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What quadrilaterals have or could have diagonals that intersect at right angles?
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∙ 10y ago
Squares
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∙ 10y ago
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Prove that a rhombus has congruent diagonals?
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 45⁰, 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 180⁰ ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).
What are angles that are formed when two lines intersect?
When two unique lines intersect, two pairs of equal angles will be formed. All four angles could be 90 degrees (right angles) if the lines are perpendicular. If the lines are oblique, the pairs of angles can vary (almost) infinitely within a given range (1 degree-179 degrees).The intersection of two lines results in two pairs of equal angles such that the sum of angles equals 360. Another way to state this is that two adjacent angles will always sum to 180 degrees. That said, the best we can do is to express one angle in terms of the other. Call a pair of adjacent angles a and b. In the case of two adjacent angles, a = 180 - b, or b = 180 - a.
Why are the diagonals of a square perpendicular bisectors of each other?
You could prove this by congruent triangles, but here are two simpler arguments: --------------- Since a square is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, then the diagonals of a square must be perpendicular bisectors of each other -------------------- A square has four-fold rotational symmetry - as you rotate it around the point where the diagonals cross, there are four positions in which it looks the same. This means that the four angles at the centre must be equal. They will each measure 360/4 = 90 degrees, so the diagonals are perpendicular. Also. the four segments joining the centre to a vertex are all equal, so the diagonals bisect each other.
How many perpendicular lines does a rhombus have?
Normally, none.
Are the diagonals of a trapezoid equal?
Not for all trap. Only for an isosceles trapezoid. Well, for an Isosceles trapezoid, it could be any of the two diagonals! In order to have an isosceles trapezoid with the diagonals equal, the isosceles part MUST be precise and specific.
What quadrilaterals bisect each other?
Quadrilaterals do not bisect each other. They could in special cases. In parallelograms (types of quadrilaterals), the diagonals bisect each other.
What quadrilateral has right angles but the diagonals do not bisect each other?
the anser to this question is a trapiezuim as it could have right angles and its diagonals definatly do not bisect each other
What is the most right angles a quadrilateral can have?
Quadrilaterals have four sides, and therefore four angles, so the most right angles it could have is four, like a square.
Is there a quadrilaterals that has no right angles?
Yes. A quadrilateral is any four sided figure, so you could draw any number of these figures that have no right angles...a parallelogram for instance.
What quadrilateral has 45 and 135 angles?
Many different quadrilaterals can have those angles. Depending upon the lengths of the sides, where the angles are and how many pairs of parallel sides it has, it could be: A cyclic quadrilateral, a trapezium, a parallelogram or a rhombus.
Prove that a rhombus has congruent diagonals?
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 45⁰, 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 180⁰ ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).
If you had a project on how all quadrilaterals are alike how would you do it?
You could say they all consist of angles, never curved, and all have 4 sides...
What is a quauratualand 2 of its angles are obtuse and 2are acute and its oppoisite angliews are not equlal?
I am assuming the question refers to a quadrilateral (quaratual) with two obtuse angles and two acute angles, and its opposite (oppoisite) angles (angliews) are not equal (equlal)! It could be a trapezium or a quadrilateral with so specific name. Quadrilaterals with equal angles or equal sides, or parallel sides are given special names but there is no specific name for quadrilaterals which do not have any of these characteristics.
How do you find the sides of a rhombus if diagagonal are given?
The diagonals bisect each other at right angles. So you could use Pythagoras on half the diagonals. So, if the diagonals are a and b units long, then half the diagonals are a/2 and b/2 units long. Then, by Pythagoras, the sides of the rhombus are sqrt[(a/2)2 + (b/2)2]
Are rhombus ever rectangles?
Yes! Since rhombuses are quadrilaterals that have four congruent sides, a square can be considered as a rhombus. And since rectangles are quadrilaterals which has four 90 degree angles, a square is also a rectangle. Thus a rhombus could be a rectangle.
What formed by two real number lines that intersect at a right angle?
It could be the horizontal x and the vertical y axes that intersect each other at right angles on the Cartesian plane at the point of origin which is (0, 0)
Which quadrilaterals have sides that could be parallel?
All quadrilaterals. That is part of the definition of being a quadrilateral.
