In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some point. Find the length of segments between this point and the vertices of the greater base.

Consider trapezoid ABCD with bases BC=11 units and AD=18 units. The lengths of legs are AB=7 units and CD=3 units. Point S is the point of intersection of the extensions of the legs AB and CD.

Let BS=x units and CS=y units.

Consider triangles BSC and ASD. By AAA theorem these triangles are similar (because ∠SAD≅∠SBC, ∠ADS≅∠BCS and ∠S is common).

Then

[tex]\dfrac{BS}{AS}=\dfrac{CS}{DS}=\dfrac{BC}{AD},\\ \\\dfrac{x}{x+7}=\dfrac{y}{y+3}=\dfrac{11}{18}.[/tex]

Therefore,

[tex]\dfrac{x}{x+7}=\dfrac{11}{18},\\ \\18x=11(x+7),\\ \\18x=11x+77,\\ \\7x=77,\\ \\x=11\ units.[/tex]

[tex]\dfrac{y}{y+3}=\dfrac{11}{18},\\ \\18y=11(y+3),\\ \\18y=11y+33,\\ \\7y=33,\\ \\y=\dfrac{33}{7}\ units.[/tex]

The lengths of segments between point S and the vertices of the greater base are

[tex]AS=AB+BS=7+11=18\ units,\ DS=DC+CS=3+\dfrac{33}{7}=\dfrac{54}{7}\ units.[/tex]