Hypotenuse and perimeter problem

Question

I need help finding the area of a triangle that has a hypotenuse of 6 and a perimeter 14 ,how can I do it ?

The cloest I've gotten is that 5=b and 3=a but that would make c=5.8 I've asked around and my professor said that it can be done but I'm looking at it the wrong way ,I am clueless any help would be appreciated

Answer 1

Method I used: We are given the hypotenuse and the perimeter. Thus, we can use this to our advantage, because note that we not only know the sum of the other two sides, which is the perimeter-hypotenuse, but we also have the Pythagorean theorem. We can use the Pythagorean theorem to derive a quadratic equation we can use to then solve for the length of our sides, which I've done so below:

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$a+b+c\quad =\quad 14\\ a+b\quad =\quad 8,\quad since\quad 14-c\quad =\quad 8\\ Let\quad a\quad =\quad 8-b\\ { (8-b) }^{ 2 }+{ b }^{ 2 }={ 6 }^{ 2 }\\ 64-16b+{ b }^{ 2 }+{ b }^{ 2 }=36\\ 64-16b+{ 2b }^{ 2 }=36\\ 28-16b+2{ b }^{ 2 }=0$

$28-16b+2{ b }^{ 2 }=0\\ 2{ b }^{ 2 }-16b+28=0\\ 2({ b }^{ 2 }-8b+14)=0\\ { b }^{ 2 }-8b+14=0\\ a=1,b=-8,c=14\\ \frac { 8\pm \sqrt { 64-4*14 } }{ 2 } =\frac { 8\pm \sqrt { 8 } }{ 2 } =4\pm \sqrt { 8 } ;\quad $

$\frac { 8\pm \sqrt { 64-4*14 } }{ 2 } =\frac { 8\pm \sqrt { 8 } }{ 2 } =4\pm \sqrt { 8 } ;\quad \\ If\quad b=4+2\sqrt { 2 } ,\\ a=8-b,\therefore a=8-(4+2\sqrt { 2 } )=4-2\sqrt { 2 } \\ If\quad b=4-2\sqrt { 2 } ,\\ a=8-b,\therefore a=8-(4-2\sqrt { 2 } )=4+2\sqrt { 2 } \\ \therefore (a,b)\quad =\quad (4-2\sqrt { 2 } ,4+2\sqrt { 2 } )\\ 1/2bh\quad =\quad 1/2a*b=1/2(4-2\sqrt { 2 } )(4+2\sqrt { 2 } )\\ \\ \\ \\ \\ \\ \\ $

If you simplify the above expression, you will get the area (which is 4).

Answer 2

Hint: The problem uses the word "hypotenuse", so the triangle is right. Let the legs be $x$ and $y$. What is the area? What is the perimeter? You have two equation in two unknowns. Or, just think about the right triangles you know.

Answer 3

Hint. You need $$a+b=8\ ,\quad a^2+b^2=36\ .$$ Since $$(a+b)^2=a^2+b^2+2ab$$ you can find the value of $ab$. And the area of the triangle in terms of $a$ and $b$ is...?

See if you can finish this.

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Comment. You also have $$a^4+a^2b^2=36a^2\ ;$$ you can substitute the value of $ab$ and solve a quadratic to find values of $a$ and $b$. But you don't actually need this in order to solve the problem asked.

Answer 4

We have $$14=6(1+\cos t+\sin t),\iff \cos t+\sin t=\dfrac43$$

Square it

Now we need $$\dfrac{6\cos t\cdot6\sin t}2$$