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John, 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 John covers in travelling from A to D?
A. 10.5 B. 12.5 C. 11.5 D. 15.5 E. 18.5
adapted from CL
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John, 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 John covers in travelling from A to D?
A. 10.5 B. 12.5 C. 11.5 D. 15.5 E. 18.5
adapted from CL
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We can visualize the information as shown below -
Attachment:
Screenshot 2023-04-20 120140.jpg
In \(\triangle CEB\)
\(\angle BEC = 90\), therefore
\(CB^2 = CE^2 + BE^2\)
\(BE^2 = (\frac{13}{10})^2 - (\frac{5}{10})^2\)
\(BE^2 = \frac{144}{100}\)
\(BE = \frac{12}{10}\)
\(\triangle EDA\) \( \approx\) \(\triangle ECB\)
Therefore
\(\frac{ED}{EC} = \frac{EA}{EB}\)
\(\frac{2.5}{0.5} = \frac{EA}{1.2}\)
EA = \(\frac{60}{10}\) = 6
Total Distance = AE + EB + BC + CE + ED = 6 + 1.2 + 1.3 + 0.5 + 2.5 = 11.5
Option C
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.