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The triangles in the figure above are equilateral and the [#permalink]
05 Apr 2008, 21:14

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Attachment:

2triangles.GIF [ 1.58 KiB | Viewed 7210 times ]

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?

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 would be my answer. 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

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

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

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 \(area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}\).

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 \(area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}\).

Re: The triangles in the figure above are equilateral and the [#permalink]
09 Jun 2013, 21:14

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This post received KUDOS

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=\(k-k/4\) = \(3/4k\) _________________

___________________________________________ Consider +1 Kudos if my post helped

Re: The triangles in the figure above are equilateral and the [#permalink]
01 Oct 2015, 22:41

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