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Why drawing a diagonal on any parallelogram will always result in two congruent triangles?
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Does a parellelogram have four congruent triangles?
Yes. Read on for why: Take a parallelogram ABCD with midpoints E and F in the bases. So something like this (forgive the "drawing"): A E B __.__ /__.__/ C F D We know that parallelogram AEFC = EBDF, since they have the same base (F bisects CD, so CF = FD), height (haven't touched that), and angles (<ACF = <EFD because they're parallel - trust me that everything else matches). We also know that every parallelogram can be divided into two congruent triangles along their diagonal. So if two congruent parallelograms consistent of two congruent triangles each, then all four triangles are congruent. So your congruent triangles are ACF, AEF, EFD, and EBD. You can further reinforce this through ASA triangle congruency proofs (as I did at first), but this is a far more concise and equally valid answer.
Proving that a parallelogram has equal pair of sides?
First draw a parallelogram. I cannot draw one here so I will have to describe the picture and you should draw it. Let ABCD be a parallelogram. I put A on the bottom left, then B on the bottom right, C on the top right and D on the top left. Of course the arguments must apply to an arbitrary parallelogram, but just so you can follow the proof, that is my drawing. Now draw a segment from A to C. It is a diagonal. AB is parallel to CD and AD is parallel to BD because a parallelogram is a quadrilateral with both pair of opposite sides parallel. Now ABC and CDA both form triangles. Let angles 1 and 4 be the angles created by the diagonal and angle BCD of the parallelogram. Angle 1 is above and angle 4 is below. Similarly, let angles 3 and 2 be created by the intersection of the diagonal and angle DAB or the original parallelogram. Now angles 1 and 2 are congruent as are 3 and 4 because if two parallel lines are cut by a transversal, the alternate interior angles are congruent. Next using the reflexive property AC is congruent to itself. Now triangle ABC is congruent to triangle CDA by Angle Side Angle (SAS). This means that AB is congruent to CD and BC is congruent to AD by corresponding parts of congruent triangles are congruent (CPCTC). So we are done!
The sum of the interior angle measures of an octagon?
To find interior angle measurements, you must divide the shape into triangles by drawing diagonal lines. The diagonal lines draw triangles, and the interior angle measure of triangles are always 180 degrees. The sum of interior angles of an octagon is 1080 degrees. How ever many triangles you have, multiply it by 180. See octagons in the link for more help,
Why is the sum of the interior angles of a quadrilateral 360 degrees?
This result follows from the theorem that the sum of the angles of a triangle is 180 degrees. Drawing a diagonal in the quadrilateral splits it into two triangles and the angles of the triangles together combine to form the angles of the quadrilateral.
A heptagon can be divided into how many different triangles by drawing all of the diagonals from one vertex?
5 triangles.
Why is drawing the diagonals of a parallelogram help full in proving many of the parallelograms properties?
It is helpful (not help full) because the two triangles formed by either diagonal are congruent.
Does Drawing one diagonal trapezoid creates two congruent triangles?
No, in general, it does not.
Drawing one diagonal trapezoid creates two congruent triangles?
The answer is: usually not.
Why drawing a diagonal on any parsllelogram will always result in two congruent triangles?
Well a parallelogram is a 4 sided shape with 2 pairs of parallel lines, hence PARALLELogram. That's the reason, because there are two pairs of parallel lines.
Does drawing one diagonal in a trapezoid create two congruent triangles?
Yes It always does because of how a trapezoid is shaped.
Does a parellelogram have four congruent triangles?
Yes. Read on for why: Take a parallelogram ABCD with midpoints E and F in the bases. So something like this (forgive the "drawing"): A E B __.__ /__.__/ C F D We know that parallelogram AEFC = EBDF, since they have the same base (F bisects CD, so CF = FD), height (haven't touched that), and angles (<ACF = <EFD because they're parallel - trust me that everything else matches). We also know that every parallelogram can be divided into two congruent triangles along their diagonal. So if two congruent parallelograms consistent of two congruent triangles each, then all four triangles are congruent. So your congruent triangles are ACF, AEF, EFD, and EBD. You can further reinforce this through ASA triangle congruency proofs (as I did at first), but this is a far more concise and equally valid answer.
The length and width of a rectangle are 8 inches and 4 inches respectively What would be the area of one of the triangles formed from drawing a diagonal in the rectangle?
A=l*w A=8*4 A=32 diagonal cuts the rectangle into two congruent triangles. 32/2 = 16
Proving that a parallelogram has equal pair of sides?
First draw a parallelogram. I cannot draw one here so I will have to describe the picture and you should draw it. Let ABCD be a parallelogram. I put A on the bottom left, then B on the bottom right, C on the top right and D on the top left. Of course the arguments must apply to an arbitrary parallelogram, but just so you can follow the proof, that is my drawing. Now draw a segment from A to C. It is a diagonal. AB is parallel to CD and AD is parallel to BD because a parallelogram is a quadrilateral with both pair of opposite sides parallel. Now ABC and CDA both form triangles. Let angles 1 and 4 be the angles created by the diagonal and angle BCD of the parallelogram. Angle 1 is above and angle 4 is below. Similarly, let angles 3 and 2 be created by the intersection of the diagonal and angle DAB or the original parallelogram. Now angles 1 and 2 are congruent as are 3 and 4 because if two parallel lines are cut by a transversal, the alternate interior angles are congruent. Next using the reflexive property AC is congruent to itself. Now triangle ABC is congruent to triangle CDA by Angle Side Angle (SAS). This means that AB is congruent to CD and BC is congruent to AD by corresponding parts of congruent triangles are congruent (CPCTC). So we are done!
The sum of interior angles of a square is?
360 degrees. Same for any other rectangle, rhombus, parallelogram, or convex quadrilateral. The reason is that any quadrilateral can be divided into two triangles by drawing a diagonal and the sum of the angles of each triangle is 180 degrees.
The sum of the interior angle measures of an octagon?
To find interior angle measurements, you must divide the shape into triangles by drawing diagonal lines. The diagonal lines draw triangles, and the interior angle measure of triangles are always 180 degrees. The sum of interior angles of an octagon is 1080 degrees. How ever many triangles you have, multiply it by 180. See octagons in the link for more help,
Why is the sum of the interior angles of a quadrilateral 360 degrees?
This result follows from the theorem that the sum of the angles of a triangle is 180 degrees. Drawing a diagonal in the quadrilateral splits it into two triangles and the angles of the triangles together combine to form the angles of the quadrilateral.
How do you prove the diagonals of a rhombus divide the rhombus into four congruent triangles?
Take a rhombus ABCD. A rhombus as 4 equal sides, thus AB = BC = CD = DA Draw in 1 diagonal AC. This splits the rhombus into 2 triangles. ABC and CDA with side AB = CD, BC = DA and AC common to both triangles. Thus ABC and CDA are congruent by Side-Side-Side. Triangles ABC and CDA are isosceles triangles since they have two equal sides (AB = BC and CD = DA) thus angles DAC = DCA = BAC = BCA. Specifically DAC = BAC. But DAC + BAC = DAB, thus DAC = BAC = ½ DAB; similarly DCA = BCA = ½ BCD = ½ DAB Drawing in the other diagonal BD, the same arguments show triangles ABD and CDB are congruent and angles ADB = CDB = ABD = CBD = ½ ABC Let the point where the diagonals meet be E. We now have 4 triangles ABE, BCE, CDE and DAE with equivalent angles and sides: Angles DAE* = BAE = BCE = DCE (= ½ DAB) Angles ABE = CBE = CDE = ADE (= ½ ABD) Sides AB = BC = CD = DA Thus the 4 triangles are congruent by Angle-Angle-Side. *Angle DAE = DAC since E lines along AC; similarly for all the other angles involving point E, ie angle BCE = BCA, ADB = ADE, etc
