# Perimeter Problem #2

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.

## 11 thoughts on “Perimeter Problem #2”

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.

• Maths says:

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.

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

2. Owen says:

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!