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ABCD is a parallelogram Which is greater a. The area of ABCD b. 24 |dw:1437774703693:dw|
do you know how to find the area of parallelogram ? |dw:1437775012406:dw|
where the area is b*h

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

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