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Ex 9.3, 3 - Show that diagonals of a parallelogram divide - Median divides triangle into two triangles of equal area

Ex 9.3, 3 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Since rectangle, square, rhombus are all parallelograms.

We can also say that

Diagonals of a rectangle divide it into 4 triangles of equal area

OR

Diagonals of a square divide it into 4 triangles of equal area

OR

Diagonals of a rhombus divide it into 4 triangles of equal area

 

 

 

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Transcript

Ex 9.3, 3 Show that the diagonals of a parallelogram divide it into four triangles of equal area. Given: A parallelogram ABCD With diagonals AC & BD To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD) Proof : ABCD is a parallelogram Diagonals of a parallelogram bisect each other ∴ O is the mid-point of BD, i.e., OB = OD & O is the mid-point of AC, i.e., OA = OC In ∆ ABC, Since OA = OC ∴ BO is the median of ∆ ABC ⇒ ar(∆ AOB) = ar(∆ BOC) In ∆ ADC, Since OA = OC ∴ DO is the median of ∆ ADC ⇒ ar(∆ AOD) = ar(∆ COD) Similarly, In ∆ABD, Since OB = OD ∴ AO is the median of ∆ ABD ⇒ ar(∆ AOB) = ar(∆ AOD) From (3) , (4) & (5) ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD) Hence proved

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