Question #5c832

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Question #5c832

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Answer 1 · 2017-10-22T17:54:59

$s=36sqrt(2)-36$

Explanation:

(PS, it's best to draw this octagon. The explanation will make much more sense).

Imagine a regular octagon $ABCDEFGH$ , where $A$ is the point at the top left and you move clockwise to letter the points. We can use the following formula to determine each angle in an octagon:

$180*("num sides"-2)/("num sides")$

$=180*(8-2)/8$

$=180*6/8$

$=135^@$

Now, on your octagon, draw the line $AF$ . Based on the problem, we know that $AF$ is $36$ inches. Let's also assume that every side length of the octagon has length $s$ . Then, draw the line from point $G$ to the line $AF$ . Call this point of intersection $I$ . Notice that $GI$ and $AF$ are perpendicular and $GI$ and $GH$ are perpendicular. This means that angle $HGI$ is $90^@$ , so angle $FGI$ is $135-90=45^@$ .

Since angle $GIF$ is $90^@$ , triangle $GIF$ is a $45-45-90$ triangle. Since $GF$ has length $s$ , we can use trigonometry to determine that $FI$ has length $s/sqrt(2)$ .

Note that if you draw the line from $H$ to $AF$ which intersect at point $J$ , then you get rectangle $GHJI$ . Since $GH$ has side length $s$ , so does $IJ$ .

Finally, we can see that triangle $HJA$ is congruent to $GIF$ , so $JA$ also has side length $s/sqrt(2)$ .

We know that $AF$ has length $36$ . But $AF$ can be broken down to three parts: $AJ$ , $JI$ , and $IF$ . Thus:

$s/sqrt(2)+s+s/sqrt(2)=36$

$s+ssqrt(2)+s=36sqrt(2)$

$s(2+sqrt(2))=36sqrt(2)$

$s=(36sqrt(2))/(2+sqrt(2))$

$s=(36sqrt(2))/(2+sqrt(2))*(2-sqrt(2))/(2-sqrt(2))$

$s=(72sqrt(2)-72)/(2)$

$s=36sqrt(2)-36$

Answer 2 · 2017-10-22T19:18:47

The sides are $14.9$ inches long.

Explanation:

The distance between two parallel sides of the octagon is given as $36$ inches.

A regular octagon can be divided up into two isosceles trapeziums and a rectangle. Therefore the distance of $36$ inches is made up of:

Height + height + side. (height being the distance between the parallel sides of the isosceles trapeziums.

As each angle of the octagon is $135°$ , the base angles of the trapezium are $135°-90° = 45°$

We therefore need to find the heights of two right-angled isosceles triangles.

With $s$ as the length of the sides of the octagon and $h$ as the equal sides of the triangles, by Pythagoras' we have:

$h^2+h^2 = s^2$

$2h^2 = s^2$

$h^2 = s^2/2$

$h = sqrt(s^2/2)$

Now make an equation from: height + height + side = $36$

$sqrt(s^2/2)+sqrt(s^2/2)+s =36$

$(2sqrts^2)/sqrt2 = 36-s$

$(2sqrts^2)/sqrt2 xxsqrt2/sqrt2= 36-s$

$(cancel2sqrt(2s^2))/cancel2=36-s" "larr$ square both sides

$2s^2 = 1296-72s+s^2$

$s^2 +72s -1296=0" "larr$ solve by completing the square

$s^2 +72s +36^2 = 1296+36^2$

$(s+36)^2 = 2592$

$s+36 = sqrt2592" "larr$ only positive root is valid

$s = 50.911688 -36$

$s=14.9$ inches

Answer 3 · 2017-10-22T20:38:31

$14.91$ (2.d.p.)

Explanation:

There is a really quick way of solving this.

$180/8=45$ This is the angle at the apex of the triangles that are form when dividing the octagon into sectors. $36/2=18$ is the length of the perpendicular bisector of the angle at the apex and meets the base line you want to find. So using this:

$tan(22.5)= 1/2$ base line/18:

So:

18tan(22.5)= half the length of octagon side:

$36tan(22.5)= 14.91$ (2.d.p.)

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