Parallelogram ABCD and rectangle ABEF are on the same base AB. If AB=14 cm, BC=12 cm, then the possible value for the perimeter of ABEF is Ans is 52. but how is solved
Asked by | 23rd Mar, 2012, 06:52: PM
Expert Answer:
IIgm ABCD and rectangle ABEF are between the same parallels AB and CF.
AB = EF (For rectangle) and AB = CD (For parallelogram)
? CD = EF ? AB + CD = AB + EF ... (1)
Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
? AF < AD
similarly we write , BE < BC ? AF + BE < AD + BC ... (2)
From equations (1) and (2), we get
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD
Now, Perimeter of parallelogram ABCD = AB + BC + CD + DA = 14 + 12 + 14 + 12 = 52
Thus, Perimeter of rectangle ABEF < 52 cm
AB = EF (For rectangle) and AB = CD (For parallelogram)
? CD = EF ? AB + CD = AB + EF ... (1)
Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
? AF < AD
similarly we write , BE < BC ? AF + BE < AD + BC ... (2)
From equations (1) and (2), we get
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD
Answered by | 25th Mar, 2012, 01:29: PM
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