You have to view it as a big triangle minus a quite-big triangle, minus a small triangle, minus a rectangle. You can work out the area of each of those from the information given (and it gives a nice round answer). Not sure how it relates to technical knowledge, though - it's all a bit Bletchley.
After faffing for a bit I went with short side as the base of a triangle to apply area = 1/2*base*perp.height.
Then turns out perp.height is just the diagonal of the middle square, and the top square is irrelevant?
Could definitely have been a tricksy-ish question on my 1985 maths O level...
I’m not a huge fan of these sorts of things, mostly because the general grasp of logic is so slim that far too many people will form the incorrect conclusion that “not being able to solve it means you’re not super smart” when the reason may actually be unfamiliarity with Euclidean geometry.
I can see to go about it. Pythag theorem… conjecture some shapes that are not marked but which are implicit… equations to reduce etc. My schools days are 40 years in my past and maths was never my forte, so I can’t do all this. But I sense that anyone good at GCSE level maths ought to be able to.
That was my general direction of thought - fill in all the rectangles, then work out which ones you need to cut in half to subtract from the 6pi minus root 5 by 6pi plus 3root 5 right angle triangle. Not so much “super smart” as good with shapes.
I can solve the problem. Work out the length of each of the sides of the squares, then use those & Pythagoras to work out the length of each side.
Then apply Heron's formula.
It's too sodding tedious to work it all through.
As a mathematician who enjoys recreational geometry, I can do it in my head in thirty seconds, but I doubt I'm any better suited than you to be a "Whitehall disrupter". Although I could shout "What's even the point of you knock-off Tories!" quite loudly, so there's that.
Oh I got one free read but I can't go back and read it again. It's one big RA triangle minus 2 triangles and a rectangle. You can work out the lengths because you know the size of each square, assuming I interpreted I correctly.
It is, i would suggest, relatively straight forward if you did well at gcse maths, given it can be answered by primarily knowing what the area of a right angled triangle and a rectangle are.
Yeah it’s not that difficult, just construct right angled triangles where the hypotenuse is one of the two long red lines, the final line length that’s smallest is 10
DSIT just also announced a cyber expert committee where half of them are frauds , total frauds from Twitter. One spends every 30 minutes posting and another is a customer service rep who moonlights running a sponsored podcast so isn’t independent. So we are getting no innovation. Just lobbying
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Then turns out perp.height is just the diagonal of the middle square, and the top square is irrelevant?
Could definitely have been a tricksy-ish question on my 1985 maths O level...
You calculate the lengths of two of the diagonals, then do one multiplication.
It's the area of two trapezoids minus a third trapezoid with known sides and height from the squares.
You don't need all the sides.
Sometimes you need concentrated effort to stick to simple tools,especially if someone tries to big up these problems.
N10 submits this problem to the FT to find geniuses one could think it is really difficult while a 10 yr old has the tools to solve it
Then apply Heron's formula.
It's too sodding tedious to work it all through.