What is the area of the large outer square?

Kindly provided by UKMT A figure in the shape of a cross is made from five 1 x 1 squares, as shown. The cross is inscribed in a large square whose sides are parallel to the dashed square, formed by four vertices of the cross.
What is the area of the large outer square?
28 Comments
Phil
3/8/2014 02:06:49 am
Exactly 9.8, I believe (I'll post an explanation later...)
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Thirumalai Thathachary
3/9/2014 06:07:00 pm
49/5
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James McE
3/8/2014 03:31:51 am
Area of cross = 5 square units ( 5 su)
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James McE
3/8/2014 03:47:23 am
OOPS!!
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Phil
3/8/2014 03:48:01 am
Hi James,
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Phil
3/8/2014 03:51:55 am
The small blue triangle has non-hypotenuse sides 1 and 0.5, so it's base is sqrt(5)/2. From similar triangles I get its altitude to be 1/sqrt(5).
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James McE
3/8/2014 03:52:49 am
Hi Phil.
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Phil
3/8/2014 03:53:18 am
I did the same last time on a much more spectacular scale!
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James McE
3/8/2014 03:53:44 am
Hi Phil.
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Charles Greene
3/8/2014 02:07:54 pm
You are assuming the angle of plcement is constant. (Sqrt(5) + 2 sin(angle acute angle to red line)) ^2 would be the area.
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3/8/2014 03:10:17 pm
I get 10.0. The problem does not specify the orientation of the cross.
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Phil
3/8/2014 04:40:16 pm
Hi Pierre,
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3/8/2014 03:22:44 pm
To get a handle on this problem, I ran an imaginary visual simulation of the bounding square flexing in size as the cross is rotated through a 1/4 turn. In my mind's eye, I saw the square grow to a maximum then shrink, so I adopted finding the maximum area of the bounding square as a solvable problem. This problem clearly benefits from visual experience using dynamic geometry software.
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David Radcliffe
3/8/2014 08:00:21 pm
I agree with Phil's answer of 9.8. We can see from the picture that if the outer square is expanded to 7x7 then the blue cross will have area 25. http://gotmath.com/images/greekcross.png
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Phil
3/8/2014 08:32:36 pm
That's a really neat way to visualise this! And it make nice sense of the rather un-obvious number 9.8, because the ratio of the area of the overall square to that of the blue squares is now quite obviously 49:5.
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4/8/2014 03:04:52 pm
Yes - I posted my solution - but this approach is even more satisfying. 3/8/2014 08:28:14 pm
Thanks, Phil....the orientation of the cross is indeed fixed by the dashed line square. I now get 9.8 as the area.
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4/8/2014 11:06:07 am
( sqrt(5)+2/sqrt(5) )^2. Smaller square area is 5 and small squares in corner have side 1/sqrt(5). Gives 9.8
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4/8/2014 12:50:59 pm
High school geometry solution:
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Paul
5/8/2014 12:57:05 am
Great responses as always.Thanks guys.
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Vickie Frazier
5/8/2014 02:44:43 pm
for those who may not understand all the vector stuff I used triangles to find the solution. You can find my solution at https://dl.dropboxusercontent.com/u/70231773/Math%20puzzle.pdf
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Seong Kim
7/8/2014 10:13:22 pm
this can be a good example for such a test as SAT or ACT;
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Seong Kim
8/8/2014 01:27:57 am
the three sides of one of the three same right triangles are 1, 1/2, and sqrt(5/4);
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Judy Wheeler
4/9/2014 11:48:23 am
Where did the area of 9/20 for the larger triangle come from? If it is from the formula using the two legs of the right triangle, then my question is: how did you derive the leg lengths? Are you using the similarity relationship of the larger triangle to the smaller triangle?
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Paul
8/8/2014 03:52:31 am
Thanks to Seong and also to Vickie, who has impressively recreated the diagram as part of her really interesting solution!
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James
11/8/2014 02:54:20 am
Interesting solutions. And they say that there is nothing new under the sun!!
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10/10/2014 01:10:53 am
area outer square = area of 9 little squares.
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Qusay
10/11/2014 09:37:14 pm
the area= (sqr(5) + sqr(3/4))^2 = 9.6
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