ABCDtrapezoid

BH is perpendicular to DC.

BH = 12 cm.

In △ BHC, BC^2 = BH^2 + HC^2 so HC = 5 cm.

therefore DC = 16 cm.

In △ ADC, AC^2 = AD^2 + DC^2 so AC = 20 cm.

Consider my diagram. First, draw the dashed line as shown. We have now a rectangle and a small triangle and a big triangle. Find the unknown $x$ in the small triangle by using pythagorean theorem, we have

$x=\sqrt{13^2-12^2}=5$

So the legs of the big triangle are $12$ and $11+5=16$ .

By pythagorean theorem again, the unknown $d$ (hypotenuse of the big triangle) is

$d=\sqrt{12^2+16^2}=\boxed{20}$

Relevant wiki: Pythagorean Theorem

Pythagoras' Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$\text{If we extent this figure to create a rectangle, we will get a new triangle in the top right with a hypotenuse of 13 and one leg which is 12.} \\ \text{We can recognize this as a 5-12-13 triangle which tells us the length of the other leg is 5. We can now use the Pythagorean Theorem to solve for d} \\ 12^2 + 16^2 = d^2 \\ \boxed{d = 20}$

Since the Quadrilateral's sides are not equal but only two sides being right angles, Taking out the part 11 x 12 First thing you have to find the extra part dimensions ( a right angle triangle with the hypotenuse 13 and height 11, the width will be 5, add this triangle as rotated 180 degrees to the upper right side of the quadrilateral - hat will make the shape perfect square -, now split the square from center by the line "d" you will have a right angled triangle with the dimensions 16 x 12. Next step is applying Pythagorean theorem to the triangle to find the hypotenuse and BOOM d=20

The diagnoal side of the quadrilateral is the long side of a pythagorean triangle and measures 13 units. We know the the longer of the other two sides measures 12 units. Thus the width of this triangle is 5cm. This means that the diagonal we are being asked to calculate is effectively the long diagonal of a 16 x 12 rectangle, or the hyopteneuse of a right-angle triangle whiose shorter sides are 12 and 16. Thus it is equal to the square root of ((12 x 12) + (16 x 16)) = the square root of (256 + 144) = the square root of 400 = 20.

Using Vector addition.

we take the top side(line) to be the vector A=(11,0), then we take the vector B of magnitude 13 to be B=(13cos(theta), 12); we add x and y components together and square root the the squares i.e. sqrt( (added x)^2+(added y)^2 )

note: some geometry similar to other answers is used to obtain the angle theta.

$d=\sqrt{12^2+(11+\sqrt{13^2-12^2})^2}=\sqrt{144+(11+\sqrt{169-144})^2}=\sqrt{144+(11+\sqrt{25})^2}=\sqrt{144+(11+5)^2}=\sqrt{144+16^2}=\sqrt{144+256}=\sqrt{400}=\color{#3D99F6}{\boxed{\large{20}}}$

If you allow for the base to be shorter than the top side then there is a second solution. Following the same logic as other solutions and dropping a 5,12,13 triangle down we get the base is 11-5=6 and by pythagoras d=6×sqrt(5)

We can make a right triangle where we use the side of length 13 as its hypotenuse. The height of this right triangle (with hypotenuse of length 13) is parallel and equal to the side with length of 12. Thus its other side is 5. Looking back at the whole picture of the trapezoid, the other base is equal to 11+5 (the given base and the base of the right triangle of hypotenuse length of 13) which is 16. We can now have the larger right triangle, where we can use 16 as its base, 12 (as given) as its height, and "d" (the diagonal of the trapezoid that we are looking for) as its hypotenuse. We can now use again the Pythagorean Theorem using the given, to find d. 12^2 +16^2 =d^2 . so we have 144+256=d^2 400=d^2 20=d

Extend the 12cm width to the 13cm side. This makes an extra triangle.

Use 180-arcsin(12/13) to then get the angle at the top right of the trapezium. It would be 112.6 degrees.

Then use Cosine rule: c^2= a^2 + b^2 - 2ab*cos(112.16)

c^2 = 400

c = 20.

This makes the length of the diagonal "d" 20cm.

$\triangle{DEB}$ is a $(5,12,13)$ right triangle and $\triangle{ACB}$ is a $(3*4,4*4,5*4) = (12,16,20)$ right triangle $\therefore BC = \boxed{20}$ .