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.
Advertisements
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, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
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).
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)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through. P, draw a line parallel to AB]
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 ______.
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).
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)
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].
