For a particular rectangle, three times the width is 23 meters longer than the length. the area of the rectangle is 400 square meters. what is the perimeter of the rectangle?

Concept:

(i) In general, the length of any rectangle is always greater than its width.

(ii) Area of rectangle (A) = L × B

(iii) Perimeter of rectangle (P) = 2 ( L + B)

Given: Area of rectangle 'A' = 400 square meters

Let 'L' and 'B' be the length and width of the given rectangle.

According to the problem,

L = 3 B - 23 ------------------(1)

Area of the rectangle = 400 m²

or, L× B = 400 m²

or, (3 B - 23 )×B = 400

or, 3 B² - 23 B - 400 = 0

Apply, discriminant method,

Here, we will calculate only the positive value of B because width or length will never be negative.

Now, using equation (1),

L = 3 B -23 = 3×16 - 23 = 25 m

Hence, length (L) = 25 m and Width (B) = 16 m

Now, we shall calculate the perimeter (P) of the rectangle

P = 2( L + B)

or, P = 2 (25 + 16 ) m

or, P = 82 m

Hence, the required perimeter of the given rectangle will be 82 m

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!