Question

Find the area of the triangular section of the rectangle solid shown in the figure .

Answer 1

Length of three diagonal A,B and C of given solid is

  $A=\sqrt{12^2+4^2}\\\\A=\sqrt{160}\\\\A=4 \sqrt{10}\\\\B=\sqrt{12^2+8^2}\\\\B=\sqrt{208}\\\\B=4\sqrt{13}\\\\C=\sqrt{8^2+4^2}\\\\C=\sqrt{80}\\\\C=4\sqrt{5}$

Area of Triangle

                  $=\frac{1}{2} \times \text {Product of two adjacent sides}\times \text{Angle between these two sides}$

$\cos B=\frac{c^2+a^2-b^2}{2*a*c}\\\\\cos B=\frac{160+80-208}{2*4\sqrt{5}*4\sqrt{10}}\\\\ \cos B=\frac{32}{32*\sqrt{50}}\\\\ \cos B=\frac{1}{\sqrt{50}}\\\\\ sin^2B=1-\frac{1}{50}\\\\ \sin B=\frac{7}{\sqrt{50}}\\\\ \text{Area}\Delta=\frac{1}{2}ac \sin B\\\\=\frac{1}{2} \times 4\sqrt{5} \times 4\sqrt{10}\times \frac{7}{\sqrt{50}}\\\\=\frac{112}{2}\\\\\text{Area}\Delta=56 \text{Square Unit}$

Answer 2

I hope this helps you