In rhombus ABCD,overline{AC} cap overline{BD} | vedclass

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(C) The diagonals of a rhombus bisect each other at right angles.This division creates four congruent triangles within the rhombus.Therefore,$\Delta O

Vedclass · Accepted Answer

$\frac{1}{4}$

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(C) The diagonals of a rhombus bisect each other at right angles.This division creates four congruent triangles within the rhombus.Therefore,$\Delta OAB \cong \Delta OCB \cong \Delta OCD \cong \Delta OAD$.Since these four triangles are congruent,their areas are equal:$	ext{Area}(\Delta OAB) = 	ext{Area}(\Delta OCB) = 	ext{Area}(\Delta OCD) = 	ext{Area}(\Delta OAD)$.The total area of the rhombus $ABCD$ is the sum of the areas of these four triangles:$	ext{Area}(\square ABCD) = 	ext{Area}(\Delta OAB) + 	ext{Area}(\Delta OCB) + 	ext{Area}(\Delta OCD) + 	ext{Area}(\Delta OAD)$.Substituting the equal areas,we get:$	ext{Area}(\square ABCD) = 4 	imes 	ext{Area}(\Delta OAB)$.Thus,$	ext{Area}(\Delta OAB) = \frac{1}{4} 	imes 	ext{Area}(\square ABCD)$.

In rhombus $ABCD$,$\overline{AC} \cap \overline{BD} = \{O\}$. Then the area of $\Delta OAB = \ldots \ldots \ldots \ldots \times$ area of $\square ABCD$.

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