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A square has diagonals that split the angles into two 45-degree parts, thus bisecting them.

Q: Which of the quadrilateral have diagonals that bisect the vertex angles?

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The diagonals of a square bisect each corner or vertex of the square.

The diagonals of a rhombus are lines drawn from one corner, or vertex, to the opposite one. They have two important properties. 1. Diagonals bisects a pair of opposite angles. 2. Diagonals are !!perpendicular!!

how many diagonals from a vertex a heptagon have

The number of Diagonals in one vertex of a Triangle is 0 (zero)..

yes it can if you can use mirror images than first you bisect the hexagon from vertex to vertex, then bisect with a perpendicular to that then WALA!

Related questions

No, but the diagonals of a square does bisects its interior angles.

Only a square and a rhombus will have all its diagonals bisecting vertices. In other shapes some - but not all - diagonals can bisect vertices.

Yes.

It depends on what you mean by bisect. All rectangles have diagonals that bisect the other one. Only certain rectangles (Squares) have diagonals that bisect its vertex, the ninety degree angle.

The diagonals of a square bisect each corner or vertex of the square.

The best classification for a parallelogram that has perpendicular diagonals is a rhombus. A rhombus has four sides that are congruent. The also diagonals bisect the vertex angles of this type of parallelogram.

Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.

A diagonal is normally defined as a straight line joining a vertex of a polygon with any vertex other than an adjoining vertex (lines joining a vertex to an adjoining vertex would simply be a side of the polygon). Since a triangle has only got adjoining vertices, it has no diagonals. Since there are no diagonals, they cannot bisect one another.

4

Construct and angle and bisect it, that should do it.

Since the diagonals of a rhombus are perpendicular between them, then in one forth part of the rhombus they form a right triangle where hypotenuse is the side of the rhombus, the base and the height are one half part of its diagonals. Let's take a look at this right triangle.The base and the height lengths could be congruent if and only if the angles opposite to them have a measure of 45â°, which is impossible to a rhombus because these angles have different measures as they are one half of the two adjacent angles of the rhombus (the diagonals of a rhombus bisect the vertex angles from where they are drawn), which also have different measures (their sum is 180â° ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).

A quadrilateral has 4 vertex, sides, angles.