Advertisements
Advertisements
प्रश्न
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
Advertisements
उत्तर
It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.
∴ AD = BC ... (1)
Similarly, for parallelograms DCEF and ABFE, it can be proved that
DE = CF ... (2)
And, EA = FB ... (3)
In ΔADE and ΔBCF,
AD = BC ................[Using equation (1)]
DE = CF ................[Using equation (2)]
EA = FB ................[Using equation (3)]
∴ ΔADE ≅ BCF (SSS congruence rule)
⇒ Area (ΔADE) = Area (ΔBCF)
APPEARS IN
संबंधित प्रश्न
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(1) ar ( ADEG) = ar (GBCD)
(2) ar (ΔEGB) = `1/6` ar (ABCD)
(3) ar (ΔEFC) = `1/2` ar (ΔEBF)
(4) ar (ΔEBG) = ar (ΔEFC)
(5)ΔFind what portion of the area of parallelogram is the area of EFG.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.
In the following figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)
ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)
If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).
[Hint: Join BD and draw perpendicular from A on BD.]
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].
