+0

# Area Puzzles

+26
1141
3

Here are 2 area puzzles. Bertie drew my attention to them.

Those of you who have been here a long time will remember Bertie.

They were in   http://fivethirtyeight.com/

Under [Culture] then under "The Riddler - Rock, Paper, Scissors, double scissors"

I have said exactly where they are as I assume there is a new puzzle in this section each week.

(no cheating)

The first one is quite simple and would be a good challenge for our very young budding mathematicians.

The second is not so simple :)

With both you are given some side lengths and some areas and you have to find a missing area.

Question 1 Question 2 Have fun Oct 4, 2017
edited by Melody  Oct 4, 2017

#1
+2

Here's the second one [ I think ]

Let the width of the top left rectangle = a  and its height  = b

So.....ab = 32   ...so its width is  a = 32/b

Let the height of the bottom right rectangle = d  and its width = 45/d

And d = 11 - b

Now....if we had a "completed" rectangle its area would be  14 * 11  = 154

And 14 =  width of the top left rectangle  + width of bottom right rectangle

So we have

(32/b +  45/d) (11) = 154     sub for d

(32/b + 45/(11-b))   = 14     simplify

32/b (11-b) + 45  = 14 (11 - b)

352/b - 32 + 45  =  154 - 14b

352/b + 13 = 154 - 14b

352/b - 141 + 14b = 0

14b^2 - 141b + 352  = 0   ⇒  b= 5.5  or b  = 32/7

If b = 5.5   then a  = 32/5.5 ≈ 5.81

If b = 32/7  then a =  32 / [32/7]  = 7    which is too large

So b = 5.5

And d = 11- b =  11- 5.5  = 5.5

a = 32/5.5   = 64/11

ad =   64/ 11 * 5.5   =   64 * (5.5 / 11)  =  64 * 1/2  = 32   Oct 5, 2017
#2
0

That is excellent Chris.  I shall have to examine what you have done.

There is another way, It is much simpler if you spot/know the trick.

Oct 5, 2017
#3
+2

I am going to give an alternate answer for Question 2 First I need to examine some properties of rectangles

First I will put in the 2 axes of symmetry of a rectangle.   The diagram is below;

It can be seen that the area of the whole rectangle is $$2x*2y=4xy\;\;\;units^2$$

The sum of the diagonally opposite rectanges inside it is   $$xy+xy=2xy=\dfrac{ \text{total area}}{2}$$ Now what happens if one of the dividing lines is an axis of symmetry but the other is moved over a little bit ?   See diagram below

The total area is still  $$4xy\;\;\;units^2$$

The sum of the diagonally opposite (green)rectanges inside it is

$$y(x+\delta x)+y(x-\delta x)=2yx=\dfrac{\text{total area}}{2}\\ \text{Which is still the same as before.}$$ So what happens if BOTH dividers are moved away from the axis of symmetry?

See diagram below:

The total area is still  $$4xy\;\;\;units^2$$

The sum of the diagonally opposite (blue) rectanges inside it is

$$(y+\delta y)(x+\delta x)+(y-\delta y)(x-\delta x)\\ =yx+y(\delta x)+x(\delta y)+(\delta x)(\delta y)\quad+\quad yx-y(\delta x)-x(\delta y)+(\delta x)(\delta y)\\ =2yx+2(\delta x)(\delta y)\\ \ne \dfrac{\text{total area of rectangle}}{2}\\$$ So what I have shown here is that if a rectangle is divided into 4 using 2 intervals, one inteval  parallel to each of the sides. Then the sum area of the diagonally opposite internal rectangles will be equal to half the area of the whole rectangle IF AND ONLY IF on of the intervals is an axis of symmetry of the original rectangle.

[maybe this could be worded better ]

----------------------------------------------------------------------

So lets look at the actual question 2 now!

Total area = 11*14 = 154 inches squared

1/2 of totall area = 77 inches squared

45+32=77 inches squared

SO one of those dividing lines MUST be an axis of symmetry of the whole rectangle.

So the question mark must equal either 45 or 32

Area of the known peices is 32+34+45 = 111   Area unaccounted for = 154-111= 43

So the question mark area must be smaller than 43.

Hence the question mark area is 32 inches squared. Oct 9, 2017