Advertisements
Advertisements
Question
ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F.
If ar.(∆DFB) = 30 cm2; find the area of parallelogram.
Advertisements
Solution
BC = CE .....( given )
Also, in parallelogram ABCD, BC = AD
⇒ AD = CE
Now, in ΔADF and ΔECF, We have
AD = CE
∠ADF = ∠ECF .....( Alternate angles )
∠DAF = ∠CEF ......( Alternate angles )
∴ ΔADF ≅ ΔECF ......( ASA Criterion )
⇒ Area( ΔADF ) = Area( ΔECF ) ....(1)
Also, in ΔFBE, FC is the median ....( Since BC = CE )
⇒ Area( ΔBCF ) = Area( ΔECF ) .....(2)
From (1) and (2)
Area( ΔADF ) = Area( ΔBCF ) ......(3)
Again, ΔADF and ΔBDF are on the base DF and between parallels DF and AB.
⇒ Area( ΔBDF ) = Area( ΔADF ) ........(4)
From (3) and (4),
Area( ΔBDF ) = Area( ΔBCF ) = 30 cm2
Area( ΔBCD ) = Area( ΔBDF ) + Area( ΔBCF ) = 30 + 30 = 60 cm2
Hence, Area of parallelogram ABCD = 2 x Area( ΔBCD ) = 2 x 60 = 120cm2.
APPEARS IN
RELATED QUESTIONS
The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]
The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that:
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.
In the following, AC // PS // QR and PQ // DB // SR.
Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.
The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.
Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
In a parallelogram ABCD, point P lies in DC such that DP: PC = 3:2. If the area of ΔDPB = 30 sq. cm.
find the area of the parallelogram ABCD.
ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = `1/8` of the area of parallelogram ABCD.
ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that the area of ∆ADX = area of ∆ACY.
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of ΔADF is 60 cm2; find
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of ΔADB.
(iii) also, find the area of parallelogram ABCD.
In ΔABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O,
prove that the ΔOBC and quadrilateral AEOF are equal in area.
