Perimeter Problem #2

tough

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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11 thoughts on “Perimeter Problem #2

      • 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

      • 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.

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