ABCD is a parallelogram with AB=15cm and bc=18cm.the height of a parallelogram from vertex A to BC is 9cm.find the height of parallelogram from vertex C to AB?
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Answered by
39
AB=15
BC=18
CO IS THE PERPENDICULAR ON AB.
CO=9
So,Area of parallelogram ABCD =AB*CO
=18*9=162
162=BC*AM
AM=162/15=10.8
BC=18
CO IS THE PERPENDICULAR ON AB.
CO=9
So,Area of parallelogram ABCD =AB*CO
=18*9=162
162=BC*AM
AM=162/15=10.8
Answered by
30
Let the height from A to BC be AE.
LET THE HEIGHT FROM C TO AB BE CF
AB=CD=15CM (OPPOSITE SIDES OF A PARALLELOGRAM ARE EQUAL)
BC=AD=18CM ( " " " " " " " )
AE=9CM
AREA OF PARALLELOGRAM=BASE X PERPENDICULAR HEIGHT FROM BASE
AREA= (18 X 9)CM²= 162CM²
II CASE
ARE OF IIGM=BASE X PERPENDICULAR TO BASE HEIGHT
AREA= (15 X CF)CM²⇒162/15=CF
CF=10.8CM
LET THE HEIGHT FROM C TO AB BE CF
AB=CD=15CM (OPPOSITE SIDES OF A PARALLELOGRAM ARE EQUAL)
BC=AD=18CM ( " " " " " " " )
AE=9CM
AREA OF PARALLELOGRAM=BASE X PERPENDICULAR HEIGHT FROM BASE
AREA= (18 X 9)CM²= 162CM²
II CASE
ARE OF IIGM=BASE X PERPENDICULAR TO BASE HEIGHT
AREA= (15 X CF)CM²⇒162/15=CF
CF=10.8CM
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