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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]

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09 Oct 2007, 22:21

ggarr wrote:

Quote:

area of middle part (quad) = B * h (6 - 2√3) * 2 = 12 - 4√3

aren't we solving what's inside the parentheses first? 6 - 2√3 = 4√3 4√3 * 2 = 8√3 should we be using the distributive property here?

6 - 2√3 does not = to 4√3!!
6√3 - 2√3 = 4√3
use a calculator and you will see the difference
and i did use distributive property
(6 - 2√3) * 2 =
6*2 - 2√3*2 =
12 - 4√3

Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]

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11 Sep 2010, 14:27

beckee529 wrote:

this is how i got 12 by finding the 3 subparts (by drawing two lines to create two triangles and quadrilateral):

30:60:90 x: √3: 2x 2: 2√3: 4

Is it mandatory to mug up this rule regarding 30:60:90 triangle? No other ways of solving this problem? Isn't it correct that opposite sides of angles 30 and 60 are in ratio 1:2?

this is how i got 12 by finding the 3 subparts (by drawing two lines to create two triangles and quadrilateral):

30:60:90 x: √3: 2x 2: 2√3: 4

Is it mandatory to mug up this rule regarding 30:60:90 triangle? No other ways of solving this problem? Isn't it correct that opposite sides of angles 30 and 60 are in ratio 1:2?

Any help is appreciated.

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

BACK TO THE ORIGINAL QUESTION:

Now, as hypotenuse PQ (the largest side) equals to 4 then the side opposite 30 degrees (smallest side, which is also the height of the parallelogram) equals to 4/2=2. Thus area of parallelogram is height*base=2*6=12.

Answer: B.

For more on this issues check Triangles chapter of Math Book (link in my signature).

Are we supposed to remeber the corelations for the standard triangles

Yes, I think it's good to know below 2 cases:

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles. • Area of a 45-45-90 triangle. As you see from the figure above, two 45-45-90 triangles together make a square, so the area of one of them is half the area of the square. As a formula \(A=\frac{S^2}{2}\). Where S is the length of either short side.

For more on this issues check Triangles chapter of Math Book (link in my signature).

Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]

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19 Nov 2014, 02:18

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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]

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21 Apr 2016, 22:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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