@Shadow
Area of a Rectangle: length times width Area of a Parallelogram: base times h
the parallelogram is 12
would the answer be 87?
Where'd you get that number from?
So you used the hypotenuse of both triangles to solve for the length and width of the rectangle?
yea
What did you get for them
12
and its a parallelogram
i did it as parallelogram and rectangle
|dw:1519345996850:dw|
I got 3.60555127546 and 10.8166538264 for the hypotenuse.
Multiply those two, then add w/12
okay im confused, i drew the two triangles to find the length and width of the rectangle and square thing, they arent part of the problem
Yeah, you draw the triangles to use Pythagoras theorem. Those lines are at a slope, so drawing a straight line and comparing it doesn't work.
okay
so was my answer right or no?
No
dang it
what did i do wrong
Say I have a crayon, and a pencil. I can slope the pencil so that it fits next to the crayon like so |dw:1519346460843:dw|
That crayon may be 5 inches, but the pencil is longer
You drew the crayon (9) up against the length of the rectangle (which is at a slope). That doesn't make the length of the rectangle 9
okay, and i got 50.99999999998581
Hmm, round it to 51
okay
is that the answer?
Yes
okay and is this one right
The answer to this one isn't 51. It's the answer to the previous one above. If you have another question, open up a new one.
yeah i got that but thats what i got for that one
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