ps test7 #16

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ps test7 #16

by dunkin77 » Fri Apr 06, 2007 1:14 pm
Hi,

The answer to the attached question is A) but I can't seem to figure why the answer is A). Can anyone help?
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by vk.neni » Fri Apr 06, 2007 4:18 pm
Hi,
From the problem statement, it is clear that each of the smaller triangles are of the same area (all are equilateral triangles and the ratio of the sides etc implies this). Since three of the four triangles are shaded, the shaded area of the bigger triangle is K*3/4.

OR (I proceeded the following way a bit before realizing the above :(

you could solve it by finding the areas of the bigger triangle and the smaller ones and doing a ratio of them.

Let 's' be the side of the bigger triangle. Since this is an equilateral, the area is = s^2 * sqrt(3)/4.

Now for the smaller triangle. we know that the side of smaller triangle is half that of the bigger one. So, area = (s/2)^2 * sqrt(3)/4.


Now substitute K for (s^2) * sqrt(3)/4 in the smaller triangle and subtract this from K to get the answer of 3K/4.


-Neni

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by dunkin77 » Fri Apr 06, 2007 4:52 pm
Thank you very much!!! - I finally figured.:)

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by Cybermusings » Fri Apr 06, 2007 11:44 pm
In the figure all 5 triangles are equilateral triangles.

The ratio of the side of the bigger triangle to that of the smaller triangles is 2:1. Which means if one side of the bigger triangle is 12 the side of each of the smaller ones is 6.

For any equilateral triangle the area = (x^2 * sqr. rt 3)/2

For a equilateral triangle with side x/2 the formula becomes {[(x/2)^2]*sqr rt. 3}/2

Now there are 3 such triangles. Hence multiply the above equation by 3

Thus the value = {[x/2]^2]*sqr rt. 3}*3/2

= 3K/4

Hence Choice A