Ex 9.3

Ex 9.3, 1
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Ex 9.3, 2 Important Deleted for CBSE Board 2022 Exams

Ex 9.3, 3 Deleted for CBSE Board 2022 Exams

Ex 9.3, 4 Deleted for CBSE Board 2022 Exams

Ex 9.3, 5 Important Deleted for CBSE Board 2022 Exams

Ex 9.3, 6 Important Deleted for CBSE Board 2022 Exams

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Ex 9.3, 8 Deleted for CBSE Board 2022 Exams

Ex 9.3, 9 Important Deleted for CBSE Board 2022 Exams You are here

Ex 9.3, 10 Deleted for CBSE Board 2022 Exams

Ex 9.3, 11 Important Deleted for CBSE Board 2022 Exams

Ex 9.3, 12 Important Deleted for CBSE Board 2022 Exams

Ex 9.3, 13 Deleted for CBSE Board 2022 Exams

Ex 9.3, 14 Deleted for CBSE Board 2022 Exams

Ex 9.3, 15 Deleted for CBSE Board 2022 Exams

Ex 9.3, 16 Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)

Serial order wise

Last updated at Aug. 25, 2021 by Teachoo

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