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Q11. In a parallelogram P is any point inside it prove that ar (AABP)+ ar(ADCP)-1/2 ar(ABCD). |
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Answer» Given,ABCD is a ||gm and P is a pt on CDR.T.P : Area of triangle APD+Area of triangle BCP = Area of triangle ABPNow,triangle APD + triangle BCP +triangle ABP = ||gm ABCD ...(i)but,by the theorem on area,13.4,which states that the area of a triangle lying on the same base and between the same parallel lines as of the ||gm has area equal to half of that of the parallelogram.=>area of triangle ABP=1/2 area of ||gm ABCD ...(ii)from (i) and (ii),we have,area of trian. APD +area of trian.BCP + 1/2 area of ||gm ABCD=area of ||gm ABCDarea of trian. APD + area of trian. BCP=(1-1/2)area of ||gm ABCD =1/2 area of ||gm ABCD =Area of triangle APB (from (ii)).Hence, proved. |
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