find the length, to the nearest tenth, of the apothem of a regular ocatagon whose sides are each 10 inches long.

write an equation of the line that passes through the point at (4,4) and is perpendicular to the line whose equation is 2x+y=7

i want to know how to do these problems so if you could show the work and explain please.

## Answers (1)

I'll do some related problems and then you try your problems:

1) The apothem of a regular pentagon with sides 14 centimeters long.

You need to know that the apothem is the perpendicular distance from the center to a side. In a regular polygon it is the perpendicular bisector of each side. If you add segments from the center to each vertex, then the polygon (with n sides) is divided into 2n identical right triangles who legs are the apothem and half the side.

For a regular pentagon, we have 10 identical right triangles, each with an angle of 360/10 = 36 degrees at the center, an opposite leg equal half the side: 14/2=7 cm, and an adjacent leg equal the apothem: x.

Tan(36 degrees) = 7/x So: x = 7/Tan(36) = (7/5)√(5(5+2√5))

2) Equation of a line through (3,9) perpendicular to the line 4x+5y=9.

A few facts you ought to know by heart:

1) Point-line form of the equation of a line:

y=m(x-a)+b where (a,b) is a point on the line and "m" is the slope.

2) Parallel lines have the same slope

3) Perpendicular lines have negative reciprocal slope. (-1/m) is perpendiculalr to m.

You also ought to be able to rearrange the given equation.

(Solve for 'y', to get the point-slope from):

y = -(4/5)x + (9/5)

Then the perpendicular slope is +5/4 and the point-slope form gives:

y = (5/4)(x-3) + 9 or 5x - 4y + 21 = 0

Ask about any parts that you do no understand.

References