मराठी

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)

Advertisements
Advertisements

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)

 

Advertisements

उत्तर

Let us join HF.

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD (Opposite sides of a parallelogram are equal)

⇒ 1/2AD = 1/2BC and AH || BF

⇒ AH = BF and AH || BF (∵ H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram.

Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,

∴ Area (ΔHEF) = 1/2Area (ABFH) ... (1)

Similarly, it can be proved that

Area (ΔHGF) = 1/2Area (HDCF) ... (2)

On adding equations (1) and (2), we obtain

Area(ΔHEF) + Area(ΔHGF) = 1/2Area (ABFH) + 1/2Area (HDCF)

                                       = 1/2[Area (ABFH) + Area (HDCF)]

⇒ Area(EFGH) = 1/2Area(ABCD)

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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)


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?


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 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.]


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.


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.


ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = `7/9` ar (XYBA)


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×