मराठी

In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Advertisements
Advertisements

प्रश्न

In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Advertisements

उत्तर

It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.

∴ AD = BC ... (1)

Similarly, for parallelograms DCEF and ABFE, it can be proved that

DE = CF ... (2)

And, EA = FB ... (3)

In ΔADE and ΔBCF,

AD = BC             ................[Using equation (1)]

DE = CF             ................[Using equation (2)]

EA = FB             ................[Using equation (3)]

∴ ΔADE ≅ BCF (SSS congruence rule)

⇒ Area (ΔADE) = Area (ΔBCF)

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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 below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)


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 ______.


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.


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.]


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].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×