Question

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.

Answer 1

`AS=18\ units,\ DS=(54)/(7)\ units.`

Explanation:

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

`(BS)/(AS)=(CS)/(DS)=(BC)/(AD),\\ \\(x)/(x+7)=(y)/(y+3)=(11)/(18).`

Therefore,

`(x)/(x+7)=(11)/(18),\\ \\18x=11(x+7),\\ \\18x=11x+77,\\ \\7x=77,\\ \\x=11\ units.`

`(y)/(y+3)=(11)/(18),\\ \\18y=11(y+3),\\ \\18y=11y+33,\\ \\7y=33,\\ \\y=(33)/(7)\ units.`

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

`AS=AB+BS=7+11=18\ units,\ DS=DC+CS=3+(33)/(7)=(54)/(7)\ units.`