The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilate

1.	The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is(a) Rectangle (b) Square (c) Parallelogram (d) Rhombus
Answer» (c) Parallelogram AB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\) BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\) CD =\(\sqrt{(4-1)^2+(3-2)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\) AD = \(\sqrt{(4-1)^2+(5-2)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\) AC = \(\sqrt{(4-4)^2+(5-3)^2}\) = \(\sqrt4\) = 2 BD = \(\sqrt{(7-1)^2+(6-2)^2}\) = \(\sqrt{36+16}\) = \(\sqrt{52}\) = \(2\sqrt{13}\) AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.

The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is(a) Rectangle (b) Square (c) Parallelogram (d) Rhombus

Answer»

(c) Parallelogram

AB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)

BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)

CD =\(\sqrt{(4-1)^2+(3-2)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)

AD = \(\sqrt{(4-1)^2+(5-2)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)

AC = \(\sqrt{(4-4)^2+(5-3)^2}\) = \(\sqrt4\) = 2

BD = \(\sqrt{(7-1)^2+(6-2)^2}\) = \(\sqrt{36+16}\) = \(\sqrt{52}\) = \(2\sqrt{13}\)

AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal.

⇒ ABCD is a parallelogram.

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