Question

find the perimeter (circumference) and the area of each figure

Explanation:

Left Figure (Circle)

Step1: Calculate the radius

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here \(A(5,3)\) and \(B(2,1)\), so \(r=\sqrt{(5 - 2)^2+(3 - 1)^2}=\sqrt{9 + 4}=\sqrt{13}\)

Step2: Calculate the circumference

The formula for the circumference of a circle is \(C = 2\pi r\). Substitute \(r=\sqrt{13}\), so \(C = 2\pi\sqrt{13}\approx2\times3.14\times3.61\approx22.67\)

Step3: Calculate the area

The formula for the area of a circle is \(A=\pi r^{2}\). Substitute \(r = \sqrt{13}\), so \(A=\pi\times(\sqrt{13})^{2}=13\pi\approx13\times3.14 = 40.82\)

Right Figure (Triangle)

Step1: Calculate the base length

The base \(AC\) with \(A(2,1)\) and \(C(10,1)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), since \(y_1=y_2 = 1\), \(b=\vert10 - 2\vert=8\)

Step2: Calculate the height

The height \(h\) is the vertical distance from \(B(8,6)\) to the line \(y = 1\). So \(h=\vert6 - 1\vert=5\)

Step3: Calculate the perimeter

First, find \(AB=\sqrt{(8 - 2)^2+(6 - 1)^2}=\sqrt{36+25}=\sqrt{61}\approx7.81\), \(BC=\sqrt{(10 - 8)^2+(1 - 6)^2}=\sqrt{4 + 25}=\sqrt{29}\approx5.39\), \(AC = 8\). Perimeter \(P=AB + BC+AC\approx7.81+5.39 + 8=21.2\)

Step4: Calculate the area

The formula for the area of a triangle is \(A=\frac{1}{2}bh\). Substitute \(b = 8\) and \(h = 5\), so \(A=\frac{1}{2}\times8\times5=20\)

Answer:

• Left Figure (Circle):
• Circumference: \(2\pi\sqrt{13}\approx22.67\)
• Area: \(13\pi\approx40.82\)
• Right Figure (Triangle):
• Perimeter: \(\sqrt{61}+\sqrt{29}+8\approx21.2\)
• Area: \(20\)