## anonymous one year ago Help

1. anonymous

ABCD is a parallelogram Which is greater a. The area of ABCD b. 24 |dw:1437774703693:dw|

2. freckles

do you know how to find the area of parallelogram ? |dw:1437775012406:dw|

3. freckles

where the area is b*h

4. freckles

we are actually given b is 4

5. freckles

can you find h?

6. anonymous

how do i find h?

7. freckles

you could use a trig ratio you know an angle and you also know the hyp of the triangle I drew inside the parallelogram

8. freckles

the opp of that angle is the thing you want to find

9. freckles

opp and hyp think sin( )

10. anonymous

wait.... I cant use sin we don't use that it has to be an alternative method...

11. freckles

ok what is the alternative method?

12. anonymous

im not sure it has to be done without sin tho...

13. freckles

oh then I don't know of another way to find the area

14. anonymous

crap... it's talking about the area of the parallelogram is equal to the length of base AD which is 4 times the height which is less than 6

15. freckles

ok

16. anonymous

Since the product of 4 and a number less than 6 must be less than 24, the area of the parallelogram must be less than 24.

17. freckles

so we don't need to find the area exactly

18. freckles

we can approximate it

19. anonymous

what does this all mean..

20. freckles

|dw:1437775540138:dw| anyways .... by Pythagorean theorem we have $h=\sqrt{36-a^2} \\ \text{ so area of parallelogram is } 4 \cdot h=4 \cdot \sqrt{36-a^2} \\ \text{ and we know that } 0<a<6 \\ \text{ so } \\ 4 \cdot h=4 \sqrt{36-a^2} <4 \sqrt{36-0^2}=4 \sqrt{36}=...$ cam you finish simplifying this

21. anonymous

24

22. freckles

right 4*h<24

23. freckles

where 4*h is the area of the parallelogram

24. freckles

didn't label what I called a by the way |dw:1437775996191:dw| I used Pythagorean theorem

25. freckles

$6^2=h^2+a^2 \\ 6^2-a^2=h^2 \\ \sqrt{6^2-a^2}=h$

26. anonymous

hmmmmm i think i get it now..

27. freckles

and I wanted the thing inside the square root to be positive and I also want a to be positive because a represents a measurement

28. freckles

$6^2-a^2>0 \text{ and } a>0 \\ 6^2>a^2 \text{ and } a>0 \\ 36>a^2 \text{ and } a>0 \\ \text{ solving } a^2<36 \text{ means we have } -6< a<6 \\ \text{ but remember we also need} a>0 \\ \text{ so we need } 0<a<6$

29. anonymous

ok

30. freckles

and if we multiply both sides by -1 we have $-6<-a<0 \\ \text{ so back to the inequality } \\ 4 h=4 \sqrt{36-a^2}<4 \sqrt{36-0^2}$

31. freckles

but yeah I would have used my trigonometry knowledge :p

32. anonymous

this is complicated I'm gonna sit and stare at this for a while.

33. freckles

$4h=4 \cdot 6 \sin(55)=24\sin(55) \approx 19.66$

34. freckles

all you have to know is that since h is between 0 and 6 (not including 6) we have 4h<4(6)

35. anonymous

hmmm... ok got it

36. freckles

how about we say something in general... |dw:1437776683255:dw| so the simplest way I can write this for your problem is: |dw:1437776770728:dw|

37. freckles

and we know a>h because in the right triangle we drew we have a is the hyp or in your example 6 is the hyp

38. freckles

anyways I think I'm done now :) unless you have questions

39. anonymous

ohhhh okay i think i understand better now

40. anonymous

needed that visual thank you ^_^

41. freckles

visuals are the secret to math

42. anonymous

agreed!!

43. freckles

I think it helps to know the hyp is the longest side in a right triangle on this question

44. anonymous

yea definitely helps

45. freckles

but in all fairness to you I think I started with a complicated reason but the last reason I gave I think is definitely less complicated

46. anonymous

yea it helped towards the end thanks again =)

47. freckles

np