Question

segment pq
$d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$m = \frac{4 - 0}{3 - 0} = \frac{4}{3}$
segment qr
$d = \sqrt{(8 - 3)^2 + (4 - 4)^2} = \sqrt{25} = 5$
$m = \frac{4 - 4}{8 - 3} = 0$
segment rs
$d = \sqrt{(5 - 8)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$m = \frac{0 - 4}{5 - 8} = \frac{4}{3}$
segment sp
$d = \sqrt{(0 - 5)^2 + (0 - 0)^2} = \sqrt{25} = 5$
$m = \frac{0 - 0}{0 - 5} = 0$
what is the best classification for quadrilateral pqrs?
rhombus
square
rectangle
parallelogram

Explanation:

Step1: Recall the properties of quadrilaterals

• Rhombus: All sides are equal, and the slopes of adjacent sides are not perpendicular (product of slopes is not - 1).
• Square: All sides are equal, adjacent sides are perpendicular (product of slopes = - 1).
• Rectangle: Opposite sides are equal, adjacent sides are perpendicular.
• Parallelogram: Opposite sides are equal and parallel (equal slopes for opposite sides).

Step2: Analyze the side - lengths and slopes

• Side - lengths: \(PQ=\sqrt{(3 - 0)^2+(4 - 0)^2}=5\), \(QR=\sqrt{(8 - 3)^2+(4 - 4)^2}=5\), \(RS=\sqrt{(5 - 8)^2+(0 - 4)^2}=5\), \(SP=\sqrt{(0 - 5)^2+(0 - 0)^2}=5\). All sides are equal.
• Slopes:
• Slope of \(PQ=m_{PQ}=\frac{4-0}{3 - 0}=\frac{4}{3}\)
• Slope of \(QR=m_{QR}=\frac{4 - 4}{8 - 3}=0\)
• Slope of \(RS=m_{RS}=\frac{0 - 4}{5 - 8}=\frac{4}{3}\)
• Slope of \(SP=m_{SP}=\frac{0 - 0}{0 - 5}=0\)
• Parallel sides: \(m_{PQ}=m_{RS}=\frac{4}{3}\), \(m_{QR}=m_{SP}=0\). So, opposite sides are parallel.
• Adjacent sides:
• Product of slopes of \(PQ\) and \(QR\): \(\frac{4}{3}\times0 = 0

eq-1\) (not perpendicular)

Answer:

Rhombus