<p>Ah I get it (the cube/pyramid problem), thanks. YES, guessed correctly.</p>
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<p>Pythagorean theorem. Base of whole triangle is 5, make two right triangles = 2.5. The hypotenuse of the right triangle is 5 (it’s an edge of the figure). 2.5^2 + b^2 = 5^2. b = 4.33 or 5root3/2</p>
<p>it was 48. because if you think about it, there are 3 more lengths of 12. there’s three sides left for both squares. side 1 + side 1 from the two squares = 12. meaning side 2 + side 2 = 12…that’s how i thought of it.</p>
<p>@futuresuccess - Visualize a cube. Yes, 6 sides. Now put a pyramid on opposite sides. The 2 sides that were once part of the 6 sides are now mashed between the cube and the pyramid(s). Thus only 4 sides of the cube and the 2 surface areas of the triangle are part of the surface area of the figure.</p>
<p>@overrated - The triangles were equilateral. It said all edges were 5 and all sides of all the triangles were on the edges.</p>
<p>All edges are length 5, so all of the triangles are equilateral. Split them down the middle and you get two 30-60-90 triangles. The sides of 30-60-90 triangles are ALWAYS in the ratio 1:√3:2.</p>
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<p>I did it algebraically. One side of the larger square = x, so one side of the smaller square = 12 - x.</p>
<p>Did anyone get the question with a larger cube made of up 27 smaller cubes. The surface of the larger cube is painted red, how many small cubes have to faces painted red?</p>
<p>I solved that by dividing the area of the larger triangle by the area of the triangles with height 4 and base 2 since 4 was 1/3 of 12 i thought the base had to be 1/3 of 6 so 2; is that even the correct way to do it? I got it right but I think I may just have gotten lucky</p>