Question

In a Trapezium ABCD, diagonals AC and BD intersect at point o and AB = 3DC, then find ratio of area of triangles COD and AOB

Answer 1

The ratio of area of triangles COD and AOB is 1:9.

Step-by-step explanation:

It is given that,

In a trapezium ABCD,  

Diagonals AC and BD intersect at point O.

AB = 3 DC ….. (i)

Let’s consider ∆AOB and ∆COD, we have

∠AOB = ∠COD ….. [since AB // CD, therefore angle AOB & angle COD are vertically opposite angles]

∠ABO = ∠CDO …… [alternate angles, AB//CD & BD is transversal line]

∴ By AA similarity, ∆AOB ~ ∆COD

We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ [Area of ∆COD] / [Area of ∆AOB] is,

= [$\frac{CD}{AB}$]²  

= [$\frac{CD}{3CD}$]² …….. [substituting from (i)]

= [$\frac{1}{3}$]²

= $\frac{1}{9}$

Thus, the ratio of the area of triangles COD and AOB is 1:9.

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Answer 2

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