LUANDATO wrote:
In the cube above, AH and BG are diagonals and the surface area of side ABFE is 16. What is the area of rectangle ABGH?
$$A.\ \ 4\sqrt{2}$$
$$B.\ \ 16\sqrt{2}$$
$$C.\ \ 16+\sqrt{2}$$
$$D.\ \ 16$$
$$E.\ \ 15\sqrt{3}$$
The OA is
B.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
If the figure is a cube, we know that all the sides are equal, (and each face is a square) so if the surface area of ABFE is 16, each side must be 4.
We also know that the diagonal of a square splits the square into two 45:45:90 triangles. If triangle ADH is a 45:45:90 triangle, and AD = 4, then we know that AH, the hypotenuse, is equal to 4*√2, as the ratio of the sides of the 45:45:9 triangle is x: x: x√2.
If AD = 4 and AH = 4*√2, then rectangle ABGH has an area of 4 * 4*√2 = 16*√2. The answer is
B
