bonanova 85 Posted July 15, 2014 Report Share Posted July 15, 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? Quote Link to post Share on other sites

0 Solution witzar 18 Posted July 15, 2014 Solution Report Share Posted July 15, 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. Quote Link to post Share on other sites

0 nana77 17 Posted July 15, 2014 Report Share Posted July 15, 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. Quote Link to post Share on other sites

0 bonanova 85 Posted July 15, 2014 Author Report Share Posted July 15, 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. Quote Link to post Share on other sites

0 nana77 17 Posted July 15, 2014 Report Share Posted July 15, 2014 Did you only want the area of overlap? Quote Link to post Share on other sites

0 bonanova 85 Posted July 15, 2014 Author Report Share Posted July 15, 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. Quote Link to post Share on other sites

0 nana77 17 Posted July 15, 2014 Report Share Posted July 15, 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. Quote Link to post Share on other sites

0 DeGe 9 Posted July 15, 2014 Report Share Posted July 15, 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 Quote Link to post Share on other sites

0 k-man 26 Posted July 15, 2014 Report Share Posted July 15, 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 Quote Link to post Share on other sites

0 k-man 26 Posted July 15, 2014 Report Share Posted July 15, 2014 witzar beat me to it Quote Link to post Share on other sites

0 bonanova 85 Posted July 15, 2014 Author Report Share Posted July 15, 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. Quote Link to post Share on other sites

## Question

## bonanova 85

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?

## Link to post

## Share on other sites

## 10 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.