Advertisements
Advertisements
Question
In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.

If the area of parallelogram ABCD is 48 cm2;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.
Advertisements
Solution
(i) Since triangle BEC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC // AD.
So Area ( ΔBEC )= `1/2 xx "Area" ( square`ABCD )
= `1/2` x 48 = 24 cm2
(ii) Area (` square "ANMD" ) = "Area" ( square` BNMC )
= `1/2"Area" ( square` ABCD)
= `1/2` x 2 x Area ( ΔBEC )
= Area ( ΔBEC )
Therefore, Parallelograms ANMD and NBCM have areas equal to triangle BEC.
APPEARS IN
RELATED QUESTIONS
In the given figure, ABCD is a parallelogram; BC is produced to point X.
Prove that: area ( Δ ABX ) = area (`square`ACXD )
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.
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.
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 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 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.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
