Question

find the perimeter and area of the figure if each unit on the graph measures 1 centimeter. round answers to the nearest tenth if necessary

Explanation:

Step1: Find the length of the sides using the distance formula

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Let's take two adjacent vertices \((- 1,1)\) and \((2,0)\):
\(d_1=\sqrt{(2 + 1)^2+(0 - 1)^2}=\sqrt{9 + 1}=\sqrt{10}\)
Take another pair of adjacent vertices \((2,0)\) and \((3,3)\):
\(d_2=\sqrt{(3 - 2)^2+(3 - 0)^2}=\sqrt{1+9}=\sqrt{10}\)
Since it's a rhombus (all sides equal), perimeter \(P = 4s\), where \(s=\sqrt{10}\)
\(P = 4\sqrt{10}\approx4\times3.16 = 12.6\)

Step2: Find the area using the formula for the area of a rhombus \(A=\frac{1}{2}d_1d_2\)

First, find the length of the diagonals.
Let the vertices be \(A(-1,1)\), \(B(2,0)\), \(C(3,3)\), \(D(0,4)\)
Length of diagonal \(d_1\) (between \(A(-1,1)\) and \(C(3,3)\)):
\(d_1=\sqrt{(3 + 1)^2+(3 - 1)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\)
Length of diagonal \(d_2\) (between \(B(2,0)\) and \(D(0,4)\)):
\(d_2=\sqrt{(0 - 2)^2+(4 - 0)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\)
\(A=\frac{1}{2}\times2\sqrt{5}\times2\sqrt{5}=\frac{1}{2}\times20 = 10\)

Answer:

Perimeter: \(12.6\) centimeters, Area: \(10\) square centimeters