What we want to prove is the algebraic identity:
$ \displaystyle (a+b)^2=a^2 + 2ab + b^2$
So only easy maths accepted,
ok here it is :
well... that probably didn't make many sense! Lets take it one step at a time.
(we don't really care how much actually a and b are)
now put a same line in 90 degrees angle next to it let's do that two more times :
a square is formed!
Next, we draw two more lines and we are done
put some color into it...
As we see, four areas were created.
Areas 1 and 2 are squares, 3 and 4 are parallelograms.
Can you see 3 and 4 have the same dimensions? If so, you can tell that they are equal.
For squares and rectangles, we can find their area by multiplying their dimensions
(just if someone doesn't remember that!)
We are finally to find the total area of the big initial square. So, the square of a plus b is equal to the square of a plus twice the a times b plus the square of b.
Hold on a second, isn't that what we have read in the very first equation?Let's have a look again: $ \displaystyle (a+b)^2=a^2 + 2ab + b^2$
Hold on a second, isn't that what we have read in the very first equation?Let's have a look again: $ \displaystyle (a+b)^2=a^2 + 2ab + b^2$
Yes, sure it is! We actually just proved it.
Now if you really understood that, you should be able to find on your own
the square of a plus b plus c.
Now if you really understood that, you should be able to find on your own
the square of a plus b plus c.
...or you can use a tiny hint!
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