Advertisements
Advertisements
Question
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.
Advertisements
Solution
Joining PC we get,
ΔABC and ΔBPC are on the same base BC and between the same parallel lines AP and BC.
∴ A( ΔABC ) = A( ΔBPC ) ....(i)
ΔBPC and ΔBQP are on the same base BP and between the same parallel lines BP and CQ.
∴ A( ΔBPC ) = A( ΔBQP ) ....(ii)
From (i) and (ii), we get
∴A( ΔABC ) = A( ΔBQP )
Hence proved.
APPEARS IN
RELATED QUESTIONS
In the given figure, if the area of triangle ADE is 60 cm2, state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.
The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]
In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
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.
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.
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.

Show that:
A diagonal divides a parallelogram into two triangles of equal area.
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.
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.
