Question

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pls answer.. tomorrow is my test

Q. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer 1

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✨ $<b>Here's ur answer$

➡️ Ans. Let parallelogram be ABCD and its diagonals AC and BD intersect each other at O.

In ABC and ADC,

AB = DC [Opposite sides of a parallelogram]

BC = AD [Opposite sides of a parallelogram]

And AC = AC [Common]

ABC CDA [By SSS congruency]

Since, diagonals of a parallelogram bisect each other.

O is the mid-point of bisection.

Now in ADC, DO is the median.

ar (AOD) = ar (COD) ……….(i)

[Median divides a triangle into two equal areas]

Similarly, in ABC, OB is the median.

ar (AOB) = ar (BOC) ……….(ii)

And in AOB and AOD, AO is the median.

ar (AOB) = ar (AOD) ……….(iii)

From eq. (i), (ii) and (iii),

ar (AOB) = ar (AOD) = ar (BOC) = ar (COD)

Thus diagonals of parallelogram divide it into four triangles of equal area.

Answer 2

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