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Suppose you drew a diagonal in each of the following quadrilaterals rectangles trapezoid parallelogram. In which figures do triangles with a different size and shape form?
When you draw a diagonal in a rectangle or a parallelogram, it divides the shape into two congruent triangles, meaning both triangles are the same size and shape. In contrast, drawing a diagonal in a trapezoid results in two triangles that can differ in size and shape, as the bases of the trapezoid are unequal. Thus, different size and shape triangles form only in the trapezoid.
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.
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 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.
Suppose you drew a diagonal in each of the following quadrilaterals rectangles trapezoid parallelogram. In which figures do triangles with a different size and shape form?
When you draw a diagonal in a rectangle or a parallelogram, it divides the shape into two congruent triangles, meaning both triangles are the same size and shape. In contrast, drawing a diagonal in a trapezoid results in two triangles that can differ in size and shape, as the bases of the trapezoid are unequal. Thus, different size and shape triangles form only in the trapezoid.
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!
Can you divide a trapezium into two congruent triangles?
No, it is not possible to divide a trapezium into two congruent triangles. A trapezium has only one pair of parallel sides, while a triangle has no parallel sides. Therefore, it is not geometrically feasible to divide a trapezium into two congruent triangles.
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.
A heptagon can be divided into how many triangles by drawing all of the diagonals from one vertex?
A heptagon has seven sides, so when drawing diagonals from one vertex, it will create five triangles. This is because each diagonal drawn from a single vertex will create a triangle until it intersects the previous diagonal. Therefore, the number of triangles formed by drawing all diagonals from one vertex in a heptagon is five.
