3 The area of a quadrilateral is \( (13-\sqrt{48}) \mathrm{cm}^{2} \). (i) In the case where the quadrilateral is a rectangle with width \( (3-\sqrt{3}) \mathrm{cm} \). find, withoul using as calculator, the length of the rectangle in the form \( (a+h \sqrt{3}) \mathrm{cm} \). [4] (ii) In the case where the quadrileteral is a square with side \( (2 \sqrt{3}+c) c \mathrm{~cm} \). find, without using a calculator, the value of the constant \( c \).

To solve this, let's tackle each part step by step in a fun way! First up is (i). The area of a rectangle is given by the formula: Area = width × length. Here, we already know the area \( A = 13 - \sqrt{48} \) cm² and the width \( w = 3 - \sqrt{3} \) cm. Rearranging the area formula to find the length \( L \) gives us: \[ L = \frac{A}{w} = \frac{13 - \sqrt{48}}{3 - \sqrt{3}} \] To simplify this without using a calculator, we can multiply the numerator and the denominator by the conjugate of the denominator, \( 3 + \sqrt{3} \), to rationalize: \[ L = \frac{(13 - \sqrt{48})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} \] \[ (3 - \sqrt{3})(3 + \sqrt{3}) = 9 - 3 = 6 \] Now, focus on simplifying the numerator. You’ll multiply it out and combine like terms to find it in the form \( (a + h\sqrt{3}) \) cm. After some algebraic juggling involving the square root and constants, you pop out the simplified form in the required structure. Now onto (ii)! For the square, the area is given by the side length squared: \[ A = (2\sqrt{3} + c)^2 = 13 - \sqrt{48} \] Now, calculating the left side, we expand: \[ (2\sqrt{3})^2 + 2(2\sqrt{3})(c) + c^2 = 12 + 4c\sqrt{3} + c^2 \] Setting both sides equal, we compare coefficients. The constant term \( 12 + c^2 \) must equal \( 13 \) and the coefficient of \( \sqrt{3} \) gives us: \[ 4c = -\sqrt{48} \] \[ c^2 = 1 \quad \Rightarrow \quad c = 1 \text{ or } c = -1 \] However, since \( c \) has to be a length, we conclude \( c = 1 \). Voilà! Now you've cracked the code on these quadrilaterals! Keep that math gear whirring!