Question

6. Prove that of all parallelograms of given sides, the rectangle has the maximum area.
7. Prove that the diagonals of a parallelogram divides it into four triangles of equal areas.
8. The diagonals AC and BD of a quadrilateral ABCD intersect at point '0' and the quadrilateral
divides into four triangles of equal areas. Prove that ABCD is a parallelogram.
9. The diagonal AC of a quadrilateral ABCD divides it into two triangles of equal areas. Prove that
diagonal AC bisects the diagonal BD.
10. Prove that the area of a rhombus is equal to half of the product of its diagonals.
CE​

Answer 1

Answer:

6th solution and 7th solution

Answer 2

Let us consider in a parallelogram ABCD the diagonals AC and BD are cut at point O.

To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Proof:

In parallelogram ABCD the diagonals bisect each other.

AO = OC

In ∆ACD, O is the mid-point of AC. DO is the median.

ar (∆AOD) = ar (COD) ….. (1) [Median of ∆ divides it into two triangles of equal arreas]

Similarly, in ∆ ABC

ar (∆AOB) = ar (∆COB) ….. (2)

In ∆ADB

ar (∆AOD) = ar (∆AOB) …. (3)

In ∆CDB

ar (∆COD) = ar (∆COB) …. (4)

From (1), (2), (3) and (4)

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

Hence proved.

${\fcolorbox{blue}{black}{\blue{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:HumanIntelligence\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}$