Finding length of diagonal in parallelogram given two sides and other diagonal A parallelogram has sides of length 11 cm and 13 cm and has one diagonal 20 cm long the length of the other diagonal is what I am supposed to find. Now my answer came out to be 13.2 cm.
Finding the area of the first triangle:
Perimeter = $11 + 13 + 20 = 44$ cm
Semi-Perimeter = $44 ÷ 2 = 22$ cm
Area = $\sqrt{p(p - a)(p - b) (p - c)} = \sqrt{22(22 - 11)(22 - 13)(22 - 20)} = \sqrt{4356} = 66 \text{cm}^2$
Find the area of the parallelogram:
Area of parallelogram = $2 \times \ $area of triangle
Area = $66 * 2 = 132 cm²$
Find the other diagonal:
Let the other diagonal be x
$1/2 (diagonal 1 * diagonal 2) = Area $
$1/2 (20x) = 132 $
$10x = 132 $
$x = 13.2 cm $
However, surprisingly my answer wasn't in the options. Now my friend sent me this attachment from the website topper and claimed that 20 is the correct answer so can you guys pls tell me whether 13.2 cm is the correct answer or 20? Do pls also tell whether what is the mistake done by the other one? Thanks in advance for your help.
 A: Obviously, both diagonals cannot be $20$, otherwise you would have a rectangle, and this would imply $11^2 + 13^2 = 20^2$, which is false.
One solution that is accessible to you is to employ Heron's formula as you have done, but extend the computation.  You already found that the area of the parallelogram is $66(2) = 132$.  However, in order to find the other diagonal, you must solve the equation $$66 = \sqrt{s(s-11)(s-13)(s-c)},$$ where $s = (11+13+c)/2$ and $c$ is the desired diagonal.  You already know $c = 20$ satisfies this equation; the goal is to find the other positive root.  Some algebra will give $$66 = \frac{1}{4} \sqrt{-2304 + 580c^2 - c^4},$$ after which obtaining the nontrivial positive root should be straightforward.
Alternatively, we can employ trigonometry:  we know by the Law of Cosines that $$20^2 = 11^2 + 13^2 - 2(11)(13) \cos C,$$ hence the other diagonal has length $$c^2 = 11^2 + 13^2 - 2(11)(13) \cos (180^\circ - C) = 11^2 + 13^2 + 2(11)(13) \cos  C.$$  In fact, this is how the diagonal identity can be shown:  for general side lengths $a, b$ with one given diagonal $d$, we find the other diagonal $c$ obeys $$c^2 + d^2 = 2(a^2 + b^2), \tag{1}$$ since we must have the simultaneous solution $$\begin{align} c^2 &= a^2 + b^2 - 2ab \cos C \\ d^2 &= a^2 + b^2 - 2ab \cos (180^\circ - C) \end{align}$$ and since $\cos (180^\circ - C) = -\cos C$, we add the two equations to get $c^2 + d^2 = 2a^2 + 2b^2$, proving Equation $(1)$.
A: 
This is one possible setup: the other setup is obtained by rotating this diagram $90º$ either way.
Extend $CD$ to $E$ so that $AE$ is perpendicular to $EC$.  Since area $\Delta ADC = 66$, $\frac{1}{2} AE \cdot DC = 66 \Rightarrow AE = 12$. By Pythagoras, $DE = 5$, and thus $DB^2 = (11-5)^2+12^2 \Rightarrow DB = \sqrt{180} = 6\sqrt5$.
A: The length of the other diagonal is $\sqrt {180}$ and can be found by the parallelogram law as suggested by @hgmaths.
$$x^2+20^2=2(11^2+13^2)$$.
Your friend's method was correct but at the end he used the same angle instead of 180 minus that angle. He therefore obtained the original diagonal! If you change the sign of his final term you will get $\sqrt {180}$.
