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 \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: