ABCD is a rectangle with O as any point in its interior.If area of triangle AOD= 3cm square and area of triangle BOC= 6cm square,then what is the area of rectangle ABCD?
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given: ABCD is a rectangle, and O is any point in its interior.
area(ΔAOD)= 3 sqcm and area(ΔBOC)=6 sq cm.
from O draw a line EF parallel to AB intersecting AD at E and BC at F.
therefore OE⊥AD and OF⊥BC.
let OE=x and OF=y
OE+OF=EF=AB
therefore x+y=AB......(1)
area(ΔAOD)+area(ΔBOC)=3+6=9 sq cm.
since BC=AD [opposite sides of a rectangle]
since EF=AB
AD*AB=18
area of the rectangle ABCD = AB*AD=18
area(ΔAOD)= 3 sqcm and area(ΔBOC)=6 sq cm.
from O draw a line EF parallel to AB intersecting AD at E and BC at F.
therefore OE⊥AD and OF⊥BC.
let OE=x and OF=y
OE+OF=EF=AB
therefore x+y=AB......(1)
area(ΔAOD)+area(ΔBOC)=3+6=9 sq cm.
since BC=AD [opposite sides of a rectangle]
since EF=AB
AD*AB=18
area of the rectangle ABCD = AB*AD=18
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Answer:
18cm^2
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