prove that the diagonals of parallelogram divides it into 4 triangle of equal area
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Answered by
2
consider a ║gm ABCD.
as we know that the diagonals of a ║gm divide it into two triangles of equal areas, so
ar. (ADC) = ar. (BDC)
now, in tri. ADC, OD is the median so,
ar. ( AOD) = ar. (AOB ) ....1
similarly, ar. (COD) = ar. (BOC) ....2
NOW, ar. (ADC) = ar. (BDC)
⇒ 1/2 ar.(ADC) = 1/2 ar. (BDC)
from 1 and 2,
ar. (AOD) = ar. (AOB) = ar. (COD) = ar. (BOC)
hence, proved.
as we know that the diagonals of a ║gm divide it into two triangles of equal areas, so
ar. (ADC) = ar. (BDC)
now, in tri. ADC, OD is the median so,
ar. ( AOD) = ar. (AOB ) ....1
similarly, ar. (COD) = ar. (BOC) ....2
NOW, ar. (ADC) = ar. (BDC)
⇒ 1/2 ar.(ADC) = 1/2 ar. (BDC)
from 1 and 2,
ar. (AOD) = ar. (AOB) = ar. (COD) = ar. (BOC)
hence, proved.
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Dakshansh:
plz mark as the brainliest....: )
Answered by
3
ABCD is a parallelogram in which the diagonals bisect each other
In ΔADC
OD is the median
∴ar DOA = ar DOC ________ (1)
similarly in ΔADB ,OA i s the median
∴ar DOA = ar BOA________________(2)
similarly we can prove
ar AOB = ar BOC_______________(3)
and
ar BOC = ar DOC _______________(4)
from 1 , 2 , 3 and 4
we get
ar (DOC)= ar(DOA)=ar(AOB)= ar(BOC)
PROVED.......
In ΔADC
OD is the median
∴ar DOA = ar DOC ________ (1)
similarly in ΔADB ,OA i s the median
∴ar DOA = ar BOA________________(2)
similarly we can prove
ar AOB = ar BOC_______________(3)
and
ar BOC = ar DOC _______________(4)
from 1 , 2 , 3 and 4
we get
ar (DOC)= ar(DOA)=ar(AOB)= ar(BOC)
PROVED.......
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