Let the diagonals of the rhombus be \(d_1\) and \(d_2\). The length of each side of the rhombus is given as \(a = 36\) cm. Step 1: Apply the formula for the area of a rhombus. The area of a rhombus in terms of its diagonals is \[ \text{Area} = \frac{1}{2} d_1 d_2. \] Given that the area is \(396\ \text{cm}^2\), we have \[ \frac{1}{2} d_1 d_2 = 396, \] which gives \[ d_1 d_2 = 792. \] Step 2: Use the relationship between diagonals and side. In a rhombus, the diagonals bisect each other at right angles. Using the Pythagorean theorem, \[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2. \] Multiplying both sides by 4, \[ d_1^2 + d_2^2 = 4a^2 = 4(36^2) = 5184. \] Step 3: Find the difference between the diagonals. Using the identity \[ (d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2, \] and substituting the known values, \[ (d_1 - d_2)^2 = 5184 - 2(792) = 5184 - 1584 = 3600. \] Taking the square root, \[ |d_1 - d_2| = \sqrt{3600} = 60. \] Hence, the required difference between the diagonals is \(60\).
In the figure, \(O\) is the centre of the circle and \(AC\) is the diameter. The line \(FEG\) is tangent to the circle at \(E\). If \(\angle GEC = 52^\circ\), find the value of \(\angle E + \angle C\).
