Question

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

Answer 1

Answer :

Area of the rectangle = 8 sq. units.

Step - by - step explanation :

ABCD is a rectangle with AD as its diagonal.

We know, the diagonal of a rectangle divides it into two congruent triangles of equal area.

∴ ABD & ACD are the two equal triangles.

Also, Ar(ABCD) = 2[Ar(ABD/ACD)]

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Now, using the formula for finding the area of a triangle :

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

We get,

$\bold{Ar_{ABD}=\frac{1}{2}(-2(2-3)+6(3-2)+-2(2-2))}$

☛ $\bold{Ar_{ABD}=\frac{1}{2}(-2(-1)+6(1)+-2(0))}$

☛ $\bold{Ar_{ABD}=\frac{1}{2}(2+6)}$

☛ $\bold{Ar_{ABD}=\frac{1}{2}\times{}(8)}$

☛ $\bold{Ar_{ABD}=4\:sq.\:units}$

∴ Area of triangle ABD = 4 sq. units.

Then, total area of the rectangle :

☛ $\bold{Ar_{ABCD}=2\times{}Ar_{ABD}}$

☛ $\bold{Ar_{ABCD}=2\times{}4\:sq.\:units}$

☛ $\bold{Ar_{ABCD}=8\:sq.\:units}$

∴ Area of the rectangle = 8 sq. units.