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
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.
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]
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.
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that
ar (BPC) = ar (DPQ).
[Hint: Join AC.]
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?
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.
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.
