QUESTION IMAGE
Question
- determine the perimeter of trapezoid abcd. round to the nearest tenth, if needed.
- First, assume the coordinates of the vertices of the trapezoid:
- Let's assume \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\), and \(D(x_4,y_4)\) based on the grid - if we assume each square of the grid has a side - length of 1 unit. Suppose \(A(- 3,-2)\), \(B(-1,2)\), \(C(2,2)\), \(D(2,-2)\).
- Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) to find the lengths of the sides.
- Calculate the length of side \(AB\):
- For \(A(-3,-2)\) and \(B(-1,2)\), \(x_1=-3,y_1 = - 2,x_2=-1,y_2 = 2\).
- \(AB=\sqrt{(-1-(-3))^2+(2 - (-2))^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\approx4.5\).
- Calculate the length of side \(BC\):
- For \(B(-1,2)\) and \(C(2,2)\), \(x_1=-1,y_1 = 2,x_2=2,y_2 = 2\).
- \(BC=\sqrt{(2-(-1))^2+(2 - 2)^2}=\sqrt{(3)^2+0^2}=3\).
- Calculate the length of side \(CD\):
- For \(C(2,2)\) and \(D(2,-2)\), \(x_1=2,y_1 = 2,x_2=2,y_2=-2\).
- \(CD=\sqrt{(2 - 2)^2+(-2 - 2)^2}=\sqrt{0^2+(-4)^2}=4\).
- Calculate the length of side \(DA\):
- For \(D(2,-2)\) and \(A(-3,-2)\), \(x_1=2,y_1=-2,x_2=-3,y_2=-2\).
- \(DA=\sqrt{(-3 - 2)^2+(-2-(-2))^2}=\sqrt{(-5)^2+0^2}=5\).
- Calculate the perimeter \(P\) of the trapezoid:
- \(P=AB + BC+CD + DA\).
- \(P\approx4.5+3 + 4+5=16.5\).
Step1: Identify the coordinates
Assume \(A(-3,-2)\), \(B(-1,2)\), \(C(2,2)\), \(D(2,-2)\) based on the grid.
Step2: Calculate length of \(AB\)
Use distance formula \(AB=\sqrt{(-1 - (-3))^2+(2-(-2))^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.5\).
Step3: Calculate length of \(BC\)
Use distance formula \(BC=\sqrt{(2-(-1))^2+(2 - 2)^2}=3\).
Step4: Calculate length of \(CD\)
Use distance formula \(CD=\sqrt{(2 - 2)^2+(-2 - 2)^2}=4\).
Step5: Calculate length of \(DA\)
Use distance formula \(DA=\sqrt{(-3 - 2)^2+(-2-(-2))^2}=5\).
Step6: Calculate the perimeter
\(P = AB+BC + CD+DA\approx4.5+3 + 4+5 = 16.5\).
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16.5