If a right triangle has a hypotenuse of 6, and a perimeter of 14, what is the area of the right triangle?

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Answer 1 · 2018-04-25T15:57:20

$color(blue)(6 \ \ \ \"units"^2)$

By Pythagoras' theorem, the square on the hypotenuse is equal to the sum of the squares of the other two sides.

Let the two unknown sides be $bba and bb(b)$

Then:

$a^2+b^2=36 \ \ \ \ [1]$

The perimeter of a triangle is the sum of all its sides.

$a+b+6=14 \ \ \ \ [2]$

Using $[2]$

$a=6-b$

Substituting in $[1]$

$(6-b)^2+b^2=36$

Expanding:

$b^2-12b+36+b^2=36$

$2b^2-12b=0$

Factor:

$2b(b-6)=0=>b=0 and b=6$

$b=0$ is not valid, we can't have a triangle with no side.

Substituting $b=6$ in $[2]$

$a+6+6=14$

$a=2$

Area of triangle is:

$1/2"base"xx"height"$

$1/2(6)*(2)=6 \ \ \ \"units^2$