TRUE/FALSE Question dealing with Math and Shapes?

Support your answers for each question. Who ever gets the most correct answer will get the points.





TRUE/FALSE


19. It is possible for diagonals of a quadrilateral to bisect each other without being a parallelogram.


A. true


B. false


20. It is possible for a quadrilateral to have perpendicular diagonals without being a rhombus.


A. true


B. false


21. It is possible for a quadrilateral to have one pair of opposite angles congruent without being a parallelogram.


A. true


B. false


22. A rhombus with congruent diagonals is a square.


A. true


B. false


23. A median of a trapezoid is perpendicular to the bases.


A. true


B. false


24. If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles.


A. true


B. falseTRUE/FALSE Question dealing with Math and Shapes?
';19. It is possible for diagonals of a quadrilateral to bisect each other without being a parallelogram';





False. Consider that the diagonals of a quadrilateral divide it into four triangles. Let ACB and DCE be two of these triangles on opposite sides of the quadrilateral, where A, B, D, and E are vertices of the quadrilateral such that A is opposite D and B is opposite E, and C is the intersection of the diagonals. Note that by our assumption that the diagonals bisect each other, AC≡DC and BC≡EC. Note also that ∠ACB and ∠DCE are vertical angles, thus ∠ACB≡∠DCE. Therefore, by SAS, ACB≡DCE, and AB≡DE. This works regardless of which pair of sides we choose, so by the same logic, AE≡BD. Therefore, if the diagonals of a quadrilateral bisect each other, then both pairs of opposing sides have the same length, from which it follows immediately that your quadrilateral is a parallelogram.





';20. It is possible for a quadrilateral to have perpendicular diagonals without being a rhombus.';





True. For instance, consider the diamond formed by connecting the endpoints of the Christian cross. This quadrilateral has perpendicular diagonals, and is most definitely not a rhombus.





By the way, this example isn't meant to signify anything religous - it simply happens to be the easiest way to get across the shape I'm referring to without resorting to ASCII art (which I suck at).





';21. It is possible for a quadrilateral to have one pair of opposite angles congruent without being a parallelogram.';





True, and the counterexample is the same shape from the last question - the left and right angle are congruent, and the shape is not a parallelogram.





';22. A rhombus with congruent diagonals is a square.';





True. Consider that by the law of cosines, D²=2S² - 2S² cos θ, where D is the diagonal, S is the length of any side (in a rhombus they're all the same length), and θ is the angle opposite the diagonal. Note that if one diagonal is determined by one angle, then the other diagonal is given by either of the adjacent angles. In a parallelogram (which all rhombuses are), adjacent angles are supplementary, which means that for the two diagonals to be equal in length, cos θ must equal cos (180°-θ). This is only possible (for angles in the range 0%26gt;θ%26gt;180°) if θ=180°-θ, which means θ=90°, 180°-θ (i.e. the other pair of angles)=90°, and you've got yourself a square.





';23. A median of a trapezoid is perpendicular to the bases.';





False. A median of a trapezoid is parallel to the bases, by definition.





';24. If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles.';





True. Let b be the length of the base, h be the height, x₁ be the horizontal distance from the lower-left vertex to the point on the base directly beneath the upper-right, and x₂ be the horizontal distance from the lower-right vertex to the point on the base directly below the upper-left. It follows that the slope of the left side is (b-x₁)/h and the slope of the right side is -(b-x₂)/h, and since the angles are being measured in oppoite directions from the base, it follows that the two angles are equal if and only if the slope of one is equal to the negative slope of the other, which means that they are equal if x₁=x₂. Now, one diagonal is the hypotenuse of the right triangle fromed by the lower-left vertex, the upper-right vertex, and the endpoint of x₁, and the other by the lower-right, upper-left, and endpoint of x₂. It follows then that the diagonal legth is d₁²=h²+x₁², and d₂²=h²+x₂². Thus, x₁=√(d₁²-h²) and x₂=√(d₂²-h²). But if the diagonals are equal, d₁=d₂, thus x₁=x₂, the lower-left angle is the same as the lower-right angle, and the trapezoid is isoceles.TRUE/FALSE Question dealing with Math and Shapes?
19. true


20. true


21. true


22. true


23. false


24. true
Give ';pascal'; the points. He deserves it!
19.False


20.False


21. False


22. False


23. True


24. True