Because CD is parallel to AB, ∠DCP=∠CAB because alternate interior angles are congruent. Similarly, ∠CDP=∠DBA. This means that triangle CDP and triangle BPA are similar by the AA postulate. Since we know the area and the base of triangle CDP, we can calculate the height by saying area=base*height/2. 5=2*height/2. Thus, the height is 5. These two triangles have a scale factor of 2/4. Therefore, we can find the height of the second triangle by proportions. 5/x=2/4. By cross multiplying, we get x=10. To find the height of the whole entire trapezoid, we add the height of the first triangle with the height of the second triangle. 5+10=15. Next, by using the area of trapezoids formula A=h(a+b)/2, we get 15(6)/2=45.
So you have 4 boxes.
In the first box, you have to choose which ball you want to put in. You have 4 choices.
In the second box, you have to choose which ball you want to put in. This time, you have 3 choices because you already put one in the first box.
In the third box, you choose a ball to put in. Now you only have 2 balls left.
In the final box, you only have one ball left.
Therefore the amount of ways you can distribute these 4 balls is 4x3x2x1=24 ways.
I hope this helps :)
|NORT is a parallelogram||Given|
|NORT is a rhombus||definition of rhombus|
|OT bisects NR||in a parallelogram, diagonals bisect|
|NH=HR||definition of segment bisect|
|OH=OH||reflexive property of equality|
|triangle NOH=triangleROH||SSS congruency theorem|
|OHN=90, OHR=90||division property of equality|
|OT perpendicular to NR||definition of perpendicular|
I changed your chart a bit to make it more specific. I myself am a geometry student as well so if anyone sees anything that's wrong, don't be afraid to tell me!
Let's say this table is Ila's working space. If half of the working space paper.
Next, a third of the paper is sticky notes...
So, we end up getting only 1/6 of the working space is covered in sticky notes. We use 1/2x1/3 to get this.
Because this triangle is a right triangle, by the Pythagorean theorem, we can figure out the hypotenuse by square rooting the sum of the square of the two legs, 9^2+12^2=sqrt225. The sqrt of 225 is 15 so x+y=15. Now we figure out the area of the big triangle. 9*12/2=54. Another way of calculating the area is by multiplying 15 by z and dividing by 2. Since we already figured out the area of the big triangle is 54, we can substitute 54 as the area and we get 54=z*15/2. Solving for z we get 7.2. x+y+z=15+7.2=22.2