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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]
09 Oct 2007, 21:21
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]
11 Sep 2010, 13:27
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?
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]
11 Sep 2010, 13:49
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Orange08 wrote:
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?
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).
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is [#permalink]
13 Sep 2010, 07:33
Expert's post
prashantbacchewar wrote:
Bunuel
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]
19 Nov 2014, 01:18
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