anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ABCD is a parallelogram Which is greater a. The area of ABCD b. 24 |dw:1437774703693:dw|
freckles
  • freckles
do you know how to find the area of parallelogram ? |dw:1437775012406:dw|
freckles
  • freckles
where the area is b*h

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freckles
  • freckles
we are actually given b is 4
freckles
  • freckles
can you find h?
anonymous
  • anonymous
how do i find h?
freckles
  • 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
freckles
  • freckles
the opp of that angle is the thing you want to find
freckles
  • freckles
opp and hyp think sin( )
anonymous
  • anonymous
wait.... I cant use sin we don't use that it has to be an alternative method...
freckles
  • freckles
ok what is the alternative method?
anonymous
  • anonymous
im not sure it has to be done without sin tho...
freckles
  • freckles
oh then I don't know of another way to find the area
anonymous
  • 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
freckles
  • freckles
ok
anonymous
  • 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.
freckles
  • freckles
so we don't need to find the area exactly
freckles
  • freckles
we can approximate it
anonymous
  • anonymous
what does this all mean..
freckles
  • 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
anonymous
  • anonymous
24
freckles
  • freckles
right 4*h<24
freckles
  • freckles
where 4*h is the area of the parallelogram
freckles
  • freckles
didn't label what I called a by the way |dw:1437775996191:dw| I used Pythagorean theorem
freckles
  • freckles
\[6^2=h^2+a^2 \\ 6^2-a^2=h^2 \\ \sqrt{6^2-a^2}=h\]
anonymous
  • anonymous
hmmmmm i think i get it now..
freckles
  • 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
freckles
  • 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
anonymous
  • anonymous
ok
freckles
  • freckles
and if we multiply both sides by -1 we have \[-6<-a<0 \\ \text{ so back to the inequality } \\ 4 h=4 \sqrt{36-a^2}<4 \sqrt{36-0^2}\]
freckles
  • freckles
but yeah I would have used my trigonometry knowledge :p
anonymous
  • anonymous
this is complicated I'm gonna sit and stare at this for a while.
freckles
  • freckles
\[4h=4 \cdot 6 \sin(55)=24\sin(55) \approx 19.66 \]
freckles
  • freckles
all you have to know is that since h is between 0 and 6 (not including 6) we have 4h<4(6)
anonymous
  • anonymous
hmmm... ok got it
freckles
  • 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|
freckles
  • 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
freckles
  • freckles
anyways I think I'm done now :) unless you have questions
anonymous
  • anonymous
ohhhh okay i think i understand better now
anonymous
  • anonymous
needed that visual thank you ^_^
freckles
  • freckles
visuals are the secret to math
anonymous
  • anonymous
agreed!!
freckles
  • freckles
I think it helps to know the hyp is the longest side in a right triangle on this question
anonymous
  • anonymous
yea definitely helps
freckles
  • 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
anonymous
  • anonymous
yea it helped towards the end thanks again =)
freckles
  • freckles
np

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