The area of rectangle ABCD with vertices A(0, 0), B(0, 4), C(x, 4) and D(x, 0) is \(16\) square units. If x > 0, what is the value of x?
From the information given above, we can say that the width of the rectangle is \(4\)
The area of a rectangle is \(length \cdot width\)
That is, \(4 \cdot length = 16\)
That gives us length = \(16/4 = 4\)
From that, we can tell that the value of \(x\) must be \(4\).
Note that a square is a special case of a rectangle....but a rectangle isn't necessarily a square
Since AB = 4 ....then it must also be that BC = 4
So....the area = (AB) (BC) = (4)(4) = 16