Question

answer this

Answer 1

find the shaded region

Answer 2

I saw this problem in The Art and Craft of Problem Solving. The shaded area is apparently 1/7 of the area of the whole triangle.

Answer 3

solve it

Answer 4

The area of an equilateral triangle with side \(s\) is \(A=\frac{s^2 \sqrt{3}}{4}\).
Since one side is 3, the area of the large triangle is \(A=\frac{9}{4}\sqrt{3}\). Thus, the area of the shaded region is \(\frac{9}{28} \sqrt{3}\).

Answer 5

where did you get 9/8squareroot of 3?

Answer 6

@blockcolder are you just basing it off the assumption that the shaded region is 1/7 of the actual triangle?

Answer 7

if you know the actual process for solving this problem, or have a link to the proof, i would love to see it

Answer 8

@completeidiot If you have time to load a pdf, check this link: http://download.sma1pekalongan.sch.id/umum/file/Media%20Pembelajaran/Matematika/EBOOKS/ADVANCED/Paul%20Zeitz%20(Author)%20-%20The%20Art%20and%20Craft%20of%20Problem%20Solving%20(2Ed,Wiley,2006).pdf
and go to page 286.

Answer 9

@blockcolder how did you arrive at 9/28squareroot of 3 ?

Answer 10

The area of the whole triangle is \(\frac{9}{4}\sqrt{3}\). The area of the shaded region is 1/7th of that, or \(\frac{1}{7} \times \frac{9}{4}\sqrt{3}\).

Answer 11

how did you know its 1/7?

Answer 12

It's in the link I posted above. The question starts at p.286, the answer is at the end of p.287.

Answer 13

@Jhannybean something you might be interested in

Answer 14

Interesting problem and solution; I figured I would try it another way, using coordinate geometry -- my favourite!

I attached a lot of the primary work I did, although I left out the part with finding the two points on the outer sides of the triangle. I just found the slope of the line between (0,0) and (3/2, 3 sqrt(3)/2), and then created the equation y = sqrt(3) x. From there, the distance from (0,0) to (x,y) is \(\sqrt{x^2 + y^2} = \sqrt{x^2 + (\sqrt{3}x)^2} = \sqrt{4x^2} = 2x\). We make the equation 2x = 2 for the left-most point and 2x = 1 for the right one (applying the transformation x to -x+3)

Once you find the intersections of the lines y1, y2, and y3; there are other ways to find the area of the inner triangle than what I did (integration). You could find the points where the intersections are, find the distances, and then apply Heron's formula; or find two side-lengths and the angle between them using slopes converted to angles / taking the difference to find the inner angle, and then applying the area using sines: A = 1/2 ab sin C.

Moral of story, though, is that the area is indeed 1/7 of the total, and there are multiple ways to show it. :)