50 PTS The length of the smaller rectangle at the right is 1 inch less than twice its width. Both the dimensions of the larger rectangle are 2 inches longer than the smaller rectangle. The area of the shaded region is 86 square inches. What is the area of the larger rectangle?

378 square inches.

Step by step explanation:

Let x = the width of the smaller rectangle. 

The length of the smaller rectangle is 2x - 1. 

Area is A = lw 

So the area of the smaller rectangle is A = (x)(2x - 1) = 2x^2 - x 

The larger rectangle's width is two inches more than the width of the smaller rectangle (x+2). 

The larger rectangle's length is two inches more than the length of the smaller rectangle: 

2x - 1 + 2 = 2x + 1 

Area is A = lw 

The area of the larger rectangle is A = (x + 2)(2x + 1) = 2x^2 + x + 4x + 2 = 2x^2 + 5x + 2. 

The area of the larger rectangle minus the area of the smaller rectangle is 86: 

(2x^2 + 5x + 2) - (2x^2 - x) = 86 

Rewrite as adding the opposite: 

(2x^2 + 5x + 2) + (-2x^2 + x) = 86 

Combine like terms: 

6x + 2 = 86 

6x = 84 

x = 14 

The area of the smaller rectangle was 2x^2 - x, so 

2(14)^2 - (14) 

2(196) - 14 

392 - 14 

378 

The area of the smaller rectangle is 378 square inches. :)

hope this helped!

length = l

width = w

smaller

l = 2w - 1

Area of small one is

A = l*w = (2w-1)(w)

A = 2w² - w

big rectangle

L = l + 2 = (2w - 1) + 2 = 2w + 1

W = w + 2

Area of big one is

A = L*W = (2w + 1)(w+2)

A = 2w² + 5w + 2

the shaded region is the region of big rectangle - small rectangle, isn't it?

so

A(big) - A(small) = 86

2w² + 5w + 2 - (2w² - w) = 86

6w + 2 = 86

6w = 84

w = 84/6 = 14

so we can find L and W

L = 2w + 1 = 29

W = 16

the area of larger rectangle is 29*16 = 464 in²

the answer will be the last one