Area of a triangle

Plug the values into Heron,s Formula: $Area=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$ So $s=\frac{7+11+14}{2}=\frac{32}{2}=16$ $Area=\sqrt{16(16-7)(16-11)(16-14)}=\sqrt{16\times9\times5\times2}=\sqrt{144}\times\sqrt{5\times2}=\boxed{12\sqrt{10}}$ So $x=12$ and $y=10$ and $x+y=12+10=\boxed{22}$

S=(7+11+14/2) Heron's formula states that the area of a triangle is = \sqrt(S(S-a)(S-b)(S-c)) Plug in the numbers and you get \sqrt((16)(9)(5)(2))=3 4 \sqrt(5*2) =12\sqrt(10) since the area of a triangle can be expressed as the square root of y times x. x=12 y=10 x+y=22

By using Heron's formula, we have $s=\frac{7+11+14}{2}=16$ , where $s$ is the semi perimeter of the triangle. Then, the area of the triangle would be

$\sqrt{(16)(16-7)(16-11)(16-14)}$

$\implies\sqrt{(2^4)(3^2)(5)(2)}$

$\implies2^2\times3\sqrt{10}$

$\implies12\sqrt{10}$

Comparing with $x\sqrt y$ , we arrive $x+y=\boxed{22}$ , which is our desired answer.

By using the Heron's Formula, we can find the area by getting $S = \frac{7+11+14}{2} = 16$

Then we get $\sqrt{16\times(16-7)\times(16-11)\times(16-14)} = \sqrt{1440}$

$\sqrt{1440} = \sqrt{144\times10}$

Since the square root of $144$ is $12$ , we can simplifly it to $12\sqrt{10}$

Therefore, $x = 12, y = 10$

$x + y = 12 + 10 = 22$