Question

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

Explanation:

Step1: Label the vertices

Let the vertices be \(A(1,2)\), \(B(4,6)\), \(C(8,2)\)

Step2: Calculate side lengths

Use distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

• \(AB=\sqrt{(4 - 1)^2+(6 - 2)^2}=\sqrt{9 + 16}=5\)
• \(BC=\sqrt{(8 - 4)^2+(2 - 6)^2}=\sqrt{16 + 16}= \sqrt{32}=4\sqrt{2}\approx5.66\) (incorrect, actually \(BC=\sqrt{(8 - 4)^2+(2 - 6)^2}=\sqrt{16 + 16}=\sqrt{32}\) is wrong, correct: \(AC=\sqrt{(8 - 1)^2+(2 - 2)^2}=7\), \(AB=\sqrt{(4 - 1)^2+(6 - 2)^2}=5\), \(BC=\sqrt{(8 - 4)^2+(2 - 6)^2}=5\))
• \(AC=\sqrt{(8 - 1)^2+(2 - 2)^2}=7\)
• \(P=5 + 5+6=16\) (using base \(AC = 7\) is wrong, correct base \(AC=6\) (from \(x = 1\) to \(x = 7\)), height \(h = 4\) (from \(y = 2\) to \(y = 6\)) )

Step3: Calculate area

Use formula \(A=\frac{1}{2}\times base\times height\)

• Base \(b=6\) (distance from \(x = 1\) to \(x = 7\) on \(y = 2\) line), height \(h = 4\) (from \(y = 2\) to \(y = 6\))
• \(A=\frac{1}{2}\times6\times4 = 12\)

Answer:

$P = 16$ units
$A = 12$ units²