Question

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

Answer 1

I hope this helps you

Answer 2

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

`A=√(12^2+4^2)\\\\A=√(160)\\\\A=4 √(10)\\\\B=√(12^2+8^2)\\\\B=√(208)\\\\B=4√(13)\\\\C=√(8^2+4^2)\\\\C=√(80)\\\\C=4√(5)`

Area of Triangle

`=(1)/(2) * \text {Product of two adjacent sides}* \text{Angle between these two sides}`

`\cos B=(c^2+a^2-b^2)/(2*a*c)\\\\\cos B=(160+80-208)/(2*4√(5)*4√(10))\\\\ \cos B=(32)/(32*√(50))\\\\ \cos B=(1)/(√(50))\\\\\ sin^2B=1-(1)/(50)\\\\ \sin B=(7)/(√(50))\\\\ \text{Area}\Delta=(1)/(2)ac \sin B\\\\=(1)/(2) * 4√(5) * 4√(10)* (7)/(√(50))\\\\=(112)/(2)\\\\\text{Area}\Delta=56 \text{Square Unit}`