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Courier Agent Distance Calculation

XAT 2021 · Quantitative Ability & Data Interpretation · Medium · Geometry

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Rajesh, a courier delivery agent, starts at point A and makes a delivery each at points B, C and D, in that order. He travels in a straight line between any two consecutive points. The following are known: (i) AB and CD intersect at a right angle at E, and (ii) BC, CE and ED are respectively 1.3 km, 0.5 km and 2.5 km long. If AD is parallel to BC, then what is the total distance (in km) that Rajesh covers in travelling from A to D?

  1. A.

    10.2

  2. B.

    12

  3. C.

    11.5

  4. D.

    5.5

  5. E.

    18

Answer

C

Explanation

In right triangle CEBCEB, CE=0.5CE = 0.5, BC=1.3BC = 1.3. EB=1.32−0.52=1.69−0.25=1.2EB = \sqrt{1.3^2 - 0.5^2} = \sqrt{1.69 - 0.25} = 1.2 km. Since AD∥BCAD \parallel BC, triangles EBCEBC and EADEAD are similar. EBEC=EAED  ⟹  1.20.5=EA2.5  ⟹  EA=6\frac{EB}{EC} = \frac{EA}{ED} \implies \frac{1.2}{0.5} = \frac{EA}{2.5} \implies EA = 6 km. AB=AE+EB=6+1.2=7.2AB = AE + EB = 6 + 1.2 = 7.2 km. CD=CE+ED=0.5+2.5=3CD = CE + ED = 0.5 + 2.5 = 3 km. BC=1.3BC = 1.3 km. Total distance = AB+BC+CD=7.2+1.3+3=11.5AB + BC + CD = 7.2 + 1.3 + 3 = 11.5 km.

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