Question

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?

Accepted Answer

C. In right triangle $CEB$, $CE = 0.5$, $BC = 1.3$. $EB = \sqrt{1.3^2 - 0.5^2} = \sqrt{1.69 - 0.25} = 1.2$ km. Since $AD \parallel BC$, triangles $EBC$ and $EAD$ are similar. $\frac{EB}{EC} = \frac{EA}{ED} \implies \frac{1.2}{0.5} = \frac{EA}{2.5} \implies EA = 6$ km. $AB = AE + EB = 6 + 1.2 = 7.2$ km. $CD = CE + ED = 0.5 + 2.5 = 3$ km. $BC = 1.3$ km. Total distance = $AB + BC + CD = 7.2 + 1.3 + 3 = 11.5$ km.

Courier Agent Distance Calculation

Explanation

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