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Posted by on Thursday, January 15, 2009 at 3:10pm.

In a regular octagon, AB is a diagonal and CD joins the midpoints of two opposite sides. The side length of the octagon is 4 cm. To the nearest tenth of a cm. find a)AB and b)CD. I'm stumped.

  • trig - , Thursday, January 15, 2009 at 3:28pm

    form a small right-angled isosceles triangle by extending the horizontal side and the vertical side at the top right of your figure.
    this triangle has a hypotenuse of 4 cm
    let each of its sides be x
    then x^2 + x^2 = 16
    x = √8 or 2√2

    I dres CD horizontally, so clearly CD = 2√2 + 4 + 2√2 = 4 + 4√2

    I then constructed a large right-angled triangle inside the octogon with AB as the hypotenuse.
    Then the long side must be the length of CD or (4+4√2) and the shorter side must be 4
    AB^2 = (4+4√2)^2 + 4^2

    Don't know if you need your answer in radical form or as a decimal, I will let you decide.

  • trig - , Thursday, January 15, 2009 at 4:00pm

    Thanks so much! If I may pick at your brain, how did you know to draw the isosceles triangle at the top and the large triangle inside the octogan to solve this problem.I didn't even consider this.

  • trig - , Thursday, January 15, 2009 at 4:09pm

    Cathy, no easy answer to your last question.
    I guess teaching math for 35 years might have had something to do with it, lol

    Generally, if I have geometry problems like this, I start by drawing all kinds of lines. Sometimes I get results, sometimes it leads to dead ends.

  • trig - , Thursday, January 15, 2009 at 4:35pm

    Thanks again. I really appreciate it.

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