Kindly solve this question. Share with your friends Share 0 Amit Kumar answered this Solution: Please find below the solution to the asked query: Let P, Q, R, S be the midpoints of sides AB, BC , CD, DA of parallelogram ABCDWe need to show that, Areaquad PQRS=12Area∥gm ABCDFirst show that PQRS is a parallelogram.Join AC and AR In △DAC, S and R are midpoints of AD and DC respectively.So by midpoint theorem, we get SR∥AC and SR=12AC ....1In △ABC, P and Q are midpoints of AB and AC respectively.So by midpoint theorem, we get PQ∥AC and PQ=12AC ....2Thus in quadrilateral PQRS, we have PQ∥SR and PQ=SR Using 1 and 2Now median AR divides △ACD into two parts with equal areaHence, areaARD=12 areaACD ....3Median RS divides △ARD into two parts with equal areaHence, areaDSR=12 areaARD ....4From 3 and 4, we get areaDSR=14 areaACDIn a similar way, areaBQP=14 areaABC Add the above two to get, areaDSR+areaBQP=14 areaACD+areaABCThis gives, areaDSR+areaBQP=14Area ∥gmABCD .....5In a similar way, we can show that areaCRQ+areaASP=14Area ∥gmABCD .....6 Add 5 and 6 to get, areaDSR+areaBQP+ areaCRQ+areaASP=12Area ∥gmABCDBut we have, areaDSR+areaBQP+ areaCRQ+areaASP=Area ∥gmABCD-areaQuad PQRSThis gives, Area ∥gmABCD-areaQuad PQRS=12Area ∥gmABCD⇒Area ∥gmABCD-12Area ∥gmABCD=areaQuad PQRS⇒1-12Area ∥gmABCD=areaQuad PQRSThis proves that, Areaquad PQRS=12Area∥gm ABCD i Mistakenly, I have considered P, Q, R,S instead of E,F,G,H which would not affect the result. I hope it helps! 0 View Full Answer