Advertisements
Advertisements
Question
In the following, AC // PS // QR and PQ // DB // SR.
Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.
Advertisements
Solution
In Parallelogram PQRS,
AC // PS // QR and PQ // DB // SR.
Similarly, AQRC and APSC are also parallelograms.
Since ΔABC and parallelogram AQRC are on the same base AC and between the same parallels, then
A( ΔABC ) = `1/2`A(AQRC) ......(i)
Similarly,
A( ΔADC ) = `1/2`A( APSC ) .......(ii)
Adding (i) and (ii), we get
Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.
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.
In the given figure, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.
Prove that: Area of ABC = Area of // gm BDEC.
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 parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
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 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.
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.
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
