In a trapezium ABCD , AB || CD , the point P and and the point Q lie on the side AB and BC respectively. PQ || AC , prove that area of trangle ADP AND TRIANGLE ACQ
Answers
Given : In a trapezium ABCD , AB || CD , the point P and and the point Q lie on the side AB and BC respectively. PQ || AC ,
To Find : prove that area of tringle ADP = area of tringle ACQ
Solution:
the point P and and the point Q lie on the side AB and BC respectively.
PQ || AC
=> AP/ PB = CQ/ QB using BPT / Thales theorem
or AP/AB = CQ/ BC
h is distance between AB and CD
Area of Δ ADP = (1/2) * AP * h
Area of Δ ABC
= (1/2) * AB * h
Area of Δ ACQ = (CQ/BC) * Area of Δ ABC
=> Area of Δ ACQ = (CQ/BC) * (1/2) * AB * h
CQ/BC = AP/AB
=> Area of Δ ACQ = (AP/AB) * (1/2) * AB * h
=> Area of Δ ACQ = (1/2) * AP * h
=> Area of Δ ACQ = Area of Δ ADP
QED
Hence proved
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Answer:
Construction: P and C are joined
^ADP = ^APC { triangles between same parallels }
^APC = ^AQC { triangles between same parallels }
^ADP = ^AQC { from eqn 1 & 2 }
Step-by-step explanation: