0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
What are the properties of the diagonals of a rectangle?
A Rectangle is a quadrilateral (four sided polygon) with two pairs of equal and parallel sides (opposite sides are parallel and equal, one pair is usually a different length from the other pair but if they are equal it is called a square), and all angles are right angles (90°). It has two diagonals which have the properties:The diagonals are always congruent (of equal length);The diagonals bisect each other (cut each other into two equal parts);The diagonals do not bisect the angles (unless the rectangle is a square when they do);The diagonals are not perpendicular (unless the rectangle is a square when they do).PROOF of the diagonals congruent:Take a rectangle ABCD with diagonals AC and BD.Using Pythagoras on the triangles ACD and BCD:AC² = AD² + CD²BD² = BC² + CD²But as ABCD is a rectangle AD = BC since they are opposite and parallel; thus:AC² = AD² + CD² = BC² + CD² = BD²Thus, as AC and BD are the diagonals, they are equal.Therefore the diagonals of a rectangle are congruent.
The length of the rectangle is 7cm more than the widthThe perimeter is 78cm Find the rectangle?
The rectangle is 23cm x 16cmHalf the perimeter is 78 / 2= 39cm (which is one side and one end).39 - 7 = 32cm32 / 2 = 16cm (one end)16 + 7 = 23cm (one side)Proof:(23 + 16) x 2 = 78cm the perimeter.
Proof kite diagonal bisector conjecture?
The diagonals of a kite are perpendicular and therefore bisect each other at 90 degrees
Are the diagonals of a square are never perpendicular?
The diagonals of a square are always perpendicular. Proof: Without loss of generality, assume the square has side length 1 and one vertex is at the origin. The square ABCD is given by: A = (0,0) , B = (1,0) , C = (1,1) , D = (0,1) The diagonals are d1=AC and d2=BD. Finding equations for each of them yields d1 = x d2 = 1-x (you can double check this) So, the relative slopes are 1 and -1. Since their product is -1, they are perpendicular.
If the Perimeter of the rectangle is 38 inches the length is 3 inches more find the width?
The formula for the perimeter of a rectangle is P=2(l + w). You know that the length is 3" more than the width. Hence you could substitute x for the width and x+3 for the length and use the formula to solve for the width: P = 2(l + w) 38 = 2(x + 3 + x) 38 = 4x + 6 32 = 4x 8 = x Hence, the width is 8" and the length is 11" Proof: P = 2(11 + 8) P = 2(19) P = 38
What are the properties of the diagonals of a rectangle?
A Rectangle is a quadrilateral (four sided polygon) with two pairs of equal and parallel sides (opposite sides are parallel and equal, one pair is usually a different length from the other pair but if they are equal it is called a square), and all angles are right angles (90°). It has two diagonals which have the properties:The diagonals are always congruent (of equal length);The diagonals bisect each other (cut each other into two equal parts);The diagonals do not bisect the angles (unless the rectangle is a square when they do);The diagonals are not perpendicular (unless the rectangle is a square when they do).PROOF of the diagonals congruent:Take a rectangle ABCD with diagonals AC and BD.Using Pythagoras on the triangles ACD and BCD:AC² = AD² + CD²BD² = BC² + CD²But as ABCD is a rectangle AD = BC since they are opposite and parallel; thus:AC² = AD² + CD² = BC² + CD² = BD²Thus, as AC and BD are the diagonals, they are equal.Therefore the diagonals of a rectangle are congruent.
Do the diagonals of a trapezoid bisect each other?
No, never. A trapezoid may have diagonals of equal length (isosceles trapezoid), but they do not intersect at their midpoints.Draw the diagonals of a trapezoid, for example, an isosceles trapezoid, thereby creating 4 triangles inside the trapezoid. Now assume the diagonals do bisect each other. The congruent corresponding sides of the top and bottom triangles with the included vertical angle would make the triangles congruent by the side-angle-side theorem. But this is a contradiction since the respective bases of the triangles, forming the top and bottom of the trapezoid are, of course, not equal. Therefore, the triangles cannot be congruent. Hence, we have given proof by contradiction that diagonals in a trapezoid cannot bisect each other.
How do you do a proof that a shape is a parrallelogram?
First of all you need to show that the figure is a quadrilateral. After that, there are many different way and these depend on the information that is given. You could show that :the two pairs of opposite sides are parallel,the two pairs of opposite sides are of equal length,the squares on the four sides equals the squares on the two diagonals,the two diagonals bisect one another,the two pairs of opposite angles are equal,any angle is supplementary to the two adjacent angles.
A rectangle has a length three times its width The perimeter is 72 inches sketch the rectangle the rectangle and then calculate its dimensions?
The rectangle is 9 inches wide X 27 inches long. Clue: width is X; length is 3 times width (3X). Perimeter is width plus length times two (X + 3X times 2 = 8X). 8X = 72; X = 9. 3X = 27. Proof: 9 + 9 (2sides) +27 + 27 (2 lengths) = 72.
The length of the rectangle is 7cm more than the widthThe perimeter is 78cm Find the rectangle?
The rectangle is 23cm x 16cmHalf the perimeter is 78 / 2= 39cm (which is one side and one end).39 - 7 = 32cm32 / 2 = 16cm (one end)16 + 7 = 23cm (one side)Proof:(23 + 16) x 2 = 78cm the perimeter.
How do you prove that the diagonals and either base of an isosceles trapezoid form an isosceles triangle?
Consider the isosceles trapezium ABCD (going clockwise from top left) with AB parallel to CD. And let the diagonals intersect at O Since it is isosceles, AD = BC and <ADC = <BCD (the angles at the base BC). Now consider triangles ADC and BCD. AD = BC The side BC is common and the included angles are equal. So the two triangles are congruent. and therefore <ACD = <BDC Then, in triangle ODC, <OCD (=<ACD = <BDC) = <ODC ie ODC is an isosceles triangle. The triangle formed at the other base can be proven similarly, or by the fact that, because AB CD and the diagonals act as transversals, you have equal alternate angles.
Proof kite diagonal bisector conjecture?
The diagonals of a kite are perpendicular and therefore bisect each other at 90 degrees
Are the diagonals of a square are never perpendicular?
The diagonals of a square are always perpendicular. Proof: Without loss of generality, assume the square has side length 1 and one vertex is at the origin. The square ABCD is given by: A = (0,0) , B = (1,0) , C = (1,1) , D = (0,1) The diagonals are d1=AC and d2=BD. Finding equations for each of them yields d1 = x d2 = 1-x (you can double check this) So, the relative slopes are 1 and -1. Since their product is -1, they are perpendicular.
What is the length of the rabbit proof fence in western Australia?
the rabbit proof fence is 4,000,020 miles in length. i would know i measured it myself.
Would mixing 80 proof and 30 proof make it 110 proof?
No. If you mixed them in equal proportions, you'd get 55 proof.
If the Perimeter of the rectangle is 38 inches the length is 3 inches more find the width?
The formula for the perimeter of a rectangle is P=2(l + w). You know that the length is 3" more than the width. Hence you could substitute x for the width and x+3 for the length and use the formula to solve for the width: P = 2(l + w) 38 = 2(x + 3 + x) 38 = 4x + 6 32 = 4x 8 = x Hence, the width is 8" and the length is 11" Proof: P = 2(11 + 8) P = 2(19) P = 38
How do you prove that the diagonals in a rhombus are perpendicular to each other?
I will outline a way to prove it for you. I will also five a simple vector proof for those that have studied vectors. For the first proof, one can often cite some of these as known facts or refer to theorems in a text. 1. First show that a rhombus is a parallelogram 2. Next, using the above, show that diagonals of the rhombus divide it into 4 congruent triangles. 3. Last, use CPCTC and not that all 4 middle angles are congruent so that are 90 degrees. From this is is easy to say that the diagonals are perpendicular. Hints. to prove 1, use the fact that all 4 sides of the rhombus are congruent and then use SSS to find two congruent triangles. Then use CPCTC to show that the angles are the same and find a transversal. Look at same side interior angles cut by that transversal and say something about them being parallel. 2. Use SSS again and find 4 congruent triangles and look at the diagonals. I will help more by giving you another proof using vectors that is really much more straightforward. A rhombus is a quadrilateral with all sides having equal length. This means that if two vectors, a and b that form the corner of a rhombus, then the magnitude of a and b are equal The diagonals of the parallelogram are precisely a+b and a-b. Now look at the dot product of a+b and a-b and see that it is zero and remember that a dot product of zero means the vectors are perpendicular or orthogonal The first part is a pure synthetic geometry approach and if anyone need more help to finish that, just ask, The second part is a vector proof which is elegant because it is so simple.
