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The triangles in the figure above are equilateral and the

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The triangles in the figure above are equilateral and the [#permalink]

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05 Apr 2008, 22:14
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The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K
[Reveal] Spoiler: OA

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Re: 1000PS: Section 7 Question 16 [#permalink]

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05 Apr 2008, 22:44
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jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

Area of shaded = Area of Large - Area of Small
Give: Area of Large = K
We need to find Area of Small in term of K.
Triangle area = 1/2 * base * height
and since a side of the large triangle is twice larger than a side of small triangle, we are dealing with (1/2) * (1/2) = 1/4 factor.
Therefore, Area of shaded = K - K/4 = 3K/4

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Re: 1000PS: Section 7 Question 16 [#permalink]

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05 Apr 2008, 22:48
jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

A.

the ratio tells you that the larger also is a equilateral triangle.
Length of small one is a
length of larger one is b
a =b/2

the larger arear = K= b^2*(square root(3)/4) I called square root(3)/4 S
the smaller area = a^2 *S = S*b^2 /4 = K/4

the shaded area = K -K/4 =3/4 K
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Re: 1000PS: Section 7 Question 16 [#permalink]

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06 Apr 2008, 08:56
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an easy way to solve this would be to visualise the below triangle instead. it shows u the answer straight away without the need for any calculations
Attachments

trig.JPG [ 3.86 KiB | Viewed 10341 times ]

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Re: 1000PS: Section 7 Question 16 [#permalink]

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03 Dec 2010, 13:06
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Another way to do this!
Theorem : If there are Two similar triangles with sides in Ratio : S1 : S2 - then their areas are in the ratio S1^2 : S2 ^2
=> Area of Larger : Area of Smaller = S1^2 : S2^2
=> K : As = 2^2 : 1
=> As = k/4

Therefore, Area of the shaded region : K-K/4 = 3k/4

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Re: 1000PS: Section 7 Question 16 [#permalink]

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04 Dec 2010, 04:07
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jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

Yes, the property given above is very useful. It states: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{S^2}{s^2}$$.

As both big and inscribed triangles are equilateral then they are similar, so $$\frac{AREA}{area}=\frac{S^2}{s^2}=\frac{2^2}{1^2}=4$$, so if $$AREA=K$$ then $$area=\frac{K}{4}$$ --> the area of the shaded region equals to $$area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}$$.

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Re: 1000PS: Section 7 Question 16 [#permalink]

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06 Dec 2010, 21:28
very useful ratio thanks guys.

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Re: The triangles in the figure above are equilateral and the [#permalink]

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18 Oct 2012, 11:08
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m not geting this onne..how K/4??

large triangle have area 4...how k/4?
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Re: The triangles in the figure above are equilateral and the [#permalink]

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18 Oct 2012, 11:16
sanjoo wrote:
m not geting this onne..how K/4??

large triangle have area 4...how k/4?

Check here: the-triangles-in-the-figure-above-are-equilateral-and-the-62201.html#p827422

We have that $$\frac{AREA}{area}=4$$. Now, since $$AREA=K$$ then $$\frac{K}{area}=4$$ --> $$area=\frac{K}{4}$$ --> the area of the shaded region equals to $$area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}$$.

Hope it's clear.
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Re: The triangles in the figure above are equilateral and the [#permalink]

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09 Jun 2013, 22:14
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jimmylow wrote:
Attachment:
2triangles.GIF
The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

The easiest way to solve this one would be by picking numbers. Lets say each side of the larger triangle is 6 and given the ratio 2:1 each side of smaller triangle then is 3. Area of an equilateral triangle can be calculated using formula: $$s^2(\sqrt{3})/4$$ where s = side. So area of large triangle = k = $$9(\sqrt{3})$$ and area of small triangle = $$9(\sqrt{3})/4$$ = $$k/4$$. So the area of the smaller triangle is 1/4 the area of the large triangle. Area of shaded region=$$k-k/4$$ = $$3/4k$$
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Re: The triangles in the figure above are equilateral and the [#permalink]

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06 Mar 2014, 02:32
Bumping for review and further discussion.

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Re: The triangles in the figure above are equilateral and the [#permalink]

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20 Jul 2016, 11:37
Bunuel wrote:
Bumping for review and further discussion.

Let side of larger eq. triangle=2X
area=sq.root3/4*(2X)^2------->sq.root3*x^2------(A)
therefore side of smaller eq. triangle=X(given ratio=2/1)
area of smaller- sq.root3*x^2/4
=sq.root3*x^2-sq.root3*x^2/4------------>sq.root3*x^2(1-1/4)
substituting from eq. (A)
Ans A

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Re: The triangles in the figure above are equilateral and the [#permalink]

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03 Oct 2017, 10:01
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Re: The triangles in the figure above are equilateral and the   [#permalink] 03 Oct 2017, 10:01
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