draw an octagon that has perpendicular lines. 8 sides

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# Brief Explanations:
To draw an octagon with perpendicular lines (a rectangle - like base with additional sides, or a regular octagon has right angles in a way when considering its structure? Wait, a regular octagon has internal angles of 135 degrees, but to have perpendicular lines, a rectangle - based octagon (a rectangle with its corners cut off at 45 - degree angles, but actually, a more straightforward way is to draw a rectangle, then extend or cut sides to make 8 sides with right angles at some vertices. Alternatively, a "rectangular octagon" where 4 of the angles are right angles (90 degrees) and the other 4 are 270 degrees? Wait, no, the sum of internal angles of an octagon is $(8 - 2)\times180=1080$ degrees. If we have 4 right angles (90 each) and 4 angles of $x$, then $4\times90 + 4x=1080$, so $360+4x = 1080$, $4x = 720$, $x = 180$ which is a straight line, so that's not possible. Wait, maybe a better approach: draw a square, then on each side, add a smaller square (or a rectangle) attached to the side, but that would make more than 8 sides. Wait, no. Let's think of a rectangle with length $L$ and width $W$, then on each of the four sides, we cut off a right - angled triangle (or add a right - angled triangle) such that the total number of sides becomes 8. For example, start with a rectangle. Then, on the top side, instead of a straight side, we have a horizontal segment, a vertical segment up, a horizontal segment, and a vertical segment down? No, that's more complicated. Alternatively, draw a regular octagon, but a regular octagon doesn't have perpendicular lines (its internal angles are 135 degrees). Wait, maybe the problem means a octagon with some sides perpendicular to each other. Let's take a simple approach: draw a rectangle. Then, on the top - left corner, instead of a right angle, we draw a horizontal line to the left, a vertical line up, a horizontal line to the right, and a vertical line down? No, that's not right. Wait, maybe the problem is to draw an octagon where at least some lines are perpendicular. Let's draw a square. Then, on each of the four sides, we extend the side by a small length and then turn 90 degrees. For example, start with a square with side length $s$. Then, on the top side, from the left end, move right by $a$, then up by $a$, then right by $s - 2a$, then down by $a$ to meet the original top - right corner? No, that's not clear. Alternatively, draw a rectangle with length $L$ and width $W$. Then, on the top edge, we have a horizontal line of length $L - 2x$, then a vertical line up of length $x$, then a horizontal line of length $2x$, then a vertical line down of length $x$ to the original top - right corner? No, this is getting too complicated. A better way: draw a "stop - sign - like" octagon but with some right angles. Wait, maybe the problem is simpler. Let's draw a rectangle. Then, on each of the four sides, we add a small square (or a rectangle) attached to the side, but that would make 12 sides. No. Wait, the key is to have 8 sides with some of them perpendicular. Let's draw a square. Then, on each of the four sides, we cut off a right - angled triangle (with legs of length $a$) from each corner. So the original square has 4 sides, and after cutting off 4 triangles (one from each corner), we have 4 new sides (the hypotenuses of the triangles) and the remaining parts of the original sides. Wait, the original square has 4 sides, each of length $s$. After cutting off a triangle with legs $a$ from each corner, the length of the remaining part of each original side is $s - 2a$, and we have 4 new sides (the hypotenuses) of length $a\sqrt{2}$. So the total number of sides is $4+(4) = 8$. Now, the original sides (the remaining parts) are parallel to each other (horizontal and vertical), so they are perpendicular to the adjacent original sides. So in this octagon, the remaining parts of the original square's sides are perpendicular to each other. For example, if the original square has horizontal and vertical sides, the remaining horizontal sides are perpendicular to the remaining vertical sides. So to draw it:

1. Draw a square.
2. On each corner of the square, draw a right - angled triangle with legs of equal length (so that the hypotenuse is at 45 degrees to the square's sides) and cut off the corner (or draw the triangle outside, depending on whether we are adding or subtracting).

So the steps to draw:

- Draw a square with side length, say, 5 units.
- On the top - left corner, from the top - left vertex, move right by 1 unit and down by 1 unit, then draw a line from that point to the top - left vertex (this is the hypotenuse of the right - angled triangle we are cutting off).
- Do the same for the top - right, bottom - right, and bottom - left corners.

Now, the resulting figure is an octagon. The sides that were part of the original square (the horizontal and vertical remaining segments) are perpendicular to each other.

# Answer:
To draw the octagon with perpendicular lines:
1. Draw a square.
2. On each of the four corners of the square, cut off a right - angled triangle (with legs of equal length) such that the remaining figure has 8 sides. The sides that were originally part of the square (the horizontal and vertical segments) will be perpendicular to each other. (You can also draw this on a piece of paper by first sketching a square, then marking equal - length segments on each side near the corners and connecting the marks to form the octagon.)

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