document.write( "Question 475218: What is the area of an isosceles triangle with sides of 7, 7, and 12 \n" ); document.write( "

Algebra.Com's Answer #325841 by jim_thompson5910(35256) $\"\"$ $\"About$

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Solved by pluggable solver: Hero's (or Heron's) Formula (Used to Find the Area of a Triangle Given its Three Sides)

In order to find the area of a triangle 'A' with side lengths of 'a', 'b', and 'c', we can use Hero's Formula:

$\"A=sqrt%28S%28S-a%29%28S-b%29%28S-c%29%29\"$ where S is the semiperimeter and it is defined by $\"S=%28a%2Bb%2Bc%29%2F2\"$

Note: \"semi\" means half. So the semiperimeter is half the perimeter.

So let's first calculate the semiperimeter S:

$\"S=%28a%2Bb%2Bc%29%2F2\"$ Start with the semiperimeter formula.

$\"S=%287%2B7%2B12%29%2F2\"$ Plug in $\"a=7\"$ , $\"b=7\"$ , and $\"c=12\"$ .

$\"S=%2826%29%2F2\"$ Add.

$\"S=13\"$ Divide.

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$\"A=sqrt%28S%28S-a%29%28S-b%29%28S-c%29%29\"$ Now move onto Hero's Formula.

$\"A=sqrt%2813%2813-7%29%2813-7%29%2813-12%29%29\"$ Plug in $\"S=13\"$ , $\"a=7\"$ , $\"b=7\"$ , and $\"c=12\"$ .

$\"A=sqrt%2813%286%29%286%29%281%29%29\"$ Subtract.

$\"A=sqrt%28468%29\"$ Multiply.

$\"A=21.6333076527839\"$ Take the square root of $\"468\"$ to get $\"21.6333076527839\"$ .

So the area of the triangle with side lengths of $\"a=7\"$ , $\"b=7\"$ , and $\"c=12\"$ is roughly $\"21.6333076527839\"$ square units.

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