yes it is
I think this is impossible the diagonals of a four sided figure will cross
Yes, they can be. Here is an example to see how this is true. Construct two perpendicular lines AB and CD that intersect at a point O. Let AO = CO, BO = DO and AO ≠ BO, then ABDC forms an isosceles trapezoid. If the lines are not perpendicular, then also ABDC is an isosceles trapezoid and it has perpendicular diagonals.
An example of a perpendicular line is a 4 way intersection at a stop sign. Another example would be a window that has 4 smaller squares in it.
__ /__\<- Here is an example of a quadrilateral. All it needs is four connecting sides and four angles. Hence, quad= four.
A quadrilateral is a parallelogram with four sides. For example: square, rectangle, parallelogram, trapezoid, rhombus, diamond...
That is true for some parallelograms but not all. For example, the diagonals of a rhombus, kite or square are perpendicular, but those of a rectangle or general parallelogram are not.
trapezoid
I think this is impossible the diagonals of a four sided figure will cross
Yes as for example the diagonals of a square
A rectangle is an example of a quadrilateral where the diagonals are congruent and bisect each other. However, a kite is a quadrilateral that can also have congruent diagonals, but they do not bisect each other. In a kite, one diagonal bisects the other at a right angle, while the other diagonal remains unequal in length. Therefore, while both shapes can have congruent diagonals, only the rectangle has diagonals that bisect each other.
By that description, all the sides are parallel and have 90 degree angles... But the sides aren't necessarily the same length. So a square is one example, but you could also have a long rectangle. apex=False
Yes, they can be. Here is an example to see how this is true. Construct two perpendicular lines AB and CD that intersect at a point O. Let AO = CO, BO = DO and AO ≠ BO, then ABDC forms an isosceles trapezoid. If the lines are not perpendicular, then also ABDC is an isosceles trapezoid and it has perpendicular diagonals.
A square is on example. The perpendicular bisectors of the sides and the two diagonals comprie four lines of symmetry.
The diagonals in a quadrilateral are the (usually imaginary) straight lines that go through the inside of the shape from one vertex to another. For example, say you have a square (a quadrilateral) consisting of points A, B, C, and D. They are all connected on edges going A to B, B to C, C to D, and D to A. Envisioning this shape in your mind, you would likely imagine the inside of the square being blank. There would be two diagonals in a shape like this (as well as in every quadrilateral) being the lines going from A, through the center of the square, to C, and the other being the line going from B, through the center, to D.
Justification is a written reason for why you made a statement. For example if you say a figure is a Square then you would say something like the diagonals are both congruent and perpendicular to each other.
|_ example of a perpendicular line is shown above:
Converse: If the diagonals of a quadrilateral are congruent and bisect each other, then the quadrilateral is a rectangle. Given: Quadrilateral ABCD with diagonals , . and _ bisect each other Show: ABCD is a rectangle Because the diagonals are congruent and bisect each other, . Using the Vertical Angles Theorem, AEB CED and BEC DEA. So ∆AEB ∆CED and ∆AED ∆CEB by SAS. Using the Isosceles Triangle Theorem and CPCTC, 1 2 5 6, and 3 4 7 8. By the Angle Addition Postulate each angle of the quadrilateral is the sum of two angles, one from each set. For example, mDAB = m1 + m8. By the addition property of equality, m1 m8 m2 m3 m5 m4 m6 m7. So by substitution mDAB mABC mBCD mCDA. Therefore the quadrilateral is equiangular. Using 1 5 and the Converse of AIA, . Using 3 7 and the Converse of AIA, . Therefore ABCD is an equiangular parallelogram, so it is a rectangle by definition of rectangle.