KCC's Quizzes AQQ261 about a rectangle area

1. Start by solving y using the smaller Rectangle's (Asmall) given Area. The Area will be:

A_{\text{SMALL}}\text{ = y\lparen y - 8\rparen}

Solving for y, we get: 

9=y^2-8y

y=9,y=-1

2. Of the two solutions for y, y=9 seems to make more sense when plugging it back. use this value for now.

3. Solve for x by using the width of the rectangle:

x=y-8+y+2

x=2y-6

substitute y's value and solve for x.

\text{ }=\text{ }2(9)-6

This yields x = 12

4. Solve for area of recatangle ABCD (Abig). The Area will be:

A_{\text{BIG}}\text{ = x\lparen x-4\rparen}

substitute x's value into the equation and solve for Abig:

A_{\text{big}}=12(12-4)

Abig = 96 sq. units

While above answer really seems to be bigger than the given rectangle area ( Asmall ) , the given figure doesn't seem to represent this rectangle. it looks more like below:

otherwise if the values should conform to the figure, then it doesn't exist.

Bingo 01Taylor10 , you get it, and with a crystal clear development!

Bingo 01Taylor10 , you get it, and with a crystal clear development!