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What else can I help you with?
Is a parallelogram inscribed in a circle always a rectangle?
Yes. The corners must be right angles for it to be inscribed on the circle.
True or false if a parallelogram inscribed in a circle it must be a rectangle?
True.
Which of the following must be an equiangular polygon?
hexagon
How do you work out angles in a semi-circle?
An inscribed angle is an angle formed by two chords in a circle which have a common endpoint. This common endpoint forms the vertex of the inscribed angle.The other two endpoints define an intercepted arc on the circle Any angle inscribed in a semi-circle is a right angle. The proof is simply that the intercepted arc is 180 so the angle must be half of that or 90 degrees.
Assume that two chords in a given circle are the same distance from the center of the circle Which of the following must also be true?
They must be congruent.
If a parallelogram is inscribed in a circle then it must be a?
If a parallelogram is inscribed in a circle then it must be a cyclic quadrilateral.
Is a parallelogram inscribed in a circle always a rectangle?
Yes. The corners must be right angles for it to be inscribed on the circle.
True or false if a parallelogram inscribed in a circle it must be a rectangle?
True.
Which of the following must be an equiangular polygon?
hexagon
Is it true an acute angle inscribed in a circle must intercept a minor arc?
No, because there is no acute angle in a circle.
Does a triangle need to be completely in the circle to be inscribed?
yes ...all the angles of the triangle must touch a spot on the circle..
How many different inscribed circles can be inscribed in a given triangle?
There is only one possible circle that can be inscribed in any triangle because all of the sides of the triangle must touch the circle at some point. Also, there is only one "incenter" of each circle. The incenter is the center of an inscribed circle.
How do you work out angles in a semi-circle?
An inscribed angle is an angle formed by two chords in a circle which have a common endpoint. This common endpoint forms the vertex of the inscribed angle.The other two endpoints define an intercepted arc on the circle Any angle inscribed in a semi-circle is a right angle. The proof is simply that the intercepted arc is 180 so the angle must be half of that or 90 degrees.
Assume that two chords in a given circle are the same distance from the center of the circle Which of the following must also be true?
They must be congruent.
Does hexagon have obtuse angle?
Yes. A hexagon must have an obtuse angle.
How do you create a hexagon that when you move the points the angles don't change on GSP?
I'm not familiar with GSP, but if you move any vertex of a hexagon then it is no longer a hexagon. In order to maintain the shape (that is, the aspect ratio) you must move all vertices at the same time, relative to the centre of the hexagon. The easiest way to calculate where each vertex should be placed is to imagine the hexagon is circumscribed with a circle such that all the vertices touch the circle. Thus the centre of the circle is also the centre of the hexagon. So when you move any vertex, you are essentially increasing or decreasing the radius of the circle and/or rotating the circle around its centre. However, the centre never moves so you only need to know the horizontal and vertical distance of the vertex you moved relative to the centre of the circle, and Pythagoras' Theorem tells you the radius of the circle (X2 + Y2 = Z2), and thus you can determine the angle of the hypotenuse relative to the horizontal or vertical plane. Knowing the position of one vertex means you can work out all 6 vertices (hint: they are 60 degrees apart).
What is the sq footage of 12' hexagon?
What is the 12 foot? It could be the radius or diameter of the inscribed or circumscribed circles. Or it could be the length of each side. There are thus three solutions depending on above parameters. The hexagon must be a regular hexagon. Let R = radius of circumscribed circle in feet Let r = radius of inscribed circle in feet Let s = length of one of the sides in feet Then area in square feet = 2.598*s2 or 2.598*R2 or 3.464*r2 (R = s = 1.155*r where 1.155 = 1 / sin 60 degr.) The shortened formula 2.598*s2 is obtained as follows: There are six equilateral triangles in a regular hexagon when three lines are drawn connecting opposite corners. Therefore the area is the sum of the six triangles. Each triangle area is 1/2 base x height. Therefore each A = 0.5s x (sqrt of 3 x 0.5s) = 0.25s2 x sqrt of 3 = 0.433s2 So 6 triangles will be 6 x 0.433s2 = 2.598s2
