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Question
ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that: (i) ΔAOB and ΔCOD (ii) If OA = 6 cm, OC = 8 cm,
Find:(a) `("area"(triangleAOB))/("area"(triangleCOD))`
(b) `("area"(triangleAOD))/("area"(triangleCOD))`
Solution
We have,
AB || DC
In ΔAOB and ΔCOD
∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate interior angles]
Then, ΔAOB ~ ΔCOD [By AA similarity]
(a) By area of similar triangle theorem
`("area"(triangleAOB))/("area"(triangleCOD))="OA"^2/"OC"^2=6^2/8^2=36/64=9/16`
(b) Draw DP ⊥ AC
`therefore("area"(triangleAOD))/("area"(triangleCOD))=(1/2xxAOxxDP)/(1/2xxCOxxDP)`
`="AO"/"CO"=6/8=3/4`
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