Question

prove: abcd is a square.
proof:

• ( ab = sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } ) so ( a b = )
• ( b c = sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } ) so ( b c = )
• ( c d = sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } ) so ( c d = )
• ( d a = sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } ) so ( d a = )

since ( a b c d ) has four angles and four sides, it is a square.

Explanation:

Step1: Calculate length of \(AB\)

For \(A(0,6)\) and \(B(6,6)\), using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we have \(x_1 = 0,y_1=6,x_2 = 6,y_2 = 6\). Then \(AB=\sqrt{(6 - 0)^2+(6 - 6)^2}=\sqrt{36+0}=6\)

Step2: Calculate length of \(BC\)

For \(B(6,6)\) and \(C(6,0)\), \(x_1 = 6,y_1=6,x_2 = 6,y_2 = 0\). Then \(BC=\sqrt{(6 - 6)^2+(0 - 6)^2}=\sqrt{0 + 36}=6\)

Step3: Calculate length of \(CD\)

For \(C(6,0)\) and \(D(0,0)\), \(x_1 = 6,y_1=0,x_2 = 0,y_2 = 0\). Then \(CD=\sqrt{(0 - 6)^2+(0 - 0)^2}=\sqrt{36+0}=6\)

Step4: Calculate length of \(DA\)

For \(D(0,0)\) and \(A(0,6)\), \(x_1 = 0,y_1=0,x_2 = 0,y_2 = 6\). Then \(DA=\sqrt{(0 - 0)^2+(6 - 0)^2}=\sqrt{0+36}=6\)

Since \(ABCD\) has four right angles (as shown by the perpendicular symbol in the graph) and four congruent sides.

Answer:

\(AB = 6\), \(BC = 6\), \(CD = 6\), \(DA = 6\), right, congruent