diagonals AC and BD of a trapezium ABCD intersects at O .prove that ar(AOD)=ar(BOC)
Answers
Answered by
6
It is given that:
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
(BY adding area(AOB) to both sides)
Area (ΔADB) = Area (ΔACB)
We know that triangles on the same base having areas equal to each other lie between the same parallels
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels i.e.,
AB || CD
Therefore, ABCD is a trapezium
HOPE IT'S HELPFULLY ^_^ ^_^
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
(BY adding area(AOB) to both sides)
Area (ΔADB) = Area (ΔACB)
We know that triangles on the same base having areas equal to each other lie between the same parallels
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels i.e.,
AB || CD
Therefore, ABCD is a trapezium
HOPE IT'S HELPFULLY ^_^ ^_^
rakshit233:
I am from vijapur
state
that only
what
Karnataka
ok thanks again
for your answer
kkk
find the ratio in which the point (-3 p) divides the line segment joing the point (-5 4) and (-2 3) and find the value of p
Can u help
Answered by
4
attachment have the solution
Attachments:
Similar questions