3 congruent rectangles are arranged as shown, such that two rectangles have a vertex pinned to the centre of a different rectangle.

Find the perimeter of the red triangle.

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Still trying to calculate the exact value, but the approximate value is 21.89 cm.

Hopefully you’ll end up with 22cm (ish). My usual answer checker is on holiday! We seem to be converging on the same result though.

I make it ½√137 + ½√(125 + 34√2) + √(99/2 + 28√2) with a mix of Pythagoras and cosine rule using the length of the diagonal of the rectangles.

I’m going to email you my calculations for this one. I’ve spent more time over it today and I get 24.04924695cm, With side 1 being 5.85235cm (same as ‘Maths’ below), side 2 being 6.578cm, and side 3 being 11.6188cm

Agree with the first 2 sides so obviously we disagree on the angle if cosine rule used

I got 98.672?

I agree with this figure

Sides being 5.852, 6.578 and 9.439 making 21.869 some rounding issues possibly

I didn’t do the same method, so I’m fairly convinced S1 and S2 are correct.

For S3, I drew the smallest possible rectangle (a new shape) around the entire shape (all 3 red rectangles). From that I deduced two sides of a new right-angled triangle, whose hypotenuse is S3. I’ll recheck my working as if I’m honest, I’m least confident about S3 in my method!

I worked out S2 by drawing a second triangle with longest side S2, and both ends of S2 joining with the middle of red rectangle 2.

S3 = 9.4391726198cm … I think!!

This is my solution: [IMG]http://oi61.tinypic.com/smv86b.jpg[/IMG]

i am not even sure how to approach this problem, i will have to wrestle with it during my prep.

It’s tough! The left side is easiest. Just use pythag by drawing in a right angled triangle. But after that you have to get very creative!