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Ex 9.3, 9 - The side AB of parallelogram ABCD is produced - Ex 9.3

Ex 9.3, 9 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.3, 9 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3

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Ex 9.3, 9 The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that ar (ABCD) = ar (PBQR). Given: A parallelogram ABCD where CP AQ & PBQR is a parallelogram To prove: ar (ABCD) = ar (PBQR) Construction : Join AC & PQ Proof : For ACQ and AQP , ACQ and AQP are on the same base AQ and between the same parallels AQ and CP. ar(ACQ) = ar (APQ) Subtracting ar(ABQ) both sides ar(ACQ) ar(ABQ) = ar(APQ) ar(ABQ) ar(ABC) = ar(PBQ) In parallelogram ABCD, AC is diagonal ABC ADC ar(ABC) = ar(ADC) ar(ABC) = ar(ADC) = 1/2 ar(ABCD) In parallelogram PBQR, PQ is the diagonal PBQ PRQ ar(PBQ) = ar(PRQ) So, ar(PBQ) = ar(PRQ) = 1/2 ar(PBQR) From (1) ar(ABC) = ar(PBQ) 1/2 ar (ABCD) = 1/2 ar (PBQR) ar (ABCD) = ar (PBQR) Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.