In the given, ABCD is a quadrilateral with diagonals AC and BD intersecting at point O. Hence, area of $Δ$ AOD $\times$ area of $Δ$ BOC = area of $Δ$ AOB $\times$ area of $Δ$ COD.

True

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False

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Solution

The correct option is A
True

Whenever, the triangles have their bases along the same line and vertices at the same point, the ratio between their areas is equal to the ratio between their bases.

Therefore, for triangles AOB and AOD,

$\frac{A r e a o f Δ A O B}{A r e a o f Δ A O D}$ = $\frac{B O}{D O}$ - (I)

And, for triangles BOC and COD,

$\frac{A r e a o f Δ B O C}{A r e a o f Δ C O D}$ = $\frac{B O}{D O}$ -(II)

Combining equations I and II, we get:

$\frac{A r e a o f Δ A O B}{A r e a o f Δ A O D}$ = $\frac{A r e a o f Δ B O C}{A r e a o f Δ C O D}$

area of $Δ$ AOD $\times$ area of $Δ$ BOC = area of $Δ$ AOB $\times$ area of $Δ$ COD

Suggest Corrections

Similar questions

Diagonals AC and BD of a trapezium ABCD with AB ll DC intersect each other at O. If area (∆AOD)=37sq cm, then area (∆BOC)= ?

\frac{Δ A O B}{Δ B O C}

=

\frac{Δ A O D}{Δ D O C}

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.Then

\frac{Area (∆ AOB)}{Area (∆ COD)}

\frac{Area (∆ AOD)}{Area (∆ COD)}

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