Advertisements
Advertisements
प्रश्न
In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)
Advertisements
उत्तर
Given that, ABCD and AEFD are two parallelograms
To prove: (1) PE = FQ
(2) `"ar (ΔAPE)"/ "ar (ΔPFA)"` = `"ar (ΔQFD)"/"ar (ΔPED)"`
(3) ar (ΔPEA) = ar (ΔQFD)
Proof: (1) In ΔEPA and ΔFQD
∠PEA = ∠QFD [ ∴ Corresponding angles]
∠EPA = ∠FQD [Corresponding angles]
PA = QD [opp .sides of 11gm]
Then, ΔEPA ≅ ΔFQD [By. AAS condition]
∴ EP = FQ [c. p. c.t]
(2) Since, ΔPEA and ΔQFD stand on the same base PEand FQlie between the same
parallels EQ and AD
∴ ar (ΔPEA ) = ar (ΔQFD) → (1)
AD ∴ ar (ΔPFA) = ar (PFD) .....(2)
Divide the equation (1) by equation (2)
`"area of (ΔPEA)"/"area of (ΔPFA)"` = `"ar Δ(QFD)"/"ar Δ(PFD)"`
(3) From (1) part ΔEPA ≅ FQD
Then, ar (ΔEDA) = ar (ΔFQD)
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).
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)
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
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]
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 ______.
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.
