Ex 9.3, 9

Transcript

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 ΔACQ and ΔAQP , ΔACQ and ΔAQP 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