Advertisements
Advertisements
प्रश्न
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.
उत्तर
Given: ABCD is a parallelogram in which BC is produced to E such that CE = BC. C is the mid-point BE and ar (ΔDFB) = 3 cm2.
In triangle, ADF and triangle EFC,
∠DAF = ∠CEF ...[Alternate interior angle]
AD = CE ...[AD = BC = CE]
∠ADF = ∠FCE ...[Alternate interior angle]
So, ΔADF ≅ ΔECF ...[By SAS rule of congruence]
Now, ΔADF ≅ ΔECF ...[By SAS rule of congruence]
DF = CF ...[CPCT]
As BF is the median of triangle BCD.
ar (ΔBDF) = `1/2` ar (BCD) ...(i) [Median divides a triangle into two triangle of equal area]
As we know that a triangle and parallelogram are on the same base and between the same parallels then area of the triangles is equal to half the area of the parallelogram.
ar (ΔBCD) = `1/2` ar (ABCD) ...(ii)
ar (ΔBDF) = `1/2 {1/2 "ar (ABCD)"}` ...[By equation (i)]
3 cm2 = `1/4` ar (ABCD)
ar (ABCD) = 12 cm2
Hence, the area of the parallelogram is 12 cm2.
APPEARS IN
संबंधित प्रश्न
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2ar (PQRS)
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.
In which of the following figures, you find two polygons on the same base and between the same parallels?
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is ______.
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).
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)
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].
