1.in a trapezium pqrs a and b are mid points of diagonals pr and qs show that ab=1/2(pqrs]
2.in the fig o is the center of circle and OLM is perpendicular to AOB prove that a] A, Q, P and M are concyclic
b]
c]P,L,O,B are con cyclic
Answers
PQ is parallel to RS in the trapezium. Given that A is mid point of PR and B is midpoint of QS. Let C be midpoint of PS and D be the midpoint of QR.
CB is line joining midpoints of two sides of the triangle PQS. Hence, CB is parallel to PQ and also PQ = 2 * CB.
Similarly, in triangle PRS, CA is parallel to RS, and RS = 2 * CA
Rs = 2 (CB + BA) = PQ + 2 BA
AB = 1/2 * (RS - PQ), we assumed that RS is bigger than PQ.
If PQ is longer then RS, then it will be AB = 1/2 (PQ - RS)
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For the question 2 :
We don't know about the locations points L, M, P and Q, You need to enclose a diagram.
Answer:Given: In trapezium PQRS , A and B are midpoints of diagonals PR and QS.
To Show: AB=1/2(PQ-SR).
Construction:draw midpoints C and D of SP and RQ respectively and join CA and BD
Proof : In triangle SPQ ,C and B are midpoints of SP and SQ
Therefore,by midpoint theorem
CB || PQ and 2CB= PQ .......1
In triangle QSR B and D are midpoints of QS and RQ
Therefore, by midpoint theorem
BD || SR and 2BD =SR......2
In triangle SPR C and A are midpoints of SP and RP
Therefore by midpoint theorem
CA = 1/2 SR .......3
From 2
2 BD = SR
BD =1/2 SR ........4
From 3 and 4
CA = BD ....5
From 1 and 2
PQ - SR = 2 CB - 2 BD
PQ - SR = 2(CB - BD)
1/2(PQ - SR) =CB - BD
1/2(PQ - SR)= CA+AB - BD
From 5
CA = BD , therefore they cancel each other
Therefore ,
1/2(PQ-SR) =AB
AB= 1/2(PQ-SR)
Hence Showed
Step-by-step explanation:
