Difficulty with circle geometry

Question

I am working through the Senior Team Regional Maths Challenge questions from the 2007 paper (answers on this page: http://furthermaths.org.uk/files/SupervisorsBooklet.pdf).

On question 10 I am really struggling. It says 'The circle in the diagram below has radius 6cm. If the perimeter of the rectangle is 28cm, what is its area?'

The diagram is of a circle with a rectangle inside, all four corners touching the circumference (sorry for the absence of a diagram, I can't find the question paper itself).

I've figured out that the distance from the centre of the rectangle to the corner must be equal to the radius of the circle (6cm). Thus the diagonal length (diameter of circle) is 12cm. Using pythagoras, $x^2+y^2=144$. However, the answer for the area is 26cm. This means that is the perimeter is 28cm, the two lengths must be 2cm and 13cm. Using these lengths in pythagoras is not possible as $2^2+13^2 \not = 12^2$. So now I am absolutely confused and have no idea how to figure this out.

Answer 1

Call $a$ and $b$ the lengths of the sides of the rectangle.

We have $2(a+b) = 28$ and $a^2 + b^2 = 12^2$ by Pythagoras' Theorem.

Therefore $a+b =14$ so $(a+b)^2 = a^2 + b^2 + 2ab = 14^2$

Thus $2ab = 14^2 - 12^2 = (14-12) (14+12) = 2 \times 26$ ie $ab = 26$

The area of the rectangle is indeed 26 $cm^2$.

Answer 2

You only need to know that half the perimeter is 14 cm!!!

Answer 3

$$ 2 x + 2 y = 28, \quad 2 \sqrt{x^2+y^2} =12,\quad x y =? $$

find $ (x-y) $

$$ (x+y)^2 + (x-y)^2 = 2 {x^2+ 2y^2} $$

and so on, and choose the sign from the two.

Answer 4

Let me work backwards from the supplied answer (as you seem to have done): if the area of the rectangle is $xy = 26$ and the diagonal is $d = 12$, so that $x^2 + y^2 = d^2 = 12^2 = 144$, then $(x + y)^2 = x^2 + y^2 + 2xy = 144 + 52 = 196 = 14^2$, so the perimeter $2x+2y$ is $2 \times 14 = 28$. There is no contradiction between $xy = 26$ and $x+y = 14$, because $x$ and $y$ are not required to be whole numbers.