Question

A parallelogram has the vertices (-1, 2), (4, 4), (2, -1) and (-3, -3). Determine what type of

parallelogram [10 points]. Find the perimeter and area [20 points].

Answer 1

See explanation

Explanation:

Consider the parallelogram ABCD (see attached diagram). First, find the lengths of all sides:

`AB=√((-1-4)^2+(2-4)^2)=√(25+4)=√(29)\\ \\BC=√((4-2)^2+(4-(-1))^2)=√(4+25)=√(29)\\ \\CD=√((2-(-3))^2+(-1-(-3))^2)=√(25+4)=√(29)\\ \\AD=√((-1+3)^2+(2-(-3))^2)=√(4+25)=√(29)`

Since all sides are of equal length, this parallelogram is rhombus.

Check if it has right angles:

`\overrightarrow{AB}=(5,2)\\ \\\overrightarrow {DA}=(2,5)\\ \\\cos\angle A=\frac{\overrightarrow {AB}\cdot \overrightarrow {DA}}{|\overrightarrow {AB}|\cdot|\overrightarrow{DA}|}=(5\cdot2+2\cdot 5)/(√(29)\cdot√(29))=(20)/(29)\\eq 0`

Angle A is not right angle, so this parallelogram is neither rectangle, nor square.

The perimeter is

`P_(ABCD)=AB+BC+CD+DA=√(29)+√(29)+√(29)+√(29)=4√(29)\ un.`

The area is

`A_(ABCD)=2A_(\triangle ABD)=2\cdot (1)/(2)\cdot AB\cdot AD\cdot \sin \angle A\\ \\\\A_(ABCD)=√(29)\cdot √(29)\cdot \sqrt{1-\left((20)/(29)\right)^2}=29\cdot(√(29^2-20^2))/(29)=21\ un^2.`