Objective-C adaptation of answer by nempoBu4:

`typedef enum { path_move_to, path_line_to} Path_command;static inline CGFloat sqr (CGFloat a) { return a * a;}static inline CGFloat positive_angle (CGFloat angle) { return angle < 0 ? angle + 2 * (CGFloat) M_PI : angle;}static void add_corner (UIBezierPath* path, CGPoint p1, CGPoint p, CGPoint p2, CGFloat radius, Path_command first_add) { // 2 CGFloat angle = positive_angle (atan2f (p.y - p1.y, p.x - p1.x) - atan2f (p.y - p2.y, p.x - p2.x)); // 3 CGFloat segment = radius / fabsf (tanf (angle / 2)); CGFloat p_c1 = segment; CGFloat p_c2 = segment; // 4 CGFloat p_p1 = sqrtf (sqr (p.x - p1.x) + sqr (p.y - p1.y)); CGFloat p_p2 = sqrtf (sqr (p.x - p2.x) + sqr (p.y - p2.y)); CGFloat min = MIN(p_p1, p_p2); if (segment > min) { segment = min; radius = segment * fabsf (tanf (angle / 2)); } // 5 CGFloat p_o = sqrtf (sqr (radius) + sqr (segment)); // 6 CGPoint c1; c1.x = (CGFloat) (p.x - (p.x - p1.x) * p_c1 / p_p1); c1.y = (CGFloat) (p.y - (p.y - p1.y) * p_c1 / p_p1); // 7 CGPoint c2; c2.x = (CGFloat) (p.x - (p.x - p2.x) * p_c2 / p_p2); c2.y = (CGFloat) (p.y - (p.y - p2.y) * p_c2 / p_p2); // 8 CGFloat dx = p.x * 2 - c1.x - c2.x; CGFloat dy = p.y * 2 - c1.y - c2.y; CGFloat p_c = sqrtf (sqr (dx) + sqr (dy)); CGPoint o; o.x = p.x - dx * p_o / p_c; o.y = p.y - dy * p_o / p_c; // 9 CGFloat start_angle = positive_angle (atan2f ((c1.y - o.y), (c1.x - o.x))); CGFloat end_angle = positive_angle (atan2f ((c2.y - o.y), (c2.x - o.x))); if (first_add == path_move_to) { [path moveToPoint: c1]; } else { [path addLineToPoint: c1]; } [path addArcWithCenter: o radius: radius startAngle: start_angle endAngle: end_angle clockwise: angle < M_PI];}UIBezierPath* path_with_rounded_corners (NSArray<NSValue*>* points, CGFloat corner_radius) { UIBezierPath* path = [UIBezierPath bezierPath]; NSUInteger count = points.count; for (NSUInteger i = 0; i < count; ++i) { CGPoint prev = points[i > 0 ? i - 1 : count - 1].CGPointValue; CGPoint p = points[i].CGPointValue; CGPoint next = points[i + 1 < count ? i + 1 : 0].CGPointValue; add_corner (path, prev, p, next, corner_radius, i == 0 ? path_move_to : path_line_to); } [path closePath]; return path;}`

Here is a way using some geometry:

- the two lines are tangent to circle inscribed
- The normal to the tangent meet at centre of the circle.
- Let angle between lines be X
- Angle subtended at centre of the circle will be K = 360-90*2-X = 180-X
- Lets decide the two point of tangents as (x1,y) and (x2,y)
- The chord joining the points has length l = (x2-x1)
- Inside the circle, the chord and two normals of length r (radius) form isosceles triangle
- The pendicular divide the traingle in to equal halved right angled triangles.
- One of angle is K/2 and side is l/2
- using properties of right angle triangle sin(K/2) = (l/2)/r
- r = (l/2)/sin(K/2)
- but K = 180-X so r = (l/2)/sin(90-X/2) = (l/2)/cos(X/2)
- hence r = (x2-x1)/(2*cos(X/2))
- Now simply draw a arc from (x1,y) to (x2,y) using radius r

### Note

The above is explained only for lines which meet at origin and where the Y axis divides the angle between them in half. But it is equally applicable for all corners that you just need to apply a rotation and translation before applying the above. Moreover you need to select some X values of intersection from where you want to draw the arc. The values should not be too far or close to origin.

## Some geometry with Paint:

0. You have a corner:

1. You know the coordinates of corner points, let it be P_{1}, P_{2} and P:

2. Now you can get vectors from points and angle between vectors:

angle = atan(P_{Y}- P_{1Y}, P_{X}- P_{1X}) - atan(P_{Y}- P_{2Y}, P_{X}- P_{2X})

3. Get the length of segment between angular point and the points of intersection with the circle.

segment = PC_{1}= PC_{2}= radius / |tan(angle / 2)|

4. Here you need to check the length of segment and the minimal length from PP_{1} and PP_{2}:

Length of PP_{1}:

PP_{1}= sqrt((P_{X}- P_{1X})^{2}+ (P_{Y}- P_{1Y})^{2})

Length of PP_{2}:

PP_{2}= sqrt((P_{X}- P_{2X})^{2}+ (P_{Y}- P_{2Y})^{2})

If segment > PP_{1} or segment > PP_{2} then you need to decrease the radius:

min = Min(PP_{1}, PP_{2}) (for polygon is better to divide this value by 2)segment > min ? segment = min radius = segment * |tan(angle / 2)|

5. Get the length of PO:

PO = sqrt(radius^{2}+ segment^{2})

6. Get the C_{1X} and C_{1Y} by the proportion between the coordinates of the vector, length of vector and the length of the segment:

Proportion:

(P_{X}- C_{1X}) / (P_{X}- P_{1X}) = PC_{1}/ PP_{1}

So:

C_{1X}= P_{X}- (P_{X}- P_{1X}) * PC_{1}/ PP_{1}

The same for C_{1Y}:

C_{1Y}= P_{Y}- (P_{Y}- P_{1Y}) * PC_{1}/ PP_{1}

7. Get the C_{2X} and C_{2Y} by the same way:

C_{2X}= P_{X}- (P_{X}- P_{2X}) * PC_{2}/ PP_{2}C_{2Y}= P_{Y}- (P_{Y}- P_{2Y}) * PC_{2}/ PP_{2}

8. Now you can use the addition of vectors PC_{1} and PC_{2} to find the centre of circle by the same way by proportion:

(P_{X}- O_{X}) / (P_{X}- C_{X}) = PO / PC(P_{Y}- O_{Y}) / (P_{Y}- C_{Y}) = PO / PC

Here:

C_{X}= C_{1X}+ C_{2X}- P_{X}C_{Y}= C_{1Y}+ C_{2Y}- P_{Y}PC = sqrt((P_{X}- C_{X})^{2}+ (P_{Y}- C_{Y})^{2})

Let:

dx = P_{X}- C_{X}= P_{X}* 2 - C_{1X}- C_{2X}dy = P_{Y}- C_{Y}= P_{Y}* 2 - C_{1Y}- C_{2Y}

So:

PC = sqrt(dx^{2}+ dy^{2})O_{X}= P_{X}- dx * PO / PCO_{Y}= P_{Y}- dy * PO / PC

9. Here you can draw an arc. For this you need to get start angle and end angle of arc:

Found it here:

startAngle = atan((C_{1Y}- O_{Y}) / (C_{1X}- O_{X}))endAngle = atan((C_{2Y}- O_{Y}) / (C_{2X}- O_{X}))

10. At last you need to get a sweep angle and make some checks for it:

`sweepAngle = endAngle - startAngle`

If sweepAngle < 0 then swap startAngle and endAngle, and invert sweepAngle:

`sweepAngle < 0 ? sweepAngle = - sweepAngle startAngle = endAngle`

Check if sweepAngle > 180 degrees:

`sweepAngle > 180 ? sweepAngle = 180 - sweepAngle`

11. And now you can draw a rounded corner:

## Some geometry with c#:

`private void DrawRoundedCorner(Graphics graphics, PointF angularPoint, PointF p1, PointF p2, float radius){ //Vector 1 double dx1 = angularPoint.X - p1.X; double dy1 = angularPoint.Y - p1.Y; //Vector 2 double dx2 = angularPoint.X - p2.X; double dy2 = angularPoint.Y - p2.Y; //Angle between vector 1 and vector 2 divided by 2 double angle = (Math.Atan2(dy1, dx1) - Math.Atan2(dy2, dx2)) / 2; // The length of segment between angular point and the // points of intersection with the circle of a given radius double tan = Math.Abs(Math.Tan(angle)); double segment = radius / tan; //Check the segment double length1 = GetLength(dx1, dy1); double length2 = GetLength(dx2, dy2); double length = Math.Min(length1, length2); if (segment > length) { segment = length; radius = (float)(length * tan); } // Points of intersection are calculated by the proportion between // the coordinates of the vector, length of vector and the length of the segment. var p1Cross = GetProportionPoint(angularPoint, segment, length1, dx1, dy1); var p2Cross = GetProportionPoint(angularPoint, segment, length2, dx2, dy2); // Calculation of the coordinates of the circle // center by the addition of angular vectors. double dx = angularPoint.X * 2 - p1Cross.X - p2Cross.X; double dy = angularPoint.Y * 2 - p1Cross.Y - p2Cross.Y; double L = GetLength(dx, dy); double d = GetLength(segment, radius); var circlePoint = GetProportionPoint(angularPoint, d, L, dx, dy); //StartAngle and EndAngle of arc var startAngle = Math.Atan2(p1Cross.Y - circlePoint.Y, p1Cross.X - circlePoint.X); var endAngle = Math.Atan2(p2Cross.Y - circlePoint.Y, p2Cross.X - circlePoint.X); //Sweep angle var sweepAngle = endAngle - startAngle; //Some additional checks if (sweepAngle < 0) { startAngle = endAngle; sweepAngle = -sweepAngle; } if (sweepAngle > Math.PI) sweepAngle = Math.PI - sweepAngle; //Draw result using graphics var pen = new Pen(Color.Black); graphics.Clear(Color.White); graphics.SmoothingMode = SmoothingMode.AntiAlias; graphics.DrawLine(pen, p1, p1Cross); graphics.DrawLine(pen, p2, p2Cross); var left = circlePoint.X - radius; var top = circlePoint.Y - radius; var diameter = 2 * radius; var degreeFactor = 180 / Math.PI; graphics.DrawArc(pen, left, top, diameter, diameter, (float)(startAngle * degreeFactor), (float)(sweepAngle * degreeFactor));}private double GetLength(double dx, double dy){ return Math.Sqrt(dx * dx + dy * dy);}private PointF GetProportionPoint(PointF point, double segment, double length, double dx, double dy){ double factor = segment / length; return new PointF((float)(point.X - dx * factor), (float)(point.Y - dy * factor));}`

To get points of arc you can use this:

`//One point for each degree. But in some cases it will be necessary // to use more points. Just change a degreeFactor.int pointsCount = (int)Math.Abs(sweepAngle * degreeFactor);int sign = Math.Sign(sweepAngle);PointF[] points = new PointF[pointsCount];for (int i = 0; i < pointsCount; ++i){ var pointX = (float)(circlePoint.X + Math.Cos(startAngle + sign * (double)i / degreeFactor) * radius); var pointY = (float)(circlePoint.Y + Math.Sin(startAngle + sign * (double)i / degreeFactor) * radius); points[i] = new PointF(pointX, pointY);}`

Here is my realisation of dbc's idea on c#:

`/// <summary>/// Round polygon corners/// </summary>/// <param name="points">Vertices array</param>/// <param name="radius">Round radius</param>/// <returns></returns>static public GraphicsPath RoundCorners(PointF[] points, float radius) { GraphicsPath retval = new GraphicsPath(); if (points.Length < 3) { throw new ArgumentException(); } rects = new RectangleF[points.Length]; PointF pt1, pt2; //Vectors for polygon sides and normal vectors Vector v1, v2, n1 = new Vector(), n2 = new Vector(); //Rectangle that bounds arc SizeF size = new SizeF(2 * radius, 2 * radius); //Arc center PointF center = new PointF(); for (int i = 0; i < points.Length; i++) { pt1 = points[i];//First vertex pt2 = points[i == points.Length - 1 ? 0 : i + 1];//Second vertex v1 = new Vector(pt2.X, pt2.Y) - new Vector(pt1.X, pt1.Y);//One vector pt2 = points[i == 0 ? points.Length - 1 : i - 1];//Third vertex v2 = new Vector(pt2.X, pt2.Y) - new Vector(pt1.X, pt1.Y);//Second vector //Angle between vectors float sweepangle = (float)Vector.AngleBetween(v1, v2); //Direction for normal vectors if (sweepangle < 0) { n1 = new Vector(v1.Y, -v1.X); n2 = new Vector(-v2.Y, v2.X); } else { n1 = new Vector(-v1.Y, v1.X); n2 = new Vector(v2.Y, -v2.X); } n1.Normalize(); n2.Normalize(); n1 *= radius; n2 *= radius; /// Points for lines which intersect in the arc center PointF pt = points[i]; pt1 = new PointF((float)(pt.X + n1.X), (float)(pt.Y + n1.Y)); pt2 = new PointF((float)(pt.X + n2.X), (float)(pt.Y + n2.Y)); double m1 = v1.Y / v1.X, m2 = v2.Y / v2.X; //Arc center if (v1.X == 0) {// first line is parallel OY center.X = pt1.X; center.Y = (float)(m2 * (pt1.X - pt2.X) + pt2.Y); } else if (v1.Y == 0) {// first line is parallel OX center.X = (float)((pt1.Y - pt2.Y) / m2 + pt2.X); center.Y = pt1.Y; } else if (v2.X == 0) {// second line is parallel OY center.X = pt2.X; center.Y = (float)(m1 * (pt2.X - pt1.X) + pt1.Y); } else if (v2.Y == 0) {//second line is parallel OX center.X = (float)((pt2.Y - pt1.Y) / m1 + pt1.X); center.Y = pt2.Y; } else { center.X = (float)((pt2.Y - pt1.Y + m1 * pt1.X - m2 * pt2.X) / (m1 - m2)); center.Y = (float)(pt1.Y + m1 * (center.X - pt1.X)); } rects[i] = new RectangleF(center.X - 2, center.Y - 2, 4, 4); //Tangent points on polygon sides n1.Negate(); n2.Negate(); pt1 = new PointF((float)(center.X + n1.X), (float)(center.Y + n1.Y)); pt2 = new PointF((float)(center.X + n2.X), (float)(center.Y + n2.Y)); //Rectangle that bounds tangent arc RectangleF rect = new RectangleF(new PointF(center.X - radius, center.Y - radius), size); sweepangle = (float)Vector.AngleBetween(n2, n1); retval.AddArc(rect, (float)Vector.AngleBetween(new Vector(1, 0), n2), sweepangle); } retval.CloseAllFigures(); return retval;}`

You are looking for an arc tangent to two connected line segments, of a given radius, given by some sequential array of points. The *algorithm* to find this arc is as follows:

For each segment, construct a normal vector.

If you are working in 2d, you can just subtract the two endpoints to get a tangent vector (X, Y). In that case, normal vectors will be plus or minus (-Y, X). Normalize the normal vector to length one. Finally, choose the direction with a positive dot product with the tangent vector of the next segment. (

*See update below*).If you are working in 3d not 2d, to get the normal, cross the tangent vectors of the two segments at the vertex you wish to round to get a perpendicular vector to the plane of the lines. If the perpendicular has length zero, the segments are parallel and no round can be required. Otherwise, normalize it, then cross the perpendicular with the tangent to get the normal.)

Using the normal vectors, offset each line segment towards the interior of the polygon by your desired radius. To offset a segment, offset its endpoints using the normal vector N you just computed, like so: P' = P + r * N (a linear combination).

Intersect the two offset lines to find the center. (This works because a radius vector of a circle is always perpendicular to its tangent.)

To find the point at which the circle intersects each segment, offset the circle center backwards to each original segment. These will be the endpoints of your arc.

Make sure the arc endpoints are inside each segment, otherwise you will be creating a self-intersecting polygon.

Create an arc through both endpoints with center & radius you determined.

I don't have any proper drafting software at hand, but this diagram sort of shows the idea:

At this point you will need either to introduce classes to represent a figure consisting of line and arc segments, or polygonize the arc to an appropriate accuracy and add all the segments to the polygon.

Update: I have updated the image, labeling the points P1, P2, and P3, and normal vectors Norm12 and Norm23. The normalized normals are unique only up to flipping direction, and you should choose the flips as follows:

The dot product of Norm12 with (P3 - P2) must be positive. If it is negative, multiple Norm12 by -1.0. If it is zero, the points are collinear and no rounded corner need be created. This is because you want to offset towards P3.

The dot product of Norm23 with (P1 - P2) must also be positive since you are offsetting towards P1.

I can offer a simple and very calculable and programmable approach that uses optimally few calculations including "only" 3 square roots and *no* inverse trigonometric functions whatsoever.

Do not feel daunted by the deliberately elaborate explanation that follows which I wrote as such in the interest of making sure the absolutely trivial (by comparison to all other solutions here at the time of writing this) algorithm can be understood. In fact, I devised it only after acknowledging the algorithmic and computational complexity the alternatives would require, what with multiple calls to inverse trigonometric functions (which hide a lot of computation complexity behind their apparently benign names) and substantially larger amount of operations.

(I have verified the validity of proposed approach by programming it with JavaScript and SVG. I will use the former programming language to aid in illustrating the approach)

Let's assume some corner you want to "round" is made up of known points *A*, *B* and *C*, with *B* being "the corner".

The solution can be described by the following steps:

Calculate the length of the

*BF*vector:The length equals to the radius (

*FO*) of your circle (which you obviously choose yourself and thus know) divided by the tangent of the angle between vectors*BF*and*BO*. This is obviously because the triangle made by points*B*,*O*and*F*is a 'right' triangle (the angle between vectors*BF*and*FO*is 90 degrees).The angle between vectors

*BF*and*BO*is half the angle between vectors*BA*and*BC*. This may or may not sound obvious, rest assured it's trivially provable but I omit the proof.The relationship between the angles is useful because there happens to be a fairly simple equation expressing the relationship between the tangent of an angle and the cosine of twice the angle:

`Math.tan(a/2) == Math.sqrt((1 - Math.cos(a)) / (1 + Math.cos(a))`

.And it so happens that the cosine of the angle between vectors

*BA*and*BC*(`Math.cos(a)`

) is the dot product of the two vectors divided by the product of their lengths (see definition of vector dot product on Wikipedia).And so, having calculated the cosine of the angle, you can then calculate the tangent of the half angle, and, subsequently, the length of

*BF*:(Legend: I model vectors (

`BA`

,`BC`

, etc) as objects with properties`x`

and`y`

for their respective coordinates in screen space (X increases to the right, Y downwards);`radius`

is the desired radius of the would-be rounded corner, and`BF_length`

is the length of*BF*(obviously))`/// Helper functionsconst length = v => Math.sqrt(v.x * v.x + v.y * v.y);const dot_product = (v1, v2) => v1.x * v2.x + v1.y * v2.y;const cosine_between = (v1, v2) => dot_product(v1, v2) / (length(v1) * length(v2));const cos_a = cosine_between(BA, BC);const tan_half_a = Math.sqrt((1 - cos_a) / (1 + cos_a));const BF_length = radius / tan_half_a;`

Compute the

*BF*vector. We know its length now (`BF_length`

above) and since*BF*lies on the same line the vector*BA*lies on, the former (and, by implication, the coordinate of the point*F*relative to point*B*) is computable by doing a scalar multiplication of the length of*BF*by the unit vector equivalent of*BA*:(Video) Circle Polygon Collisions Using Separating Axis Theorem - Let's Make a Physics Engine [07]`/// Helper functionsconst unit = v => { const l = length(v); return { x: v.x / l, y: v.y / l };};const scalar_multiply = (v, n) => ({ x: v.x * n, y: v.y * n });const BF = scalar_multiply(unit(BA), BF_length);`

Now that you have coordinates of

*F*from the prior step, you calculate the*FO*vector, or the*O*coordinate. This is done by rotating some vector of length`radius`

that lies on the same line that the vector*BA*lies on, both vectors pointing in the same direction, by 90 degrees, and moving it so it starts at*F*.Now, whether the rotation is clockwise or counter-clockwise depends on the

*sign*of the angle between the vectors BA and BC, more concretely if the difference between the respective angles (each counted against the same reference, in this case the X axis) of*BA*and*BC*is positive then the rotation is counter-clockwise, otherwise it's clockwise.We don't want to calculate

*angles*if we can avoid it -- it's the sign of the difference we want, after all. Long story short the sign of the angle (`sign`

) can be calculated with the expression`Math.sign(BA.x * BC.y - BA.y * BC.x)`

.Here is computation of coordinates of

*O*(`O`

), with`F`

being the coordinate of well,*F*:`/// Helper functionsconst add = (v1, v2) => ({ x: v1.x + v2.x, y: v1.y + v2.y });const rotate_by_90_degrees = (v, sign) => ({ x: -v.y * sign, y: v.x * sign });const sign = Math.sign(BA.x * BC.y - BA.y * BC.x);const O = add(F, rotate_by_90_degrees(scalar_multiply(unit(BA), radius), sign));`

That's all -- since you've obtained the point *O* with coordinates in the same space as those of your original points (*A*, *B* and *C*), you can just put a circle of the used radius with *O* as its centre.

## Optional

Computing the corresponding circular arc from points *F* and some *F'* (its equivalent on the `BC`

vector) should be fairly easy, everything considered, but I am not including it unless someone expresses a wish for it.

## Terminology

This may be obvious to most making use of this answer, but to be on the safe side: please keep in mind that in this answer I would normally refer to vectors and coordinates as the same kind of measure -- a vector has *arity* which is amount of components it has; for a 2-dimensional system of coordinates, the arity is obviously 2. A vector object thus does not specifically encode its "start", only "end" -- since there is just two components, the implication is that the vector "starts" at the coordinate system origin. The vector `BA`

, for instance is indeed the vector between points `B`

and `A`

, but since the program stores only two components for the vector (`x`

and `y`

in the snippets), it is as if the vector was moved so that the point `B`

is now at the origin of the coordinate system. A point also consists of two components, so "vector" and "point" are interchangeable. You have to understand this very clearly, otherwise some calculations I have offered may seem strange at times. It may be easier if you just think of vectors in this answer as "one dimensional" arrays with two elements each. In fact this is how I originally programmed these, but I switched to objects with `x`

and `y`

properties for the sake of illustrating the solution with code

This is a bit of a deviation from the other answers but it's also a lot simpler.

We'll deal with angles a lot so a lot of arctan2 to calculate the angles with tolerances, so know how to use that. You could also do the determinate since near 0 the lines are near parallel.

Find the angles you want to simplify, say any beyond a given threshold of some angle. Perform the following three operations.

- Segmentize: Divide the two segments that connect this angle into arbitrarily 50 small lines. You could also just segmentize the part near the sharp angle.
- Smooth: Repeatedly apply smooth to control the amount of smoothing you want of for this sharp angle. Take each series of 3 points. And move the middle-point closer to the midpoint of the first-point and the last-point. Note: closer means
`towards`

not`to`

. So`amount * (p1 - p2) + p2`

where amount is adjustable. - Simplify: Find any points whose angles have not exceeded some threshold let's say 1% and go ahead and delete those points (you can also check the deviation of the midpoint from middle-point). This would typically undo segmentize but we smoothed it and the original sharp angle propagated into a curve.

You can note that the number of times you apply smooth will propagate the angle only that many segments. So If you segmentize to 50. If you only apply smooth 10 times then only the 10 lines closest to the corner could have a modified angle.

So you just pick some values, how many lines do you segment into? How much smoothing do you apply? How many times do you apply the smoothing? And how much deviation is enough to preserve the angle when you're simplifying things back down.

I'm looking for an algorithm that allows me to create rounded corners from a polygon.

I have an array of points that represents the polygon (outlined in red) and on output I want an array of points that represents the polygon with rounded corners (outlined in black).

I would also like to have a way to control the radius of each corner.

I tried to use Bézier curves and subdivision but it's not what I'm looking for. Bézier curves and subdivision are smoothing the polygon.

What I want is to only make the corners rounded.

Does somebody know any good algorithm for doing that?

I'm working with C# but the code has to be independent from any .NET libraries.

## FAQs

### Can a polygon have rounded corners? ›

**Yes, they are**. The rounded polygon does not go trough the (xn, yn) points.

**Which of the following CSS3 code statements is used to create rounded corners? ›**

The **border-radius** CSS property rounds the corners of an element's outer border edge. You can set a single radius to make circular corners, or two radii to make elliptical corners.

**Which snippet can be used to form rounded corners? ›**

With the **CSS3 border-radius property**, you can give any element "rounded corners".

**How do I make rounded corners in UIView? ›**

You can give it round corners by **changing the cornerRadius property of the view's layer** . and smaller values give less rounded corners. Both clipsToBounds and masksToBounds are equivalent. It is just that the first is used with UIView and the second is used with CALayer .

**Can a regular polygon be curved? ›**

A polygon is a flat two-dimensional shape with straight sides that are fully closed. **The sides must be straight, not curved**.

**What is a polygon with curved sides called? ›**

The **Reuleaux triangle** is the first of a sequence of Reuleaux polygons whose boundaries are curves of constant width formed from regular polygons with an odd number of sides. Some of these curves have been used as the shapes of coins.

**How do you make a shape have rounded corners in CSS? ›**

You can **use border-radius directly to create rounded corners**. You can also set different values to each corner. If you only want one corner to be rounded, for example, you can set one of the following CSS properties.

**Which CSS property would you use to configure rounded corners? ›**

The CSS property **border-radius** adds rounded corners on images. You can round all of the image's corners or just select corners, vary the radius on different corners or display an image in the shape of an oval or a circle.

**How do I make an image have rounded corners in CSS? ›**

**CSS Rounded Corners**

- Tip: This property allows you to add rounded corners to elements! ...
- Four values - border-radius: 15px 50px 30px 5px; (first value applies to top-left corner, second value applies to top-right corner, third value applies to bottom-right corner, and fourth value applies to bottom-left corner):

**What command will create rounded corners in the drawing? ›**

An inside corner is called a fillet and an outside corner is called a round; you can create both using the **FILLET command**. These object types can be selected to define a fillet or round: 2D polylines.

### What tool is used to measure rounded corners? ›

Measuring rounded edges or corners of an object can be a difficult task, but it's often a necessary element of a successful industrial design for new products. Most people approximate by using a **ruler or caliper** to measure where the curve starts and where it ends as seen in the first picture with distance A.

**Which CSS syntax is used to make the rounded borders or rounded corners around an element? ›**

Explanation: The **border-radius** CSS property sets the rounded borders and provides the rounded corners around an element, tags, or div. It defines the radius of the corners of an element.

**Which class is used to add rounded corners to an image *? ›**

img-rounded − You can make rounded corners to an image by using **.** **rounded** class.

**Can a polygon be rounded? ›**

No. A polygon by definition has straight line segments as its sides. **If a geometrical figure has curved sides, it is not considered a polygon**.

**Is the polygon tool is used to make a curved line? ›**

The Polygon tool draws paths with straight segments. The Smooth Polygon tool draws paths with smooth line segments. **The Curve and Auto Curve tools can draw paths with straight and curved segments**.

**What is a 360 sided polygon called? ›**

Several philosophers have used the regular myriagon to illustrate issues regarding thought. Regular myriagon. A regular myriagon. Type. Regular polygon.

**What is a polygon vs monogon? ›**

**A monogon is a one-sided polygon**. It is a polygon with one edge and one vertex. It is considered "regular" since it has only one side and one vertex.

**Can an element be styled to have rounded corners? ›**

**You can give any element “rounded corners” by applying a border-radius through CSS**. You'll only notice if there is a color change involved. For instance, if the element has a background-color or border that is different than the element it's above.

**What is CSS rounded corners? ›**

CSS3 Rounded corners are **used to add special colored corner to body or text by using the border-radius property**.A simple syntax of rounded corners is as follows − #rcorners7 { border-radius: 60px/15px; background: #FF0000; padding: 20px; width: 200px; height: 150px; }

**Which of the following class is used to add rounded corners? ›**

The class **.** **img-rounded** is used to add rounded corners to an image ( IE8 does not support rounded corners).

### Which property allows you to add rounded corners to elements? ›

The **border-radius** property defines the radius of the element's corners. Tip: This property allows you to add rounded corners to elements!

**What CSS property is used to add rounded corners to a button? ›**

To add rounded corners to a button, use the **border-radius** property.

**How do you make an object round in CSS? ›**

To create a circle we can **set the border-radius on the element**. This will create curved corners on the element. If we set it to 50% it will create a circle. If you set a different width and height we will get an oval instead.

**How do you round an image without stretching CSS? ›**

To fill the image size without stretching, first, **add an image with the help of the “<img>” tag along with the class attribute.** **Then, access the image using the class name and apply the “object-fit” CSS property as “cover”**. Furthermore, apply other CSS properties for styling, such as “width”, “height”, and “margin”.

**Which tool can be used to draw round cornered rectangle or shape? ›**

Description. The **Rounded Rectangle tool** creates vector rectangles with rounded corners.

**What command would you use to draw a round shape? ›**

The **circle command** is used to draw a circle by specifying the center point and radius. Let's understand by two examples. Type Circle or C on the command line and press Enter.

**Which tool creates a rectangle or square with rounded corners? ›**

The **Rectangle tool** creates rectangles with square or rounded sides.

**What is the best tool for measuring corners? ›**

**A protractor** is one of the most common tools to measure angles. When you know how to use a protractor, you can measure both small and large angles. Protractor types are categorized according to their tasks. The most common kind of protractor is found in the school supply section of big box retailers.

**How do you measure a rounded object? ›**

Diameter is the length of a straight line across the center of the circle. For this example, we have a 6” Diameter x 4” Height box. **For the Diameter: Measure, from bottom or top of the box, across the center of the circle.** **For the Height: Measure from the bottom to the top**.

**Which tool is used to draw a square with rounded corners? ›**

The **Rectangle tool** creates rectangles with square or rounded sides.

### Why do designers use rounded corners? ›

Rounded corners are also more effective for maps and diagrams. They **allow our eyes to easily follow the lines as curves are better suited to the natural movement of the head and eyes**. Sharp corners throw your eyes off the path of the line so when it changes direction you experience abrupt pauses.

**Which option of the rectangle command allows the user to create rounded edges? ›**

Type **F or Fillet** and press Enter. (Fillets are used to create round edges.) Specify the fillet radius for the rectangle.

**What command can extend and or trim lines to a sharp corner or rounded corner? ›**

to select Fillet and Trim. FILLET & TRIM draws an arc entity at the intersection of lines, circles, arcs, ellipses and splines for a smoothly rounded corner. This command will trim or extend the selected lines or arcs to meet the ends of the fillet arc.

**Which tool allows you to round the edges of a solid body by adding material to interior edges? ›**

**The Fillet command** rounds the edges of a solid body in Fusion 360 by removing material from exterior edges or adding material to interior edges.

**Which of the following CSS3 code snippets can be used to create rounded corners for any HTML element? ›**

With the CSS3 **border-radius** property, you can give any element "rounded corners".

**What CSS property must be used if you want to add rounded corners for your text inputs? ›**

The "**border-radius**" property is used to create rounded corners in CSS.

**Which option of CSS is used to set the border of the element? ›**

The **border shorthand CSS property** sets an element's border. It sets the values of border-width , border-style , and border-color .

**How would you style an image or element to have rounded corners? ›**

**CSS Rounded Corners**

- Tip: This property allows you to add rounded corners to elements! ...
- Four values - border-radius: 15px 50px 30px 5px; (first value applies to top-left corner, second value applies to top-right corner, third value applies to bottom-right corner, and fourth value applies to bottom-left corner):

**Which of the following class is used to make the rounded corners images and it automatically adjusts to fit the size of the screen? ›**

**img-responsive** class to the <img> tag to create responsive images. The responsive images adjust themselves automatically to fit according to the screen size.

**What tool is used to smooth rounded corners? ›**

The easiest way is to use **a router** – a tool specifically designed for this job. However, you can also round wood edges without a router by using a hand plane, chisel, wood file, sandpaper, or even a jig saw.

### Does a polygon always have straight sides? ›

A polygon is a flat two-dimensional shape with straight sides that are fully closed. **The sides must be straight, not curved**. However, polygons can have any number of sides.

**Are all sides of a polygon straight? ›**

Polygons have at least three sides and three angles, and **their sides must be straight**, as seen below. Examples of polygons. Notice their straight sides and closed endpoints. A polygon cannot have any curve sides.

**Can a polygon only have straight sides? ›**

**Polygons must have sides that are straight**. This means that no arcs or curves can be present in the shape if it is a polygon. This sadly eliminates circles, semicircles, crescents, and other such shapes. Polygons must have a continuous perimeter.

**What is the polygon sides rule? ›**

Polygon Formula

**The sum of interior angles of a polygon with “n” sides =180°(n-2)** Number of diagonals of a “n-sided” polygon = [n(n-3)]/2. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n. The measure of exterior angles of a regular n-sided polygon = 360°/n.

**Can a polygon have unequal sides? ›**

**Irregular polygons are polygons that do not have equal sides and equal angles**. Common examples of irregular polygons are a scalene triangle, kite, rectangle.

**Do polygons always equal 360? ›**

Exterior Angles

Because of the congruence of vertical angles, it doesn't matter which side is extended; the exterior angle will be the same. **The sum of the exterior angles of any polygon (remember only convex polygons are being discussed here) is 360 degrees**.

**Can a polygon have different side lengths? ›**

**A polygon can have a certain number of sides, but the sides do not necessarily have to be the same length**. Both of the polygons below are pentagons because they both have five angles and sides, but look at the differences.

**What is a polygon called when it has no equal sides? ›**

There are two main types of polygon - regular and irregular.

A regular polygon has equal length sides with equal angles between each side. Any other polygon is an **irregular polygon**, which by definition has unequal length sides and unequal angles between sides.

**What do you call a polygon whose sides are not? ›**

**Irregular polygons** are those types of polygons that do not have equal sides and equal angles. In other words, irregular polygons are not regular.

**Can a polygon have two side yes or no? ›**

An enclosed figure with only straight sides as edges is called a polygon. Polygons must have at least three sides, thus polygons with the fewest number of sides are triangles. Hence a polygon with two sides is **not possible**.

### Can a polygon have infinitely many sides? ›

In geometry, **an apeirogon (from Ancient Greek ἄπειρος apeiros 'infinite, boundless', and γωνία gonia 'angle') or infinite polygon is a polygon with an infinite number of sides**. Apeirogons are the two-dimensional case of infinite polytopes.

**Can a polygon have open sides? ›**

Polygons can be classified as open or closed. **A polygon is open when the segments do not all connect at the beginning and end**.

**Can a polygon have curved lines True or false? ›**

**True** a polygon is a simple closed curve made up of line segment.

**What is polygon side formula? ›**

The sum of the measures of the interior angles of a polygon with n sides is given by the general formula (n–2)180.

**What is the polygon law equation? ›**

**AC′ = AB + BC′ = AB + (– BC′) = a – b**. The difference between vectors a and b is represented by the vector AC′.

**What polygon has 1000000000 sides? ›**