Question

1. sheila is building an addition to a house. the points ( eleft(-1 \frac{1}{2},-2 \frac{1}{2}

ight), fleft(4 \frac{1}{2},-2 \frac{1}{2}
ight) ), ( gleft(4 \frac{1}{2}, 3 \frac{1}{2}
ight) ), and ( hleft(-1 \frac{1}{2}, 3 \frac{1}{2}
ight) ) are the points she plotted on a coordinate plane to draw the new room plan. what is the shape of the addition to the house? what is the perimeter in units?

Explanation:

Step1: Determine the shape

• Calculate the distance between \(E(-1\frac{1}{2},-2\frac{1}{2})\) and \(F(4\frac{1}{2},-2\frac{1}{2})\):
• Using the distance formula for points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1 = y_2\), \(d=\vert x_2 - x_1\vert\).
• \(d_{EF}=\vert4\frac{1}{2}-(-1\frac{1}{2})\vert=\vert4.5 + 1.5\vert = 6\)
• Calculate the distance between \(F(4\frac{1}{2},-2\frac{1}{2})\) and \(G(4\frac{1}{2},3\frac{1}{2})\):
• Using the distance formula for points \((x_1,y_1)\) and \((x_2,y_2)\) with \(x_1 = x_2\), \(d=\vert y_2 - y_1\vert\).
• \(d_{FG}=\vert3\frac{1}{2}-(-2\frac{1}{2})\vert=\vert3.5+2.5\vert = 6\)
• Calculate the distance between \(G(4\frac{1}{2},3\frac{1}{2})\) and \(H(-1\frac{1}{2},3\frac{1}{2})\):
• Using the distance formula for points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1 = y_2\), \(d=\vert x_2 - x_1\vert\).
• \(d_{GH}=\vert-1\frac{1}{2}-4\frac{1}{2}\vert=\vert-6\vert = 6\)
• Calculate the distance between \(H(-1\frac{1}{2},3\frac{1}{2})\) and \(E(-1\frac{1}{2},-2\frac{1}{2})\):
• Using the distance formula for points \((x_1,y_1)\) and \((x_2,y_2)\) with \(x_1 = x_2\), \(d=\vert y_2 - y_1\vert\).
• \(d_{HE}=\vert-2\frac{1}{2}-3\frac{1}{2}\vert=\vert-6\vert = 6\)
• Since \(EF = FG=GH = HE = 6\) and adjacent sides are perpendicular (horizontal and vertical lines), the shape is a square.

Step2: Calculate the perimeter

• For a square with side length \(s\), the perimeter formula is \(P = 4s\).
• Here \(s = 6\), so \(P=4\times6=24\)

Answer:

The shape is a square and the perimeter is \(24\) units.