It is known that the area of the largest equilateral triangular section of a cube is 140sq cm. What is the side length of the cube? The distances between the centres of two adjacent faces of another cube is 8cms. What is the side length of this cube? Another cube has an edge length of 12cm. At each vertex a tetrahedron with three mutually perpendicular edges of length 4cm is sliced away. What is the surface area and volume of the remaining solid?
From the measurements and the clue given find the area of the square that is not covered by the triangle and the circle.
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Using the cosine rule to find $\angle AOD$ we have $AC=CD=1$ and so $AD = \sqrt 2$ and hence
$$\cos \angle AOD = {{OA^2 + OD^2 -AD^2}\over 2OA.OD} = {1 \over \sqrt 2}.$$
Hence $\angle AOD = 45$ degrees.
The area of the triangle $AOD$ is
$${1\over 2} OA \times OD \times \sin \angle AOD = {\sqrt 2 \over 4}(2 + \sqrt 2).$$
To find the area of the minor segment $AD$ we subtract the area of triangle $AOD$ from the area of sector $AOD$ which gives
$${\pi \over 8}(2 + \sqrt 2) - {\sqrt 2 \over 4}(2 + \sqrt 2).$$
To get the total red shaded area we now add the area of triangle $ACD$ and multiply by $4$ which gives:
$$4[{\pi \over 8}(2 + \sqrt 2) - {\sqrt 2 \over 4}(2 + \sqrt 2)+ {1\over 2}]={\pi \over 2}(2 + \sqrt 2)- 2\sqrt 2 \approx 2.535 \rm {\ sq. \ units} .$$