Question

prove that the area of any quadrilateral with perpendicular diagonals =1/2*product of diagonals

Answer 1

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A quadrilateral ABCD in which  the diagonals AC and BD intersect at O assume that AO = OC , BO = OD . AC perpendicular BD.

To prove : ABCD is a rhombus.
Proof : Since the diagonals  AC and BD of quadrilateral ABCD bisect each other  at right  angles.
Therefore,  AC is the perpendicular bisector of the segment BD.
A and C both are equidistant from B and D.
AB = AD and CB = CD        ... (1)
Also , BD is the perpendicular bisector of line segment  AC.
B and D both  are equidistant from A and C.
AB = BC and AD = DC        ... (2)
From (1) and (2), we get
AB = BC = CD = AD
Thus , ABCD is a quadrilateral  whose diagonals bisect each other  at right angles and all four sides are equal.
Hence , ABCD is a rhombus.

now it is proved tha ABCD is a rombus and area of Rombus =1/2 product of diagonals.
Hence, proved

Answer 2

a quadrilateral ABCD In which diagonal AC And BC intersect at O assume that AO=OC And BO = 0D.

TO PROVE:

Area of ABCD =

$\frac{1}{2} \times ac \times bd \:$

Area of triangle (∆ABC)=

$\frac{1}{2} \times ac \times ob = (1)$

Area of (∆ADC) =

$\frac{1}{2} \times ac \times od = (2)$

$ii)ar(abcd) = ar(abc) + ar(adc)$

$|| area \: of \: equillateral \: triangle \: =$

$\frac{ \sqrt{3} }{4} \times (side) {}^{2} = (\frac{ \sqrt{3} }{4 {}^{2} } ) {}^{2}$

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