The area of the considered rhombus ABCD is given by: Option: B: 65 square units.
How to find the area of a rhombus from the length of its diagonals?
Suppose that a given rhombus has its diagonals of length [tex]d_1[/tex] and [tex]d_2[/tex] units respectively.
Then, its area is given by:
[tex]A = \dfrac{d_1 \times d_2}{2} \: \rm unit^2[/tex]
The reason for this is because diagonals of a rhombus are perpendicular bisector of each other, so they divide each other in half and are standing vertical on each other. Plus, either side of triangles formed from its diagonals are symmetrical, thus having same area.
Now, one triangle's area = half of its height times base
and since height of a triangle obtained from splitting the rhombus by a diagonal, the other diagonal's half will be its height and that splitting diagonal can be its base), thus,
one triangle's area= [tex]\dfrac{1}{2} \times d_1 \times \dfrac{d_2}{2} = \dfrac{d_1 \times d_2}{4}[/tex] (in sq. units).
Thus, rhombus's area = twice of that triangle's area = [tex]\dfrac{d_1 \times d_2}{2} \: \rm unit^2[/tex]
For the given case, the rhombus in consideration have its diagonals as of 10 units and 13 units respectively.
Thus, its area is evaluated as:
[tex]A = \dfrac{d_1 \times d_2}{2} = \dfrac{10 \times 13}{2} =65\: \rm unit^2[/tex]
Thus, the area of the considered rhombus ABCD is given by: Option: B; 65 square units.
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