The side AB of a parallelogram ABCD is produced to any point P. A line through A parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Show that ar(parallelogram ABCD) = ar(parallelogram BPRQ).

Asked by Sthitaprajna Mishra | 4th Feb, 2014, 08:22: PM

Expert Answer:

Consider the above figure.
 
Triangles on the same base and between the same parallels have equal areas.
 
Thus, a r left parenthesis triangle A C Q space right parenthesis equals a r left parenthesis triangle A P Q right parenthesis
Subtract the area of triangle ABQ from both the sides, we have
 
a r open parentheses triangle A C Q close parentheses minus a r open parentheses triangle A B Q close parentheses equals a r open parentheses triangle A P Q close parentheses minus a r open parentheses triangle A B Q close parentheses rightwards double arrow a r open parentheses triangle A B C close parentheses equals a r open parentheses triangle Q B P close parentheses space space space space space space.... left parenthesis 1 right parenthesis
 
Since AC and PQ are the diagonals of the parallelograms ABCD and PBQR, we have
 
a r open parentheses triangle A B C close parentheses equals 1 half cross times a r open parentheses square A B C D close parentheses.... left parenthesis 2 right parenthesis a r open parentheses triangle P B Q close parentheses equals 1 half cross times a r open parentheses square P B Q R close parentheses.... left parenthesis 3 right parenthesis
 
Thus from equations (1), (2) and (3), we have
 
1 half cross times a r open parentheses square A B C D close parentheses equals 1 half cross times a r open parentheses square P B Q R close parentheses rightwards double arrow a r open parentheses square A B C D close parentheses equals a r open parentheses square P B Q R close parentheses

Answered by  | 6th Feb, 2014, 10:01: AM

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