Question

A parallelogram has vertices at (-2, -1), (4, -1), (0, 2) and (6, 2). The smaller congruent sides of this parallelogram each have a length of 3.6 units. Calculate both the area and perimeter of the parallelogram. Explain or show how you got your answers.

Answer 1

Given:

Vertices of Parallelogram – A(-2,-1), B(4,-1), C(0,2) and D(6,2).

The smaller coungruent sides of Parallelogram is having length of 3.6 units.

To Find:

Area of Parallelogram & Perimeter of Parallelogram.

Solution:

Diagonals of Parallelogram divides it into two congruent triangles, which are ∆ ABC and ∆ ACD.

Therefore,

Area of Parallelogram = Area of ∆ ABC + Area of ∆ ACD.

Formula for calculating Area of Triangle

$= \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

$\bf \underline{ For \: \triangle ABC :}$

$(x_1 , y_1) = A(-2,-1),$ $(x_2,y_2)=B(4,-1),$ $(x_3,y_3)=C(0,2).$

→ Area of ∆ ABC

$= \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

$= \frac{1}{2} \small { | -2 (-1 - 2) + 4(2 - (-1)) + 0(-1 - (-1)) |}$

$= \frac{1}{2} | -2 (-3) + 4(2+1) + 0(-1 +1) |$

$= \frac{1}{2} | 6 + 4(3) + 0(0) |$

$= \frac{1}{2} | 6 + 12 + 0 |$

$= \frac{1}{2} | 18 |$

$= \frac{18}{2}$

$= \cancel{\frac{18}{2}}$

$\therefore \fbox {Area \: of \: ∆ ABC = 9 \: units.}$

$\bf \underline {For \: \triangle ACD :}$

$(x_1 , y_1) = A(-2,-1),$ $(x_2,y_2)=D(6,2),$ $(x_3,y_3)=C(0,2).$

→ Area of ∆ ACD

$= \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

$= \frac{1}{2} \small { | -2 (2 - 2) + 6(2 - (-1)) + 0(-1 - 2) |}$

$= \frac{1}{2} | -2 (0) + 6(2+1) + 0(-3) |$

$= \frac{1}{2} | 0 + 6(3) + 0 |$

$= \frac{1}{2} | 0 + 18 + 0 |$

$= \frac{1}{2} | 18 |$

$= \frac{18}{2}$

$= \cancel{\frac{18}{2}}$

$\therefore \fbox {Area \: of \: ∆ ACD = 9 \: units.}$

$\Rightarrow$ Area of Parallelogram = Area of ∆ ABC + Area of ∆ ACD.

$\therefore$Area of Parallelogram = 9 units + 9 units

$\therefore \fbox {Area \: of \: Parallelogram = 18 \: units.}$

______________________

Perimeter of Parallelogram = 2(a + b)

where a = side and b = base of Parallelogram.

a = AD = 3.6 units (given), b = AB = ?

We can find base by using distance formula:

$\bf \underline{ For \: line \: segment \: AB :}$

$(x_1 , y_1) = A(-2,-1),$ $(x_2,y_2)=B(4,-1),$

$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$AB = \sqrt{(4-(-2))^2 + (-1-(-1))^2}$

$AB = \sqrt{(4+2)^2 + (-1+1)^2}$

$AB = \sqrt{(6)^2 + (0)^2}$

$AB = \sqrt{36+0}$

$\therefore \fbox{AB = 6 units = b}$

$\Rightarrow$ Perimeter of Parallelogram = 2(a + b)

= 2(3.6 + 6) units

= 2(9.6) units

$\therefore \fbox{ Perimeter \: of \: Parallelogram = 19.2 \: units}$

Hence, the area of Parallelogram is 18 units and Perimeter of Parallelogram is 19.2 units.