Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2021-04-11 00:16:58

clara95
Guest

Math, Reviews about finding areas of polygons

Hello,

I've been having problems with a particular exercise about finding the area of a Hexagon being divided into two trapezoids, each side measuring 8 cm. Step 1 was to find the area of a hexagon using the triangle method which was quite easy. This is the exercise:

1. Find the area of a regular hexagon with 8 feet sides.  Round the answers to the nearest whole number. (2 pts.)

    Part 1: Show your work using the triangle method taught in Lesson 21.

    Part 2: Find the area of the same hexagon using two trapezoids.  Confirm that the answers to Part 1 and Part 2 are equivalent.

    Part 3:  Construct a prism from the above regular hexagon.  The height of the hexagonal prism is 5 feet. Find the surface area and the volume of this prism.

First, I drew a diagram
Forming  6 congruent triangles
draw the height or apothem of one of the triangles
each central angle is 360/6=60°
the apothem forms two 30-60-90 right triangles
the side opposite the 30° angle is half the hypotenuse
you have a right triangle with an 8-foot hypotenuse and one leg being 4 feet
using the Pythagorean Theorem...
8^2=4^2+h^2
64=16+h^2
h^2=48
h=4√3
Area=(1/2)*apothem*perimeter
A=(1/2)(4√3)(48)
A=2√3*48
A=96√3
A=96*1.7320508
A=166.277 square feet

So the area of the hexagon is 166.277 square feet using the triangle method, now I need to do the same thing only I need to divide it into two trapezoids. So first I did so and then I drew the height which created two little right triangles. Sum of total angle of a hexagon = (6–2) x 180 = 720.
Angle of each side = 720 / 6 = 120. And I tried finding the area of the triangle and later on finding the area of the trapezoid for the last step I added the two trapezoids areas to find my hexagon's area but no matter how much I try it won't measure 166.277 square feet as it did with the triangle method.

Can you please point at what I'm doing wrong? It's been 4 days and I'm still stuck at this, I have no clue how to solve it. I'm not asking anyone to do my homework, just kindly explain it to me, please!

Thanks in advance,
Emily

#2 2021-04-11 19:15:31

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Math, Reviews about finding areas of polygons

hi clara95

Welcome to the forum.

Your triangle method and answer are correct.  I'm not quite sure what you're doing for the trapezoid method.  The usual formula for the area of a trapezoid  is sum of the parallels divide by 2 and multiply by the distance between them.  The parallels are 16 and 8; the distance between is the same as the apothem you have already worked out so the calculation becomes

(16+8)/2 x 4√3 x 2 as there are two such trapezoids.

Hope that helps,

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#3 2021-04-22 11:19:51

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Math, Reviews about finding areas of polygons

Bob wrote:

hi clara95

Welcome to the forum.

Your triangle method and answer are correct.  I'm not quite sure what you're doing for the trapezoid method.  The usual formula for the area of a trapezoid  is sum of the parallels divide by 2 and multiply by the distance between them.  The parallels are 16 and 8; the distance between is the same as the apothem you have already worked out so the calculation becomes

(16+8)/2 x 4√3 x 2 as there are two such trapezoids.

Hope that helps,

Bob

Problems like the ones listed here require a picture for a better, deeper, fuller grasp. Agree?

Offline

Board footer

Powered by FluxBB