Find area of two triangles with known full perimeter

Question

I have following problem that I cannot solve... I have a triangle with sides $a$, $b$, and $c$ which is split into two smaller triangles, $E$ and $F$, like this.

I need to find the area of $F$.

I also know that the perimeter of $F$ equals the perimeter of $E$.

I tried to equate the perimeters of $E$ and $F$, but I could not figure it out.

Thank you so much for any hint!

Answer 1

If the perimeters of $E$ and $F$ are the same:

$x + y + 20 = x + 13 + (21-y)$

Solving this, you get $y = 7$.

Use Heron's formula to find the Area of the entire triangle:

$S = \frac{A+B+C}{2}$, $A = \sqrt{S(S-A)(S-B)(S-C)}$:

$S= \frac{20+13+21}{2}=27$, $A=\sqrt{27(27-20)(27-13)(27-21)}=126$

Now use $A=\frac12bh$.
$h = 2\frac Ab = 2 \frac{126}{21}=12$

Finally, $Area\; of \;F$ $= \frac12bh=\frac12 7*12=42$

Answer 2

we have $$x^2=y^2+20^2-4y\cos(\beta)$$ $$x^2=b^2+(c-y)^2-2b(c-y)\cos(\alpha)$$ and for $$\alpha,\beta$$ we have $$a^2=b^2+c^2-2bc\cos(\alpha)$$ and $$b^2=a^2+c^2-2ac\cos(\beta)$$

Answer 3

First find angles.

To find left part $u$ of $E$ side

$$ 13+u= 21-u +20 \rightarrow u=14$$

Now apply $\frac12 bc \sin A$ formula to triangle parts.