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02 Nov 2012, 21:09
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Some doubts and new properties (for me) I found regarding the properties of some geometric shapes:
1. The median of a trapezoid is the line joining the midpoint of the two legs (the non-parallel sides). Yet, it can be calculated as the average of the two bases (the parallel sides).
2. The diagonals of a square bisect the angles. Does this apply to rhombus and rectangle too (or, parallelograms in general)?
3. The diagonals of a square and a rhombus bisect each other at 90 degrees. Does this apply to a rectangle too (or, parallelograms in general)?
4. The diagonals of a square are equal. If the diagonals of a rhombus are equal, it must be a square. Does this apply to a rectangle too (or, parallelograms in general)?
5. Does the altitude of an isosceles triangle bisect the base (the non-equal side)?
6. Does the altitude of the base (non-equal side) of an isosceles triangle bisect the vertex?
7. Do (5) and (6) apply to an equilateral triangle?
I would be thankful if the doubts were cleared off (or is it cleared away?).
1. The median of a trapezoid is the line joining the midpoint of the two legs (the non-parallel sides). Yet, it can be calculated as the average of the two bases (the parallel sides).
2. The diagonals of a square bisect the angles. Does this apply to rhombus and rectangle too (or, parallelograms in general)?
3. The diagonals of a square and a rhombus bisect each other at 90 degrees. Does this apply to a rectangle too (or, parallelograms in general)?
4. The diagonals of a square are equal. If the diagonals of a rhombus are equal, it must be a square. Does this apply to a rectangle too (or, parallelograms in general)?
5. Does the altitude of an isosceles triangle bisect the base (the non-equal side)?
6. Does the altitude of the base (non-equal side) of an isosceles triangle bisect the vertex?
7. Do (5) and (6) apply to an equilateral triangle?
I would be thankful if the doubts were cleared off (or is it cleared away?).
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02 Nov 2012, 23:32
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Will have a go at these. But this is no expert opinion.
1. The median of a trapezoid is the line joining the midpoint of the two legs (the non-parallel sides). Yet, it can be calculated as the average of the two bases (the parallel sides). What the question is is not clear.
2. The diagonals of a square bisect the angles. Does this apply to rhombus and rectangle too (or, parallelograms in general)? In the case of a rhombus, the answer should be yes. In the case of a rectangle, I should think not. To bisect the 90 degree of the rectangle, the diagonal has to be at 45 degrees. For the diagonal to be at 45 degress the base and height of the rectangle should be equal. In which case the rectangle would become a square.
3. The diagonals of a square and a rhombus bisect each other at 90 degrees. Does this apply to a rectangle too (or, parallelograms in general)? No. For the diagonals to bisect at right angles, the sum of oppostie sides should be equal. So for a rectangle or parallelogram, the diagonals do not intersect at right angles.( This is just my theory) So the only quadrilaterals which satisfy the rule would be rhombuses, kites, squares and some trapezoids.
4. The diagonals of a square are equal. If the diagonals of a rhombus are equal, it must be a square. Does this apply to a rectangle too (or, parallelograms in general)? Yes. If diagonals of a parallelogram are equal then it is a rectangle.
5. Does the altitude of an isosceles triangle bisect the base (the non-equal side)? Yes. Since both sides are equal, they meet at a point directly above the midpoint of the base.
6. Does the altitude of the base (non-equal side) of an isosceles triangle bisect the vertex? Yes. This is just an extension of of 5.
7. Do (5) and (6) apply to an equilateral triangle? Yes. Every equilateral triangle is also an iscoceles triangle.
Kudos Please... If my post helped.
1. The median of a trapezoid is the line joining the midpoint of the two legs (the non-parallel sides). Yet, it can be calculated as the average of the two bases (the parallel sides). What the question is is not clear.
2. The diagonals of a square bisect the angles. Does this apply to rhombus and rectangle too (or, parallelograms in general)? In the case of a rhombus, the answer should be yes. In the case of a rectangle, I should think not. To bisect the 90 degree of the rectangle, the diagonal has to be at 45 degrees. For the diagonal to be at 45 degress the base and height of the rectangle should be equal. In which case the rectangle would become a square.
3. The diagonals of a square and a rhombus bisect each other at 90 degrees. Does this apply to a rectangle too (or, parallelograms in general)? No. For the diagonals to bisect at right angles, the sum of oppostie sides should be equal. So for a rectangle or parallelogram, the diagonals do not intersect at right angles.( This is just my theory) So the only quadrilaterals which satisfy the rule would be rhombuses, kites, squares and some trapezoids.
4. The diagonals of a square are equal. If the diagonals of a rhombus are equal, it must be a square. Does this apply to a rectangle too (or, parallelograms in general)? Yes. If diagonals of a parallelogram are equal then it is a rectangle.
5. Does the altitude of an isosceles triangle bisect the base (the non-equal side)? Yes. Since both sides are equal, they meet at a point directly above the midpoint of the base.
6. Does the altitude of the base (non-equal side) of an isosceles triangle bisect the vertex? Yes. This is just an extension of of 5.
7. Do (5) and (6) apply to an equilateral triangle? Yes. Every equilateral triangle is also an iscoceles triangle.
Kudos Please... If my post helped.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
