In a trapezium, the two non-parallel sides are of length \(13\) cm each. The distance between the parallel sides is \(12\) cm. If the smaller of the parallel sides is \(11\) cm, then the sum of the lengths of the two diagonals, in cm, is
Show Hint
In isosceles trapezium, first find the horizontal projections of equal sides using Pythagoras.
Since both non-parallel sides are equal (\(13\) cm), it is an isosceles trapezium.
Height:
\[
h=12
\]
Side:
\[
13
\]
Using Pythagoras for side projection:
\[
x=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5
\]
Difference of parallel sides:
\[
2x=10
\]
Smaller side:
\[
11
\]
So larger side:
\[
11+10=21
\]
Now diagonal forms a right triangle with:
Horizontal length:
\[
5+11=16
\]
Vertical height:
\[
12
\]
So diagonal:
\[
d=\sqrt{16^2+12^2}=\sqrt{256+144}=\sqrt{400}=20
\]
In isosceles trapezium, both diagonals are equal.
Thus:
\[
d_1+d_2=20+20=40
\]
Hence,
\[
\boxed{40}
\]