An 8x5 rectangle. If the bottom left corner is considered (0, 0), then two lines are drawn within the rectangle, from (0, 4) to (8, 1) and from (1, 5) to (7, 0). The smaller two regions of the four these lines cut the rectangle into are shaded. What is their combined area?
this was my approach
detailed annotation for screen readers here: 4 lines have been added to the original depiction. the first set of 2 lines make a triangle shape using coordinates (4, 2.5) to (4, 0) to (7, 0). the second set of 2 lines make a triangle shape using coordinates (4, 2.5) to (8, 2.5) to (8, 1)
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( (4x2.5)- (((2.5x3)/2)+((4x1.5)/2)) )x2 = 6.5
That’s what I did too
It’s 6.5
| 0 5 1 | | 0 4 1 | | 0 4 1 | + | 4 2.5 1 | = 6.5 | 1 5 1 | | 1 5 1 |Determinants go brrr
6.5 square units
The description doesn’t match the image. The image shows two lines that intersect. The description’s lines would not intersect as stated. My answer is for the image, not the description.The first line’s coordinates should be 0,4 to 8,1. The second should be 1,5 to 7,0.I see you corrected it.
Huh. I got 7.5 square units. Now I’m rechecking my calculations frantically. Lol.
I guess there’s no rule against posting your work.
I divided it in half down the middle and made a note to double my answer at the end.
The whole half rectangle has an area of 20. The trapezoid below has an area of
(2.5+4)*4/2=13. The triangle above has an area of3*2.5/2=3.25.20-13-3.25=3.75. Multiplying that by 2 (because that 3.75 is only the area of the top left half of the shaded portion) gives 7.5.Edit: Bah! Never mind. Found my mistake.
3*2.5/2=3.75, not 3.25. With that fix I get 6.5.Posting work is encouraged 🙂
You’ve got it, it’s 6.5. I actually posted this problem because I originally found the answer using trig, which seemed a bit too brute-force-y, especially considering the original source for this problem - I wanted to see if others could/would find the simpler solution that I assumed existed. And you did ^_^
I tried it just using the area of a kite, but ended up with sqrt(2)*sqrt(22.25) and got ~6.67 I made a mistake somewhere I guess.
The issue there is it isn’t a kite: The two longer sides don’t have equal length - they’re sqrt(73)/2 and sqrt(61)/2. So it’s a decent approximation but not quite exact.
Oh jeez. Of course. Thanks.
One doesn’t really see this if the image is oriented like that. Rotating your phone or your head so that the vertical axis matches the long diagonal of the ‘kite’ makes this difference more obvious.
13/4 was my original guess, when I checked I realised I’d forgotten to multiply by 2.
I used the same method as clara (and my solution would have matched _)
(Spent a lot of time convincing myself they met in the center)
I love doing things the wrong way, so
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Also got 6.5. Calculated one section of white space as a sum of a square and two triangles. Then doubled it and subtracted from overall.
