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KCC's Quizzes AQQ261 about a rectangle area

A new challenge is there to enjoy your coming week-end! This time an easy one open to all of us (thus not only for electronical engineers)!

1. First, we start with the quote of the week: "You know you've reached middle age when you're cautioned to slow down by your doctor, instead of by the police" - Joan Rivers.

2. Quiz AQQ261 about a rectangle area puzzle:

Question:

With the information provided, calculate the area of the big rectangle ABCD.

Good luck!

P.S. Don't hesitate to forward those quizzes to colleagues and friends around you; we need more (active) participants!

Thanks

Kuo-Chang

• The area of the small rectangle is y*(y-8)

y*(y-8) = 9

y^2 -8y = 9

y^2 -8y -9 = 0

Factoring:

(y - 9)(y + 1) = 0

So y has to be 9 or -1, but in this case has to be the positive integer

y + 2 = 11 so the length of side AD, x = 12

The length of side AB = x - 4 = 8

So the area of the big rectangle ABCD = 8*12 = 96

• 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.

• the Area9 is: (y-8)*y=9 => y=9
because x=y-8+y+2=2*y-6 => x=12
AreaABCD = |AB|*|BC|=(x-4)*x=8*12=96

• (x-4) * x = area ABCD

x = (y-8) + (y+2) = 2y-6

(y-8) * y = 9

y^2-8y-9 = 0

(y-9)(y+1) = 0

y = 9 or -1 *** (can't be -1; therefore, y = 9)

x = 2y-6 = 2*9 - 6 = 12

area ABCD = (12-4)*12 = 96

• The area of ABCD is 96.

You see that (y-8)*y=9, from which you get y=9 (the other solution of the quadratic equation is negative). Then x= (y-8)+(y+2)-->x=12. So BC=12 and AB=8, from which you get the result.

Figure is not consistent with the data.

• Fantastic Martin! You are, again, the first one, congratulations!

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

• Bravo  , correct (too easy for engineers like you...) but congratulation anyhow!

• Félicitation  correct! (such quiz is too easy for people like you...)

• Area = 48