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As the title says this is my question :

The side of a square exceeds the side of other square by 4 cm and the sum of the areas of the two squares 400 sq.cm. Find the dimensions of the square.


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Try to put up a system of two equations, than subsitute one variable into another one. –  user21385 Dec 20 '11 at 17:24

3 Answers 3

up vote 1 down vote accepted

First, draw a picture and label the unknowns. This is the most important step; introduce a variable whose value is what you seek.

You have two squares. Let's call the side length of the smaller square $S$. Then the side length of the other square is $S+4$. Note that this falls in line with the statement given in the problem:

"The side length of one square exceeds the side length of the other square by 4cm"

as the above can be written

"the side length of one square is equal to 4 plus the side length of the other square"

Ok, so one square has side length $S$ and the other has side length $S+4$.

Well, that's dandy; but we need to figure out what $S$ is exactly. Towards this end, we need to find an equation involving $S$.

What do we know? Well, the problem tells us the sum of the areas of the squares is 400. This is another important thing to do: make sure you use all the information given in the problem

Write down mathematically that the sum of the areas of the two squares is equal to 400.

The area of the smaller square is $$\tag{1} S\cdot S$$

The area of the other square is $$\tag{2}(S+4)(S+4)$$

So, if we add the epressions in (1) and (2), we get 400: $$ S\cdot S+(S+4)(S+4)=400. $$ Rewriting the above gives $$ S^2+(S+4)^2=400. $$

The above is a quadratic equation whose positive solution is the side length of the smaller square. Once you've found the positive solution, you can find the dimensions of the squares (recall here that the side length of the larger square is $4+S$).

I'll leave it to you to solve the equation. Once you've done so, be sure to remember to explicitly state the answer: "the smaller square has dimensions..., and the larger square has dimensions...".

So, in summary:

1) Introduce variable(s) as determined from the problem.

2) Write down an equation (or equations) that relate the variables.

3) Solve the equations.

4) State the answer.

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Hint: if the smaller side is $s$ what is the area of the smaller square? Then the side of the larger is $s+4$, so what is the area? Add these together and get $400$, solve for $s$.

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Is it possible to show me the full procedures please. Thanks a lot in advance. –  Netizen110 Dec 20 '11 at 17:25
@Netizen, I don't doubt it is possible for Ross to do your homework for you and thereby make sure your opportunity for actually learning something is lost -- but why would he? He's not usually that mean. –  Henning Makholm Dec 20 '11 at 17:36
It's actually not my homework. :P I tried solve this but could proceed that's why I landed up here. –  Netizen110 Dec 20 '11 at 17:49

A quite detailed set of instructions on how to proceed has been given by Ross Millikan and David Mitra. The following approach is algebraically more complicated, but introduces some ideas that may be useful to you later.

Let $x$ be the side of the larger square, and let $y$ be the side of the smaller square. We were told that $$x-y=4. \qquad\qquad (1)$$ We were also told that the sum of the areas is $400$. So $$x^2+y^2=400. \qquad\qquad (2)$$ One way to continue is to note that from Equation $(1)$ we have $x=y+4$. Substitute $y+4$ for $x$ in Equation $(2)$, expand. We get a quadratic equation in $y$. Solve for $y$.

Another way is to square both sides of $x-y=4$. We get $$x^2-2xy+y^2=16. \qquad\qquad (3)$$ From Equations $(2)$ and $(3)$, we get $2xy=384$, and therefore $$(x+y)^2=x^2+2xy+y^2=400+384=784.$$

Thus $x+y=\sqrt{784}=28$. But $x-y=4$. Adding, we obtain $2x=32$, so $x=16$ and $y=12$.

Comment: The idea was to wait before "breaking symmetry." Two steps could have been collapsed into one by noting that $(x+y)^2=2(x^2+y^2)-(x-y)^2$. The above procedure (of course done verbally, without symbols) seems to have been the one used in late Babylonian times to solve problems that we now think of as quadratic equation problems.

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Thanks a lot ! you rock ! Are you mathematician btw ? –  Netizen110 Dec 20 '11 at 19:05
Netizen10: Sure, there are lots of mathematicians on this site, including all the others who answered your question. –  André Nicolas Dec 20 '11 at 19:11

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