trigonometry word problem

Question

a theater has 17 seats in first row and 22 rows in all each succesive row contains 2 more how many seats are in the theater?

 

 

 

.. pls i really need help im begging

Answer 1

Hello Daisy,

 

First, I don't see any need for trig in this problem. Be "creatively lazy" and use a shortcut. Take the "rectangle" of seats out of the equation...the grid of 17 seats per row going back 22 rows. That's 374 seats. Now as to the rest, imagine you're adding one seat per row to either side of the middle block of 374 seats...giving you a theater shaped like a trapezoid.

 

So you have two right triangles on either side of the middle block of 374 seats. Each right triangle is 21 rows deep, and begins with one seat (in the second row) and ends with 21 seats (in the 22nd row).

 

Before you sum up 1 + 2 + ... + 21, and then double it, not that the two  right triangles are congruent! So  flip one over in your mind and combine them to yield another rectangle, this one 21 seats deep (for the 21 rows behind the first row) and 22 seats across. This is 21*22 seats or 462 more seats.

 

Your total seats are thus 374 + 462 = 836 seats. Note that the seats added to the rows total substantially more than the seats in the middle rectangle of 374 seats. This surprised me, but my math is correct.

 

I'm not sure at all how you'd use trigonometry for this problem, since trig uses continuous functions and this problem deals with countable, discrete data.

 

To check your answer, what is the area of a trapezoid 22 units high, with bases of 17 units and 59 units? The area of a trapezoid is height * average base length, in this case area = 22 * ((17+59)/2) = 22 * 38 = 836. Now pretend this trapezoid is your theater, and one "unit" is a seat...you've figured that out already!

 

Rest easy, my friend. And get seats in the tenth row center (any closer you miss stuff).

 

Eric Moline

Tutor and Top Trombonist (I know all about good seats for a show)

McMinnville OR

Answer 2

This asks for the sum of an arithmetic series.

 

S_n = (n/2)[2a₁ + (n - 1)d], where S = sum, a₁ = first term, n = number of terms, d = common difference

 

S₂₂ = (11)[2(17) + (22 - 1)2]

S₂₂ = (11)[34 + 21(2)]

S₂₂ = (11)[34 + 42]

S₂₂ = (11)(76)

S₂₂ = 836