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CBSE Class 9 Answered

ABCD is trapezium in which AB is parallel to DC and AB=70 DC=50 if P and Q are respectively mid point of AD and BC prove that PQ=60 cm Are of trapezium DCQP=9/11 AR of trapezium ABQP
Asked by Sahilraipure | 08 Mar, 2019, 05:04: PM
answered-by-expert Expert Answer
Figure shows the trapezium as given in the question. Let us join BD.
 
In ΔADB, P is midpoint of side AD. A line drawn through the mid point side AD parallel to base AB  meets opposite side BD at R,
so that DR = RB  and PR = (1/2)AB  .  Hence PR = 35 cm.
 
Similarly , QR is parallel to CD and Q is a mid point of BC. Hence QR = (1/2)CD = 25 cm
 
Hence PQ = PR+RQ = 35+25 = 60 cm
 
since P and Q are midpoints of non-parallel sides of trapezium, distance between parallel lines CD and PQ is same as
distance between parallel lines PQ and AB.
 
area of trapezium DCQP = (1/2)h(60+50) = 55h
area of trapezium ABQP = (1/2)h(60+70) = 65h
 
area of trapezeium DCQP = (11/13) area of trapezium ABQP
Answered by Thiyagarajan K | 09 Mar, 2019, 01:17: PM

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