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QUESTION IMAGE

find the perimeter of the following shape, rounded to the nearest tenth:

Question

find the perimeter of the following shape, rounded to the nearest tenth:

Explanation:

Step1: Find coordinates of points

Assume the coordinates of \(A(5,3)\), \(B(7, - 1)\), \(C(3,-3)\), \(D(1, - 1)\)

Step2: Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

For \(AB\): \(x_1 = 5,y_1 = 3,x_2=7,y_2=-1\)
\(AB=\sqrt{(7 - 5)^2+(-1 - 3)^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.5\)
For \(BC\): \(x_1 = 7,y_1=-1,x_2 = 3,y_2=-3\)
\(BC=\sqrt{(3 - 7)^2+(-3 + 1)^2}=\sqrt{16 + 4}=\sqrt{20}\approx4.5\)
For \(CD\): \(x_1 = 3,y_1=-3,x_2 = 1,y_2=-1\)
\(CD=\sqrt{(1 - 3)^2+(-1 + 3)^2}=\sqrt{4 + 4}=\sqrt{8}\approx2.8\)
For \(DA\): \(x_1 = 1,y_1=-1,x_2 = 5,y_2=3\)
\(DA=\sqrt{(5 - 1)^2+(3 + 1)^2}=\sqrt{16 + 16}=\sqrt{32}\approx5.7\)

Step3: Calculate the perimeter \(P\)

\(P=AB + BC+CD + DA\)
\(P=\sqrt{20}+\sqrt{20}+\sqrt{8}+\sqrt{32}\approx4.5+4.5 + 2.8+5.7=17.5\)

Answer:

\(17.5\)