Question 878714
GUESS AND CHECK STRATEGY:
If you are not that deep into algebra and calculus, you are expected to try whole numbers for the lengths of the sides in centimeters.
The area given is a clue. It is related to the product you get by multiplying the legs' lengths in centimeters.
For a right triangle, you can take the length of one leg as the base, and the length of the other leg as the height, so
{{{area}}}{{{"="}}}{{{(leg[1])(leg[2])/2}}}
{{{(leg[1])(leg[2])/2=84}}} --> {{{(leg[1])(leg[2])=84*2}}} --> {{{(leg[1])(leg[2])=168}}}
We look for pairs of factors of {{{186}}} that could be the legs' lengths in centimeters.
We know that {{{186=2*84=2*7*12=2*7*2*2*3=2^3*3*7}}}
{{{186=7*(2*12)=7*24}}} sounds like a good possibility, because a hypotenuse so much longer that the "one leg" means that the other leg must also be much longer than the "one leg".
In a right triangle with legs measuring {{{7}}} and {{{24}}} , the hypotenuse measures
{{{sqrt(7^2+24^2)=sqrt(49+576)=sqrt(625)=25}}}
Is "four times" {{{7}}} "3 more than" {{{25}}} ?
YES. {{{4*7=28=25+3}}}
So the lengths of the sides are {{{highlight(7cm)}}} , {{{highlight(24cm)}}} , and {{{highlight(25cm)}}} .
 
Another way (no calculations involved, but your teacher may consider it cheating) would be to look up a list of Pythagorean triples. Those are sets of whole numbers that could be the lengths of the sides of a right triangle. (The squares of the first two numbers add up to the square of the third number).
One of those triples is 7 24 25, and it is the only one where 7 appears as a factor, and the two first numbers are factors of 186.
 
NOT GUESS AND CHECK?
Then you may have to start writing equations and may loose sight of the triangle issue.