RRB JE practice question
In trapezium ABCD, AB is parallel to CD, with AB = 14 cm and CD = 34 cm. A line PQ is drawn parallel to AB and CD, meeting leg AD at P and leg BC at Q, such that AP : PD = 3 : 2. Find the length of PQ.
From a RRB JE Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (D) 26 cm
When a line parallel to the two parallel sides of a trapezium divides a leg AD in the ratio AP:PD = 3:2 (measuring from the side AB), its length is the weighted average PQ = (PD-part × AB + AP-part × CD)/(AP+PD) = (2×AB + 3×CD)/(3+2). So PQ = (2×14 + 3×34)/(5) = (28 + 102)/5 = 130/5 = 26 cm.
हिंदी में प्रश्न
समलंब ABCD में, AB, CD के समांतर है, जहाँ AB = 14 सेमी और CD = 34 सेमी है। एक रेखा PQ, AB और CD के समांतर खींची जाती है, जो भुजा AD को P पर तथा भुजा BC को Q पर इस प्रकार काटती है कि AP : PD = 3 : 2 है। PQ की लंबाई ज्ञात कीजिए।
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सही उत्तर: (D) 26 cm
जब समलंब की दोनों समांतर भुजाओं के समांतर कोई रेखा भुजा AD को AP:PD = 3:2 (AB की ओर से मापने पर) के अनुपात में बाँटती है, तो उसकी लंबाई भारित औसत (weighted average) होती है: PQ = (2×AB + 3×CD)/(3+2)। अतः PQ = (2×14 + 3×34)/5 = 130/5 = 26 सेमी।
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