Users browsing this forum: No registered users and 2 guests 

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum

