[tex]PQRS[/tex] is a rhombus since [tex]\overline{QR} \cong \overline{RS}[/tex] and [tex]m\angle R \neq 0^{\circ}[/tex], [tex]m\angle R \neq 90^{\circ}[/tex] and [tex]m\angle R \neq 180^{\circ}[/tex]. [tex]\blacksquare[/tex]
How to demonstrate that a given figure represents a rhombus
In this question we must prove that a given parallelogram is a rhombus. A rhombus is quadrilateral with four sides of equal length and two pairs of angles of equal measure, different to each other. We must demonstrate that [tex]\overline {QR} \cong \overline {RS}[/tex] and [tex]m\angle R \neq 90^{\circ}[/tex] and [tex]m\angle R \neq 0^{\circ}[/tex] and [tex]m\angle R \neq 180^{\circ}[/tex] by geometric and algebraic definitions and theorems:
- [tex]Q(x,y) = (8,5)[/tex], [tex]R(x,y) = (5,1)[/tex], [tex]S(x,y) = (2,5)[/tex] Given
- [tex]QR = \sqrt{(5-8)^{2}+(1-5)^{2}} = 5[/tex] Pythagorean theorem/Line segment length formula
- [tex]RS = \sqrt{(2-5)^{2}+(5-1)^{2}} = 5[/tex] Pythagorean theorem/Line segment length formula
- [tex]QR = RS[/tex] (2) and (3)
- [tex]\overrightarrow {QR} \,\bullet\,\overrightarrow{RS} = (-3)\cdot (-3) + (-4)\cdot (4) = -7[/tex] Dot product
- [tex]\theta = \cos^{-1}\left(\frac{\overrightarrow{QR}\,\bullet\,\overrightarrow{RS}}{QR\cdot RS} \right) = \cos^{-1}\left(-\frac{7}{25} \right) \approx 106.260^{\circ}[/tex] Dot product
- [tex]\overline{QR} \cong \overline {RS}[/tex] (4) and (6)
- [tex]PQRS[/tex] is a rhombus. (6) and (7)/Result
[tex]PQRS[/tex] is a rhombus since [tex]\overline{QR} \cong \overline{RS}[/tex] and [tex]m\angle R \neq 0^{\circ}[/tex], [tex]m\angle R \neq 90^{\circ}[/tex] and [tex]m\angle R \neq 180^{\circ}[/tex]. [tex]\blacksquare[/tex]
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