What is the area of a rectangle with vertices at $(2,3), (7,3), (7,10),$ and $(2,10)$ ?

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Answer 1 · 2018-04-07T05:28:10

$35" sq. unit"$ .

Let us name the given points as

$A(2,3), B(7,3), C(7,10) and D(2,10)$ .

Using the distance formula, we have,

$AB^2=(2-7)^2+(3-3)^2=25,$

$BC^2=(7-7)^2+(3-10)^2=49$ ,

$CD^2=(7-2)^2+(10-10)^2=25$ , and,

$DA^2=(2-2)^2+(10-3)^2=49$ .

Also, $AC^2=(2-7)^2+(3-10)^2=25+49$ , showing that,

$AC^2=AB^2+BC^2$ .

We conclude that $ABCD$ is, indeed, a rectangle, having sides

$AB=5, BC=7$ .

Hence, the reqd. area $=5xx7=35" sq. unit"$ .

What is the area of a rectangle with vertices at (2,3), (7,3), (7,10),(2,3), (7,3), (7,10), and (2,10)(2,10)?