Advertisements
Advertisements
Question
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).
Advertisements
Solution
Given: In a parallelogram PSDA, points Q and R are on PS such that
PQ = QR = RS and PA || QB || RC.
To prove: ar (PQE) = ar (CFD)
Proof: In parallelogram PABQ,
And PA || QB ...[Given]
So, PABQ is a parallelogram.
PQ = AB ...(i)
Similarly, QBCR is also a parallelogram.
QR = BC ...(ii)
And RCDS is a parallelogram.
RS = CD ...(iii)
Now, PQ = QR = RS ...(iv)
From equations (i), (ii), (iii) and (iv),
PQ || AB ...[∴ In parallelogram PSDA, PS || AD]
In ΔPQE and ΔDCF,
∠QPE = ∠FDC ...[Since, PS || AD and PD is transversal, then alternate interior angles are equal]
PQ = CD ...[From equation (v)]
And ∠PQE = ∠FCD ...[∴ ∠PQE = ∠PRC corresponding angles and ∠PRC = ∠FCD alternate interior angles]
ΔPQE = ΔDCF ...[By ASA congruence rule]
∴ ar (ΔPQE) = ar (ΔCFD) ...[Since, congruent figures have equal area]
Hence proved.
APPEARS IN
RELATED QUESTIONS
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.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2ar (ABCD)
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)
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?
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 ______.
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.]
