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# if a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, then which statement would be true? the figure must be an isosceles trapezoid because it has 2 congruent base angles. the figure must be a rectangle because all rectangles have exactly 2 lines of symmetry. the figure could be a rhombus because the 2 lines of symmetry bisect the angles. the figure could be a square because the diagonals of a square bisect the right angles.

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### James

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get if a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, then which statement would be true? the figure must be an isosceles trapezoid because it has 2 congruent base angles. the figure must be a rectangle because all rectangles have exactly 2 lines of symmetry. the figure could be a rhombus because the 2 lines of symmetry bisect the angles. the figure could be a square because the diagonals of a square bisect the right angles. from EN Bilgi.

## [Expert Verified] If a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors,

If a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, then which statement would b… Get the answers you need, now! 06/19/2020 Mathematics Middle School

If a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, then which statement would be true?

The figure must be an isosceles trapezoid because it has 2 congruent base angles.

The figure must be a rectangle because all rectangles have exactly 2 lines of symmetry.

The figure could be a rhombus because the 2 lines of symmetry bisect the angles.

The figure could be a square because the diagonals of a square bisect the right angles.

Ace 13.3K answers 71.5M people helped

Option C is correct. , if a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, the figure could be a rhombus because the 2 lines of symmetry bisect the angles.

Quadrilaterals are figures with four sides and angles. Examples of quadrilateral include:

Rectangle Square rhombus

A line of symmetry is a line that divides a shape such that they become mirror images of each other.

Hence, if a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, from the options, the figure could be a rhombus because the 2 lines of symmetry bisect the angles (divides the angles equally).

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Helping Hand 2 answers 161 people helped

the figure could be a rhombus because the 2 lines of symmetry bisect the angles

Step-by-step explanation:

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4.9 (75 votes) summer school huh? Yup ^^^^ :(

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## If a quadrilateral has exactly 2 lines of symmetry ## If a quadrilateral has exactly 2 lines of symmetry, and both are angle bisectors, then which staternent would be true? The figure must be an isosceles trapezoid because it has 2 congruent base angles. The figure must be a rectangle because all rectangles have exactly 2 lines of symmetry. The figure could be a rhombus because the 2 lines of symmetry bisect the angles. The figure could be a square because the diagonals of a square bisect the right angles.

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If a quadrilateral has exactly lines of symmetry, and both are angle bisectors, then which staternent would be true?

The figure must be an isosceles trapezoid because it has congruent base angles.

The figure must be a rectangle because all rectangles have exactly lines of symmetry.

The figure could be a rhombus because the lines of symmetry bisect the angles.

The figure could be a square because the diagonals of a square bisect the right angles. Good Question (109) Answer 4.8 (997) votes Help me a lot (96) Correct answer (85) Write neatly (81)

Clear explanation (63)

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### Gauthmathier0218

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Check the full answer on App Gauthmath

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## Rhombuses_Kites_and_Trapezia

The Improving Mathematics Education in Schools (TIMES) Project

RHOMBUSES, KITES AND TRAPEZIA

Geometry and measurement : Module 21Years : 9-10

June 2011

Assumed Knowledge Motivation Content

Symmetries of triangles, parallelograms and rectangles

Rhombuses Squares Kites Trapezia Links Forward

History and Applications

The material in this module is a continuation of the module, Parallelograms and Rectangles, which is assumed knowledge for the present module. Thus the present module assumes:

Confidence in writing logical argument in geometry written as a sequence of steps, each justified by a reason.

Ruler-and-compasses constructions.

The four standard congruence tests and their application to:

proving properties of and tests for isosceles and equilateral triangles,

proving properties of and tests for parallelograms and rectangles.

Informal experience with rhombuses, kites, squares and trapezia.

Logical argument, precise definitions and clear proofs are essential if one is to understand mathematics. These analytic skills can be transferred to many areas in commerce, engineering, science and medicine but most of us first meet them in high school mathematics.

Apart from some number theory results such as the existence of an infinite number of primes and the Fundamental Theorem of Arithmetic, most of the theorems students meet are in geometry starting with Pythagoras’ theorem.

Many of the key methods of proof such as proof by contradiction and the difference between a theorem and its converse arise in elementary geometry.

As in the module, Parallelograms and Rectangles, this module first stresses precise definitions of each special quadrilateral, then develops some of its properties, and then reverses the process, examining whether these properties can be used as tests for that particular special quadrilateral. We have seen that a test for a special quadrilateral is usually the converse of a property. For example, a typical property−test pair from the previous module is the pair of converse statements:

If a quadrilateral is a parallelogram, then its diagonals bisect each other.

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Congruence is again the basis of most arguments concerning rhombuses, squares, kites and trapezia, because the diagonals dissect each figure into triangles.

A number of the theorems proved in this module rely on one or more of the previous theorems in the module. This means that the reader must understand a whole ‘sequence of theorems’ to achieve some results. This is typical of more advanced mathematics.

In addition, two other matters are covered in these notes.

The reflection and rotation symmetries of triangles and special quadrilaterals

are identified and related to congruence.

The tests for the kite also allow several important standard constructions to be explained very simply as constructions of a kite.

SYMMETRIES OF TRIANGLES, PARALLELOGRAMS AND RECTANGLES

We begin by relating the reflection and rotation symmetries of isosceles triangles, parallelograms and rectangles to the results that we proved in the previous module, Paralleograms and Rectangles.

The axis of symmetry of an isosceles triangle In the module, Congruence, congruence was used to prove that the base angles of an isosceles triangle are equal. To prove that B = C in the diagram opposite, we constructed the angle-bisector AM of the apex A, then used the SAS congruence test to prove that

ABM ACM

This congruence result, however, establishes much more than the equality of the base angles. It also establishes that the angle bisector AM is the perpendicular bisector of the base BC. Moreover, this fact means that AM is an axis of symmetry of the isosceles triangle.

These basic facts of about isosceles triangles will be used later in this module and in the module, Circle Geometry:

Theorem In an isosceles triangle, the following four lines coincide:

The angle bisector of the apex angle.

The line joining the apex and the midpoint of the base.

The line through the apex perpendicular to the base.

The perpendicular bisector of the base.

This line is an axis of symmetry of the isosceles triangle. It has, as a consequence, the interesting property that the centroid, the incentre, the circumcentre and the orthocentre of ABC all tie on the line AM. In general, they are four different points. See the module, Construction for details of this.

Extension − Some further tests for a triangle to be isosceles

The theorem above suggests three possible tests for a triangle to be isosceles. The first two are easy to prove, but the third is rather difficult because simple congruence cannot be used in this ‘non-included angle’ situation.

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EXERCISE 1

Use congruence to prove that ABC is isosceles with AB = AC if:

Source : amsi.org.au

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James 17 day ago

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