Geometry

Find the area of a hexagon with the indicated apothem:

6 sqr root 3

108 sqr root 3 in.^2
432 sqr root 3 in.^2
96 sqr root 3 in.^2
216 sqr root 3 in.^2

Last question, don't know. Please help? Thanks

asked by SkatingDJ
  1. Area = 1/2 x perimeter x apothem
    = 1/2 (6s)(6√3)

    once you tell me what the side (s) of the hexagon is, we an finish the question

    posted by Reiny
  2. They don't tell you the sides, the apothem is all they give you.

    posted by SkatingDJ
  3. So, I guess we can't find the area, since we need the side

    posted by Reiny
  4. Hm, I mean, there has to be a way to figure the side out with just the apothem. Problem is that I'm not good with formulas or square roots.
    First off, what is the apothem? What is 6 square root 3?

    posted by SkatingDJ
  5. ok, ok

    I stand corrected, there is a way, we have an angle.
    The apothem is the distance from the centre to the midpoint of a side

    Let's look at one of the triangles of the hexagon.
    It has a central angle of 60
    so in a right-angled triangle formed by the apothem and half the base, I'll call it x
    x/6√3 = tan 30
    x = 6√3tan30
    base = 12√3tan30= 12

    area = Area = 1/2 x perimeter x apothem
    = (1/2)(72)(6√3)
    = 216√3 units^2

    posted by Reiny
  6. Unit 5: Area
    Lesson 4: Perimeters and Areas of Similar Figures
    Perimeters and Areas of Similar Figures Quiz
    1) A - 14.04cm^2
    2) A - 34cm^2
    3) A - 144.5cm^2
    4) D - 12cm
    5) A - 144 + 72squareroot3
    6) B - 15.9ft^2
    7) D - 72squareroot3
    8) B - 73.5squareroot3cm^2
    9) C - 5 : 4 and 25 : 16
    10) C - 3181m^2
    11) A - $81.25
    12) D - the area would not change
    13) D - 216squareroot3in^2
    100%

    posted by Raaayleeee

Respond to this Question

First Name

Your Response

Similar Questions

  1. Algebra2/Trig

    sqr(x^2-4)+(x^2)/x^2+1 I got ((x^2+1)*sqr(x^2-4)+x^2)/x^2+1 But my teacher gave (sqr(x^2-4)+x^2)/x^2+1 I did the problem over again but I can't figure out why my teacher got the answer. Can someone confirm that my teacher is right
  2. Calc

    How close is the semi circle y= sqr.root of 16-x^2 to the point (1, sqr.root 3)? using Optimization
  3. Math

    Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My Answer: sqr(x+1)/sqr( x-1) sqr(x^2-1)/ (x-1) However, I also have to state the restrictions to the domain and range, which I do not know how to do. Could
  4. Math

    Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My Answer: sqr(x+1)/sqr( x-1) sqr(x^2-1)/ (x-1) However, I also have to state the restrictions to the domain and range, which I do not know how to do. Could
  5. Algebra 2

    Ok, so I'm trying to find the inverse of each function. We're learning about Inverse Functions and Relations. The problem is f(x)=sqr.root of x/6 The sqr root is confusing me.
  6. algebra

    sqr root of 3a + 9(sqr root of 27a^3)
  7. algebra

    (sqr root of 2 + 3*sqr root of 5)(sqr root of 2-3*sqr root of 5)
  8. math

    I have a question I have been working on since yesterday and I am not making this up. I couldn't get the right answer. If sin theta = -2/3, which of the following are possible? A: cos theta= -the sqr rt of 5/3 and tan theta =2/3.
  9. calculus

    Is this the correct answers for these questions Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a) b.f(x)=x^3=x-4 on [-2,3] c= square root 7/3 c. f(x)= x^3 on
  10. Physical chemistry

    help! I am trying to figure out how to convert from equation one to equation two. I have spent two hours on this and cannot figure it out. I would greatly appreciate if someone could help me out on this. Equation1: v= sqr root

More Similar Questions