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Hi. Just offering another perspective here from how I view it. I was also thinking along the same line but more like in between both these two methods (Sushi88 and Ace Starling).

It is less obvious to me that the little blue triangles above are forming little squares  with other trapezium part.

For me, 4 smaller right-angled triangles (like ABP, BCQ, CDR and DAO) and 1 small square (OPQR) fill up the same old big square. Then two of these new right-angled triangles form two small squares so total is 5 small squares=1 big square. 

Solving will be more like the way above.

And I like your options to “no calculator” way of solution.


To those who don’t mind a nagging mum: I sometimes push my kids to work without calculator especially during side by side practice together (no other chance when I’m not around). Then I can see how sharp (or not so sharp… while they are only kids with less years of doing Maths) their mental work is. It can built the kids’ speed, confidence and “sensitivities” with numbers which is getting lesser with more use of calculators these days. While guiding our kids, the process is as important, if not more important than the final answer. Have fun doing math!

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AO=OP=BP=PQ=CQ=QR=DR=RO= u

Area of one of the identical triangles = 1/2 (u) (2u) = u2

Area of 4 identical triangles = 4u2

Area of middle small square = Area of big Square – Area of 4 identical triangles

u2 = 19 x 19 – 4u2

5u2 = 361

u2= 361/5 = 72.2

u = 8.497cm

Hence length of side of small square is  8cm (to the nearest cm)

 

 

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