If a rectangle's length is y−2 and the width is 3y+2 write an expression for the perimeter and an expression for the area.
Accepted Solution
A:
Answer:Area = [tex]3y^2 - 4y - 4[/tex] units square.Perimeter = 8y units.Step-by-step explanation:Area of a rectangle = length * width.Perimeter of a rectangle = 2*(length + width).It is given that length = y−2 units and width = 3y+2 units. To find the area and the area of the rectangle in terms of y, simply put the length and the width in the above area and perimeter equations.Area = length * width = (y-2)*(3y+2).Expanding the expression gives:Area = 3y^2 + 2y - 6y - 4 = (3y^2 -4y - 4) units square.Similarly,Perimeter = 2*(length + width) = 2*(y-2 + 3y+2) = 2*4y = 8y units.Therefore, the expression for the area and the perimeter of the given rectangle is ([tex]3y^2 - 4y - 4[/tex]) units square and (8y) units respectively!!!