Question

A rectangle has an area of 78 square centimeters. If its length is decreased by three, and width increased by four, it becomes a square. What are the dimensions of the rectangle?

Answer 1

Oh I like this one.
So we will say the Area of Rectangle, \(A_R\) is equal to it's length, \(l_R\), times its width, \(w_R\). So:
\(A_R=l_R*w_R\)

We also know that if the length is decreased by 3, it will be equal to the width plus 4 since the sides will form a square and a square will have ALL equal sides

Answer 2

Let sides of the square be a cm long. When i decrease the length by 3 i find a, so length is a+3. When i increase the width by 4 i find a, so width is a-4.

Area = (a+3)*(a-4) = 78 = 13*6
a+3 = 13 and a-4 = 6 --> a = 10

(a+3) = 13 and (a-4) = 10 are the dimensions

Answer 3

(a-4) = 6 (not 10)

Answer 4

the length should be a-3, and width being b+4
how'd you come up with the dimensions?

Answer 5

OH YEAAH, okay @zarax , i'm sorry I buffered understanding yer point.

Answer 6

So we can make the following equation:
\[(l_R-3)=(w_R+4)\]
And solve for one variable. I will do \(l_R\)
\[l_R=w_R+7\]
So we can substitute it into the original equation:
\[\eqalign{
&A_R=l_R*w_R \\
&A_R=(w_R+7)*w_R \\
&A_R=(w_R)^2+7w_R}\]

We can substitute \(A_R=78\) and obtain:
\[\eqalign{
&A_R=(w_R)^2+7w_R \\
&78=(w_R)^2+7w_R \\
&0=(w_R)^2+7w_R-78 \\}\]

And solve for \(w_R\) by the quadratic formula:
\[w_R=\frac{-7\pm\sqrt{49+312}}{2}=\frac{-7\pm19}{2}\]

We can exclude the negative:
\[w_R=\frac{-7+19}{2}=\frac{12}{2}=6\]

And since we know \(l_R=w_R+7\) we can obtain that \(l_R=13\)

So the dimensions of the rectangle are 6*13

Answer 7

@KeithAfasCalcLover That is a solution for over academic calculus textbooks :D

Answer 8

Haha
"OHHH OF COURSE. MY BAD."

Answer 9

Is it still ok Hannah? :)

Answer 10

the length should be a-3, and width being b+4
how'd you come up with the dimensions?

Answer 11

Actually, @zarax , @KeithAfasCalcLover 's solution actually illustrated the problem-solving technique that the question is obviously testing: set up and solve a system of 2 equations in 2 unknowns.

The problem with your method is that it worked HERE because the product was easily factored into 13*6 so it was easy enough to tell, by some intuition and maybe trial-and-error, how to arrive at the answer.

Suppose the problem had been:
A rectangle has an area of 8245 square centimeters. If its length is decreased by 42, and width increased by 17, it becomes a square. What are the dimensions of the rectangle?
(NOTE: I have no idea how that problem would work out, I just pulled the numbers out of thin air to illustrate the point.)

My point is this: as a teacher, I often write problems that are computationally "easy" in order to illustrate and test the CONCEPT, in this case, setting up the 2 equations and solving. I don't need a computationally difficult and nasty problem to TEST that CONCEPT, but I DO want the problem done by the method that was taught. (In fact, I would probably explicitely say, "set up and solve a system of 2 equations").

Answer 12

OMG, thank you @KeithAfasCalcLover , now i can finish my homework :D hahah! yo, you made my day. Thanks.

Answer 13

Haha, as always @DebbieG, your words are very heart warming, humbling, and appreciated. Im just happy to contribute in which way I can :)

And @han_nah Anytime! Glad to have made your day ;)