A rectangle has a perimeter of 30 feet in area 50 feet what are the dimensions of the rectangle

Question

can u please help me bc i am having trouble in algebra 1 thanks

Answer 1

Let L be length and W be width

Perimeter of a rectangle is 30 feet

Perimeter is 2L + 2 W

So the equation will be

2L + 2W = 30 (equation 1)

I will simplify the equation by dividing with two

The new equation will be L + W = 15

I can say that

L = 15 - W

Area of rectangle is 50

Area is L * W

So the equation will be L * W = 50

I will substitute the L in area equation with the L from perimeter equation

It will be

L * W = 50

(15 - W) * W = 50

15W - W^2 = 50 solve this equation with rearrangement and factorization

W^2 - 15W + 50 = 0

(W - 10) (W - 5) =0

W-10 = 0 OR W-5 =0

W =10 W = 5

L = 15 - W L = 15 - W

= 15 - 10 = 15 - 5

= 5 = 10

Answer 2

Base = x ... Height = y

Perimeter ... x + x + y + y = 30 ..or.. 2x + 2y = 30

Area ... x • y = 50

Think of two numbers that multiply to give you 50 then plug in to the perimeter problem to see if they work. 5 and 10 multiply to give 50 and..

5 + 5 + 10 + 10 = 30 ..or.. 2(5) + 2(10) = 30