ABCD is a parallelogram in which the side BC is produced to E such that CE = BC.AE intersects CD at F.If area triangle DFB is equal to 3 cm², find the area of parallelogram ABCD.
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Given:
ABCD is a parallelogram
BC=CE
arΔDBF = 3cm²
To Find:
ar(║gm ABCD)
Steps are as follow:
∵AD║BC
∴AD║BE
In ΔADF and ΔCEF,
∠AFD = ∠EFC (vertically opposite angles)
∠DAF = ∠CEF (alt. angles equal when AD║BE cut by transversal AE)
AD = EC (∵AD=BC and BC=EC ∴AD=EC)
∴ By AAS Congruence Criterion,
ΔADF is congruent to ΔCEF
∴By CPCT, DF=FC
∴F bisects DC
Since DF = 1/2DC
∴ The area of ΔDFB = 1/2(1/2DC×h) (area of Δ = 1/2×base×height)
=1/4 DC×h
= 1/4 ar(║gm ABCD) (area of ║gm = base×height)
But, arΔDFB = 3cm²
∴ 3cm² = 1/4 ar(║gm ABCD)
∴12cm² = ar(║gm ABCD)
The Figure for the question is attached.
ABCD is a parallelogram
BC=CE
arΔDBF = 3cm²
To Find:
ar(║gm ABCD)
Steps are as follow:
∵AD║BC
∴AD║BE
In ΔADF and ΔCEF,
∠AFD = ∠EFC (vertically opposite angles)
∠DAF = ∠CEF (alt. angles equal when AD║BE cut by transversal AE)
AD = EC (∵AD=BC and BC=EC ∴AD=EC)
∴ By AAS Congruence Criterion,
ΔADF is congruent to ΔCEF
∴By CPCT, DF=FC
∴F bisects DC
Since DF = 1/2DC
∴ The area of ΔDFB = 1/2(1/2DC×h) (area of Δ = 1/2×base×height)
=1/4 DC×h
= 1/4 ar(║gm ABCD) (area of ║gm = base×height)
But, arΔDFB = 3cm²
∴ 3cm² = 1/4 ar(║gm ABCD)
∴12cm² = ar(║gm ABCD)
The Figure for the question is attached.
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