1) Area of trapezoid =1/2(base 1 +base 2)* height

$A= \frac{23 \frac{1}{3}+4 \frac{2}{3} }{2} *(12 \frac{3}{4} ) = \frac{27 \frac{3}{3} }{2} *( \frac{51}{4} ) = \frac{28}{2} * \frac{51}{4} = \frac{7*51}{2} = \frac{357}{2} =178 \frac{1}{2} x^{2}$

5 0

3 years ago

If the diagonals of a quadrilateral bisect the angles, is the quadrilateral always a parallelogram? Explain your answer.

Liula [17]

Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.

Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).

Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.

6 0

2 years ago

NOW ANSWER PLS<br> SLOPE

algol13

Answer: For first page only.

1. 3

2. 1/2

3. -5

Step-by-step explanation:

8 0

2 years ago