Question

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(3,−2), B(6,−2), C(6,5) and, D(3,5) What is the area of rectangle ABCD?

Answer 1

Given

• ABCD is a Rectangle
• $\sf{Vertices \: of \: \boxed{ \: \: \: \: \: } \: \: A(3,−2), B(6,−2), C(6,5) \: and, D(3,5)}$

To Find

• Area of rectangle ABCD

Solution

_____________________

We know that

Difference between two points is given by

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

$~~~~~~ \sf \mapsto DC = \sqrt{(6-3)² +(5-5)²}$

$~~~~~~ \sf \mapsto DC = \sqrt{(3)² +(0)²}$

$~~~~~~ \sf\mapsto DC = \sqrt{9}$

$~~~~~~ \sf \mapsto DC =3$

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

$~~~~~~ \sf \mapsto AB= \sqrt{(3)² +( 0)²}$

$~~~~~~ \sf \mapsto AB= \sqrt{9}$

$~~~~~~ \sf \mapsto AB=3$

$~~~~~~ \sf \mapsto AD = \sqrt{(3 - 3)² +( - 2 - 5)²}$

$~~~~~~ \sf \mapsto AD = \sqrt{(0)² +( - 7)²}$

$~~~~~~ \sf \mapsto AD = \sqrt{49}$

$~~~~~~ \sf\mapsto AD = 7$

$~~~~~~ \sf\mapsto BC = \sqrt{(6 - 6)² +( 5 + 2)²}$

$~~~~~~ \sf\mapsto BC = \sqrt{(0)² +( 7)²}$

$~~~~~~ \sf \mapsto BC = \sqrt{49}$

$~~~~~~ \sf\mapsto BC = 7$

${\bf \: \: AB = DC ~~~~~~~,~~~~~~~ AD = BC }$

So, ABCD is a Parallelogram

Now

In Diagonal BD and AC

$~~~~~~ \sf \mapsto BD = \sqrt{(6 - 3)² +( - 2 - 5)²}$

$~~~~~~ \sf \mapsto BD = \sqrt{(3)² +( - 7)²}$

$~~~~~~ \sf \mapsto BD = \sqrt{9 + 49}$

$~~~~~~ \sf \mapsto BD = \sqrt{58}$

$~~~~~~ \sf \mapsto AC = \sqrt{(6 - 3)² +( 5 + 2)²}$

$~~~~~~ \sf \mapsto AC = \sqrt{(3)² +( 7)²}$

$~~~~~~ \sf \mapsto AC = \sqrt{9 +49}$

$~~~~~~ \sf \mapsto AC = 58$

Clearly,

${\sf DB²=DC²+BC²~~~~~~,~~~~~~ AC²= AB²+BC²}$

Aslo We know that

• If diagonal are Equal then the Figure is Rectangle .

$\sf Area ~of ~rectangle \: ABCD = Length (AB) × Breadth (BC)$

$\sf Area ~of ~rectangle \: ABCD = 3 \times 7$

$\bf \underbrace{ \pink{Area ~of ~rectangle \: ABCD = 21 \: sq. \: units}}$

Answer 2

Answer:

See the attachment photos.

Area of rectangle ABCD = 21 sq. units.

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