Question 1105914
If the lengths of the bases of an isosceles trapezoid are 10 cm and 22 cm,
and if one of the legs' length is 10 cm,
then the other leg's length is also 10 cm,
or else it is not really isosceles.
The trapezoid is sketched below.
{{{drawing(600,300,-12,12,-2,10,
green(rectangle(-5,0,5,8)),
green(rectangle(-5,0,-5.4,0.4)),green(rectangle(5,0,5.4,0.4)),
line(-11,0,11,0),line(-5,8,5,8),
line(-11,0,-5,8),line(11,0,5,8),
locate(-11.5,0,A),locate(11.2,0,D),
locate(-5.5,8.5,B),locate(5.2,8.5,C),
locate(-5.1,0,F),locate(4.9,0,E),
locate(-8,4,10cm),locate(6.8,4,10cm),
locate(-0.5,8,10cm),locate(-0.5,0.6,10cm),
locate(-8.5,0.6,6cm)locate(7.5,0.6,6cm),
locate(-4.9,4.4,green(h)),locate(5.1,4.4,green(h)),
locate(-0.5,-0.7,22cm),arrow(-1,-1,-11,-1),arrow(1,-1,11,-1)
)}}} The green lines, perpendicular to the bases,
split the isosceles trapezoid into rectangle BCEF,
and right triangles ABF and CDE.
We know the length of all the segments in the sketch, except for the green ones.
Their length, according to the Pythagorean theorem, is
{{{green(h)=sqrt((10cm)^2-(6cm)^2)=8cm}}} .
That allows us to calculate the length of diagonal AC
which is the hypotenuse of right triangle ACE.
We use the Pythagorean theorem again, to calculate that length as
{{{sqrt((AE)^2+(CE)^2)=sqrt((16cm)^2+(10cm)^2)=sqrt(356)}}}{{{cm=4sqrt(89)}}}{{{cm}}} or approximately {{{highlight(18.9cm)}}} .