Posted 15 Jul 2014 A triangle with vertices (6,5), (8,-3) and (9,1) is reflected about the line x = 8 to create a second, overlapping triangle with vertices at (10,5), (8,-3) and (7,1). What is their combined area? 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 spoiler alert Flipping about line x creates 2 more points for a 5 point polygon (since we are only interested in total area and not the overlapping area). The (8,-3) point remains the same after the flip. The two new points are (7,1) and (10,5). We can simplify this 5 point polygon into a normal triangle by realizing that the points (7,1) and (9,1) lie exactly along the lines created by the points along y=5 and y=-3. Now just solve for that area. Simplify further by splitting that triangle into 2 right angle triangles with a new point at (8,5) a=1/2 bh, and since there are two, the total area is base times height. b=4, h=8, area = 32. 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 That is the area of a rectangle that contains all the vertices: x in [6, 10] and y in [-3, 5]. The area we need is less than that. 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 Did you only want the area of overlap? 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 No. It should be included, but only once. The two triangles (the original and its mirror image) partially overlap, making a four-sided figure with vertices at (6,5) (8,-3) (10,5) and a point that is the intersection of lines segments from (6,5) to (9,1) and from (10,5) to (7,1). We want the area of that four-sided figure. 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 Oh. I had a scatter plot to see it and was giving the area of everything in the plot. But there is a piece to subtract out of that. 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 Lets say the area of the triangle is A The area of reflected triangle in also A Lets say area of overlap is O The area we are looking for is 2A - O Area O is a kite with diagonals 2 and 5 (haven't drawn it but only pictured it... so not sure if this is correct) Therefore area (d1d2/2) of O = 5 The area we want is 2A - 5 Now for the aha moment: 2A - 5 = A + 1/4A = 5/4A 3A/4 = 5 A = 20/3 So the area we want (2A - O) = 25/3 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 To get from (7,1) to (10,5) you need to take 3 steps right and 4 steps up, so a straight line through these points croses line x=8 4/3 units above (7,1) (since it's 1 step right). Therefore the coordinates of 4th vertex of the quadrilateral are (8, 1+4/3). Given this caclulation of area of the quadrilateral is elementary: 4*(8-(2+2/3))/2 = 2*(5+1/3) = 10+2/3. 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 32/3 The interesting point of intersection is on x=8 line. The y coordinate of that point is 7/3. Due to the symmetry along the x=8 line we can only consider the are of the left side and multiply it by 2. The area of that triangle is half of the area of a rectangle 16/3 x 2. So the area we're looking for is the area of that rectangle and is 32/3 0 Share this post Link to post Share on other sites

0 Posted 15 Jul 2014 NaNa claimed enough real estate to include everything. DeGe's kite was slightly under-sized. k-man had it, but was still typing when witzar hit the Post button witzar by a nose! Good job everyone. 0 Share this post Link to post Share on other sites

Posted

A triangle with vertices (6,5), (8,-3) and (9,1) is reflected about the line

= 8xto create a second, overlapping triangle with vertices at (10,5), (8,-3) and (7,1).

What is their combined area?

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