Need help with my geometry

   / Need help with my geometry #31  
Math my way. :)

find_x_300_2285.jpg


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Bruce
 
   / Need help with my geometry #33  
There is enough info. Look at it as 2 right triangles plus a rectangle in the middle. Find height by using Pythagoras's theorem. At 81 I don't recall the math.

Yet another way to solve it. Now it is all Pathagorean.View attachment 465621

Correct. And since both right triangles are identical to each other, you could combine them to get a rectangle. That rectangle can be combined to the rectangle in the middle giving you just one larger rectangle. The dimensions of that rectangle are 20 ft x 21.81742 ft. The area is 436.348 sq ft.
 
   / Need help with my geometry #34  
Yep...we already told them that, but did they believe us? NO! :thumbsup:
 
   / Need help with my geometry #35  
It should be 24x24 - 10x10

Yep. Didn't pass the funny look test. Leg cannot be longer than hypotenuse. Leg (height of the trapezoid) is 21.8.

So 21.8 x 10 (2 triangles) + 10 x 10 (rectangle in between) = 318.2 sq.ft.

Somehow thought about this in doctor's office this morning.

Ralph
 
   / Need help with my geometry #36  
Sorry I did not get in on this in time to work this one. I love working these. a^2 +b^2=c^2 or in words, the square of the hypotenuse equals the sum of the squares of the other two sides. Must be a right triangle of course. Once a nerd, always a nerd I guess.
 
   / Need help with my geometry #37  
Sorry I did not get in on this in time to work this one. I love working these. a^2 +b^2=c^2 or in words, the square of the hypotenuse equals the sum of the squares of the other two sides. Must be a right triangle of course. Once a nerd, always a nerd I guess.

I hear ya! I felt the same way, still had to do the math, found the answer had been given, and was a little sad I didn't get to help. :)
 
   / Need help with my geometry #38  
Using common core I say its 615.1. You just can't argue with common core. I can't be wrong, It'll hurt my feelings.
 
   / Need help with my geometry #39  
   / Need help with my geometry #40  

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