The answer is: usually not.
It is helpful (not help full) because the two triangles formed by either diagonal are congruent.
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.
It's a type of drawing. I know that anime is animated and manga is the drawing/book version.
you have to go to a wizard shop and get some magic dust. Then you go to your drawing and sprinkle some on top of it and tada your drawing will come to life!
The answer is: usually not.
Yes It always does because of how a trapezoid is shaped.
A=l*w A=8*4 A=32 diagonal cuts the rectangle into two congruent triangles. 32/2 = 16
It is helpful (not help full) because the two triangles formed by either diagonal are congruent.
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.
To create two right triangles and an isosceles trapezoid by drawing two straight lines through a square, draw one line to be one of the diagonals of the square. Draw the other line parallel to the first. The three pieces shown are two right triangles and an isosceles trapezoid.
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.
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.
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,
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.
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
Three triangles