Lets understand the Area of Triangle with the help of Numerical. Click on the Link to Watch the VIDEO explanation: Watch Video
Numerical
In a trapezium ABCD, AB is parallel to CD and AB is equal to 2CD. If the diagonals intersect at O, show that the area of triangle AOB is equal to four times the area of triangle COD.
Solution:
In triangles AOB and COD,
angle AOB is equal to angle DOC since the vertically opposite angles.
angle OAB is equal to angle OCD since the alternate angles.
Therefore, triangle AOB is similar to triangle COD because of AA similarity.
From the Theorem on areas we have area of triangle AOB by area of triangle COD is equal to AB square by CD square.
this is equal to 2CD the whole square by CD square. Since AB is equal to 2CD given equals 4CD square by CD square which is equal to 4.
Therefore, area of triangle AOB by area of triangle COD is equal to 4 by 1.
Therefore, area of triangle AOB is equal to 4 times the area of triangle COD
In a trapezium ABCD, AB is parallel to CD and AB is equal to 2CD. If the diagonals intersect at O, show that the area of triangle AOB is equal to four times the area of triangle COD.
Solution:
In triangles AOB and COD,
angle AOB is equal to angle DOC since the vertically opposite angles.
angle OAB is equal to angle OCD since the alternate angles.
Therefore, triangle AOB is similar to triangle COD because of AA similarity.
From the Theorem on areas we have area of triangle AOB by area of triangle COD is equal to AB square by CD square.
this is equal to 2CD the whole square by CD square. Since AB is equal to 2CD given equals 4CD square by CD square which is equal to 4.
Therefore, area of triangle AOB by area of triangle COD is equal to 4 by 1.
Therefore, area of triangle AOB is equal to 4 times the area of triangle COD
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