# Difficulty with circle geometry

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.

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 $cm^2$.

• I understand everything up to the 5th line, but where did (a+b)^2 come from on the sixth line? After that I kind of understand that you just have to solve the equation as a difference of two squares. Oct 13 '16 at 6:18
• Are you referring to the third line ? $(a+b) = 14$ thus $(a+b)^2 = 14^2$ Oct 14 '16 at 17:40
• Yes, that's what I meant. I've figured it out now, thank you for your answer :) Oct 14 '16 at 18:04

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

find $(x-y)$

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

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

• Seems a bit cryptic at first sight, but actually everything is here :) Oct 12 '16 at 20:16
• Make them to guess tease think :) Oct 12 '16 at 20:33

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

• Where did 7+or-root23 come from? Did you substitute in the equation and solve as a quadratic? Oct 13 '16 at 6:22
• @kashveyron, yes like solving two simultaneous equations. Oct 13 '16 at 10:51
• Oh ok, that makes better sense now. Thank you. Oct 13 '16 at 14:51

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.