Answer
The areas of the four resulting smaller triangles is one fourth of the area of the given rhombus.
Work Step by Step
Draw diagonals MP and QN. the intersection of diagonals is named as O.
Let us compare $\triangle$ OPN and $\triangle$ OQP
we know in rhombus diagonals are perpendicular bisectors to each other
So OP = OM and OQ = ON
$\angle$ PON = $\angle$ POQ = $\angle$ MON = $\angle$ MOQ = 90$^{\circ}$
In $\triangle$ OPN and $\triangle$ OQP
PN =QP(in rhombus opposite sides are equal)
OP = OP (common)
$\angle$ PON = $\angle$ POQ = 90$^{\circ}$
By SAS rule $\triangle$ OPN is congruent to $\triangle$ OQP
Therefore area of $\triangle$ OPN = area of $\triangle$ OQP
Since In rhombus sides are equal
Similarly
area of $\triangle$ OMN = area of $\triangle$ OMQ
area of $\triangle$ PON = area of $\triangle$ OMN
area of $\triangle$ OQP = area of $\triangle$ OQM
so
area of $\triangle$ OQP = area of $\triangle$ OPN = area of $\triangle$ OQM = area of $\triangle$ OQN
Therefore area of MNPQ = area of $\triangle$ OQP + area of $\triangle$ OPN + area of $\triangle$ OQM + area of $\triangle$ OQN.
Therefore area of MNPQ = 4 * area of $\triangle$ OQP
area of $\triangle$ OQP = $\frac{area of MNPQ}{4}$.
Therefore area of smaller triangle is one fourth of the area of rhombus MNPQ.
