Question

Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-2, 2)A(−2,2)A, left parenthesis, minus, 2, comma, 2, right parenthesis, B(6, 2)B(6,2)B, left parenthesis, 6, comma, 2, right parenthesis, C(6, 3)C(6,3)C, left parenthesis, 6, comma, 3, right parenthesis, and D(-2, 3)D(−2,3)D, left parenthesis, minus, 2, comma, 3, right parenthesis.
What is the area of rectangle ABCDABCDA, B, C, D?

Answer 1

The area of the rectangle ABCD will be 8 sq. units.

Step-by-step explanation:

The rectangle ABCD has vertices A(-2,2), B(6,2), C(6,3) and D(-2,3).

Now, length of side AB = [6 - (- 2)] = 8 units.

{Since the line AB is parallel to the x-axis, so the length of the segment AB will be the difference between the x-coordinates of the points A and B}

Again, line BC is parallel to y-axis and hence the length BC = (3 - 2) = 1 units.

Therefore, the area of the rectangle ABCD will be (8 × 1) = 8 sq. units. (Answer)

Answer 2

Given

• ABCD is a Rectangle
• $\sf{Vertices \: of \: \boxed{ \: \: \: \: \: } \: \: A( - 2,2), B(6,2), C( - 2,3) \: and, D(6,3)}$

To Find

• Area of rectangle ABCD

Solution

_____________________

We know that

Difference between two points is given by

$\: \: \: \: \: \sf \sqrt{(Diff.~~ of ~~ abscissae )² +( Diff.~~ of ~~ ordinates)²}$

$~~~~~~ \sf \mapsto AB= \sqrt{(6 + 2)² +( 2 - 2)²}$

$~~~~~~ \sf \mapsto AB= \sqrt{(8)² +( 0)²}$

$~~~~~~ \sf \mapsto AB= \sqrt{64}$

$~~~~~~ \sf \mapsto AB= 8$

$~~~~~~ \sf\mapsto BC = \sqrt{(6 - 6)² +(3 - 2)²}$

$~~~~~~ \sf\mapsto BC = \sqrt{(0)² +(1)²}$

$~~~~~~ \sf\mapsto BC = \sqrt{1}$

$~~~~~~ \sf\mapsto BC = 1$

Now

$\sf Area ~of ~rectangle \: ABCD = 8 \times 1$

$\bf \underline{ Area ~of ~rectangle \: ABCD = 8sq. \: units}$