The rest of the question is the attached figure
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solution:
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As show in the attached figure
∠M = ∠R = 54.4°
∠N = ∠T = 71.2°
∠O = 180° - (∠M + ∠N) = 180° - (54.4°+71.2°) = 54.4°
∠S = 180° - (∠R + ∠T) = 180° - (54.4°+71.2°) = 54.4°
∠O = ∠S = 36°
∴ Δ MNO is similar to Δ RTS
So, the correct statement:
The triangles each have two given angle measures and one unknown angle measure.
So hmm notice the picture below
the pyramid itself, is really, just one regular hexagon, at the bottom
and 6 triangles, stacked up at each other at the edges
now, if you just get the area of the regular hexagon, and the 6 triangles, add them up, that'd be the total surface area of the pyramid then
now, for the triangles, well, area of a triangle is 1/2 bh, as you'd know, and you have both
2(2y^4 + 7y^3 - 4y^2 -20)
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Y^2
4y^2 - 40/y^2 + 14y - 8
2(-2y^4 - 7y^3 + 4y^2 + 20)
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Y^2
9. -1,07, 1/2. 3........it is all I could see
10. -V25= -5
-5, -4,3, 0, V15, 4,2
