The original question includes a diagram: A trapezium PQRS with the long parallel side PQ (12 cm) at the bottom and the short parallel side SR (3 cm) at the top; the diagonals PR and SQ cross at X, and a horizontal segment UT through X joins the two slanted sides.
In the figure, PQRS is a trapezium with PQ parallel to SR. The diagonals of the trapezium meet at X. U lies on SP and T lies on RQ such that UT is a line segment through X parallel to PQ.
The length of PQ is 12 cm and the length of SR is 3 cm.
What, in centimetres, is the length of UT?
- A4.2
- B4.5
- C4.8
- D5.25
- E6
Show the answer and worked solution
answer · C
- A4.2
- B4.5
- C4.8
- D5.25
- E6
Start with the diagonals. Triangles XSR and XQP are similar with ratio SR : PQ = 3 : 12 = 1 : 4, so SX : XQ = 1 : 4 and X lies one fifth of the way along SQ. Now look at triangle SPQ: UX is parallel to PQ with SU : SP = SX : SQ = 1 : 5, so UX = 15× 12 = 2.4, and the same argument on triangle RPQ gives XT = 2.4. Hence UT = 4.8. Note this is the harmonic mean 2× 12× 312+3 of the parallel sides, not their average, which is why 7.5 is not on offer.