Question

an isosceles tringle has each of its equal sides equal to 25.The area of the tringle is 168.find the length of the altitude drawn to the base.

Answer 1

$area = \frac{1}{2} \times base \times height \\ let \: \: height \: \: be \: \: \: h \\ base = 2 \times \sqrt{ {25}^{2} - {h}^{2} } \\ thus \: \: \sqrt{ {25}^{2} - {h}^{2} } \times h = 168 \\ {h}^{2} ( {25}^{2} - {h}^{2} ) = {168}^{2} \\ {h}^{4} - 625 {h}^{2} + 28224 = 0 \\ ( {h}^{2} - 576)( {h}^{2} - 49) = 0 \\ {h}^{2} = 576 \: \: or \: \: 49 \\ h = 24 \: \: or \: \: 7$