anonymous one year ago How do you find the area of this triangle?

1. anonymous

|dw:1438116778704:dw|I need to find the area of this. How do I do that?

2. anonymous

|dw:1438116845340:dw| I forgot the right angle

3. phi

area is 1/2 base * height

4. anonymous

Really? That's it?

5. phi

yes

6. anonymous

5x3/2?

7. phi

yes

8. anonymous

9. anonymous

|dw:1438116970196:dw| What about that part? WHen they ask find the area which triangle am i finding it for exactly?

10. phi

You found the area of the triangle. we can prove it

11. ChillOut

You're finding the are for the full lines one.

12. anonymous

I found the area of which triangle?

13. ChillOut

area*

14. ChillOut

|dw:1438117079817:dw|

15. phi

|dw:1438117108451:dw|

16. anonymous

Ooh, so the |dw:1438117132170:dw| is giving me the height for the triangle?

17. ChillOut

Absolutely.

18. anonymous

Ooh, Brainfart!

19. anonymous

Since I am taking geometry during summerschool I expected it to be some complicated formula. At first I ruled out that it could be that...

20. ChillOut

There are other formulas involving angles and stuff. But leave that for later :)

21. phi

here is a proof. We assume we know the area of a rectangle is width * height |dw:1438117298679:dw|

22. anonymous

So if I answer using 1/2BxH on a graded assessment it will be correct?

23. phi

now put in a diagonal. the two triangles are congruent, so the area of each is 1/2 of the rectangle's|dw:1438117369207:dw|

24. anonymous

So if I answer using 1/2BxH on a graded assessment it will be correct?

25. anonymous

They're not expecting a different method?

26. phi

now make an obtuse triangle |dw:1438117452871:dw|

27. anonymous

@phi Ooh, I see. That makes sense...

28. anonymous

ooh. C is the height of |dw:1438117583062:dw|

29. phi

if we subtract of the small triangle on the left we are left with the obtuse triangle that is $\frac{1}{2} (a+b)\cdot c - \frac{1}{2} b c \\\frac{1}{2} ac + \frac{1}{2} bc - \frac{1}{2} bc \\ =\frac{1}{2} ac$

30. phi

hopefully you can follow the logic

31. anonymous

I can, so you just proved 1/2BxH right?

32. phi

yes, that the formula works for obtuse triangles.

33. anonymous

Ok thanks. Can I ask another quick question?

34. anonymous

How do you find the area of a kite?

35. phi

1/2 the product of the diagonals http://www.mathopenref.com/kitearea.html

36. anonymous

Thanks!!!!

37. anonymous

Would the diagonals be these? |dw:1438117923899:dw|

38. anonymous

@phi

39. phi

the "crossbars" |dw:1438118022805:dw|

40. anonymous

Ooh, ok sorry but: |dw:1438118151545:dw| It would be 1+1 x 2 + 3/2 right? @phi

41. anonymous

OOps I mean 1+2 x 3+3/2

42. phi

it would be except that would not be a kite (the diagonals of a kite bisect each other) see http://www.coolmath.com/reference/kites

43. phi

correction: the longer diagonal bisects the shorter diagonal

44. anonymous

Ooh, so what i said was right? I have a problem asking for the area where they give me the length of half of both lines.

45. anonymous

Like they give me this...

46. anonymous

|dw:1438118486114:dw|

47. anonymous

I just add the 2's and 1's and use the 1/2bxh formula right?

48. phi

yes

49. anonymous

Ooh ok. Thanks.

50. phi

1/2 d1 * d2 (though you can think of the diagonals as the width and height)

51. anonymous

Ooh, thank you so much!!!!!!!!!!!!!!