Some Geometry Help

I've been given an extra credit page in geometry and I need some help. Here are a few of the questions I need help on.

1) The angles of a hexagon have degree measures that are consecutive odd integers. What is the degree measure of the largest angle?

2) When the length of each of the edges of a cube is increased by 1 cm., the cube's total surface area increases by 78 square centimeters. What is the length of an edge of the original cube?

3) If the perimeter of an isosceles triangle is 36 cm. and if the altitude to its base is 12 cm., what is the triangles area?

Any help would be appreiciated.

1) the six angles of an hexagon have to add up 720º (this is easily demonstrated by dividing the hexagon in four triangles, drawing three diagonals from the same vertex, each has to have 3 angles that add up 180º). So, if a = largest angle, then the angles would be a, a - 2, a - 4, a - 6, a - 8, a - 10.

So (a) + (a-2) + (a-4) + (a-6) + (a-8) + (a-10) = 720 => (implies)
6a - 30 = 720 = >
6a = 750 =>
a = 125

the largest angle is 125º.



2) Let "a " be the lenght of one edge of the original cube, let "S" be the surface area of this cube...

first: S = 6 a^2 (6 sides of a cube) : (1)

now, if a is increased by 1, S is increased by 78, which is written:

S + 78 = 6 (a+1)^2 which implies that:

S + 78 = 6a^2 + 12a + 6, now substitute (1)

S + 78 = S +12a +6, so:
12a = 72

a = 6



3) if 2b is the base of the triangle, and the other 2 sides are a, you can divide this isoceles triangle in two rectangular ones, the sides of this rectangular triangles would be: 12, a, and b.
Pitagoras: a = square root ( 144 + b^2) : (2)

on the other hand, the perimeter is 36, so: 2b + 2a = 36 substituting (2) here leads to:

b + squareroot (144 + b^2) = 18
so:
144 + b^2 = 324 -36b + b^2...
36b = 180

b = 5

So the area would be 2b * 12 / 2 = 60 cm ^2

check them, I might have made a mistake..

Wow!!! Thanks so much Rock! I owe you one. It's not much, but if you want a sig I'll make you one.

thanks, someday I'll change my name and avatar and ask you for a sig... (I like rock lee, but I'm getting a little tired of being him)...

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