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 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.
ABCD is a trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N.
Prove that: area of Δ ADM = area of Δ ACN.
In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:
(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).
In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.
ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.
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.
In the following figure, BD is parallel to CA, E is mid-point of CA and BD = `1/2`CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )
The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB = 1:2
Find The area of Δ APD.
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.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
