Question

find the perimeter and area of this figure.
p = ? units
a = units²

Explanation:

Step1: Find the lengths of the sides

Let the vertices be \(A(2,2)\), \(B(8,2)\), \(C(5,9)\).
Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(AB\): \(x_1 = 2,y_1 = 2,x_2 = 8,y_2 = 2\)
\(AB=\sqrt{(8 - 2)^2+(2 - 2)^2}=\sqrt{36+0}=6\)
For \(AC\): \(x_1 = 2,y_1 = 2,x_2 = 5,y_2 = 9\)
\(AC=\sqrt{(5 - 2)^2+(9 - 2)^2}=\sqrt{9 + 49}=\sqrt{58}\approx7.62\)
For \(BC\): \(x_1 = 8,y_1 = 2,x_2 = 5,y_2 = 9\)
\(BC=\sqrt{(5 - 8)^2+(9 - 2)^2}=\sqrt{9+49}=\sqrt{58}\approx7.62\)

Step2: Calculate the perimeter \(P\)

\(P=AB + AC+BC\)
\(P=6+7.62+7.62 = 21.24\)

Step3: Calculate the area \(A\)

Base \(b = AB = 6\), height \(h\) is the vertical distance from \(C\) to \(AB\). Since \(AB\) is on \(y = 2\) and \(C\) has \(y = 9\), \(h=9 - 2=7\)
Using the formula \(A=\frac{1}{2}\times b\times h\)
\(A=\frac{1}{2}\times6\times7 = 21\)

Answer:

\(P = 21.24\) units, \(A = 21\) units\(^2\)