Powerprep problems

**hotsummertime** · 10-01-2006, 10:48 PM

1. in a class there are 46 girls and 66 boys. Two students are picked at random. What is the probability that first a boy is chosen and second a girl is chosen?

2. The area of parallelogram ABCD is 80. If the edge equal to 1/4th of the sides of the paralleogram is cut off, what is the area of the cutout piece? Please refer to the attached diagram.

**Mystery Man** · 10-02-2006, 04:24 AM

Ans 1

Probability = 66/112 * 46/111 ... Please simplfy the fractions

Ans 2

The base is now 3/4. Now draw the heights form the top right corner of the diagram for both the big and the small parallelograms. 2 triangles will be formed and you will find they are similar.

So by similarity if the hypotneus are reduced to 3/4 the heights are also reduced by 3/4. Now

Ht and base of new parallelogram = 3/4b * 3/4 h = 9/16 b*h = 9/16*80 = 45

Area of cutout = 80-45 = 35

**daageep** · 10-02-2006, 04:33 AM

#1 i agree with mystery man.

#2. the solution is wonderful, mystery man. i didn't even think of using similar triangles.

any idea if similar triangles is a frequent topic?

**een** · 10-02-2006, 04:41 AM

Hi, hotsummertime!!
In my humble opinion -for ques.2) The answer is 35.
S_0(the area of the cutout piece)=80-S_1
If we consider parallelogram ABCD (AB=a, BC=b), S of parallelogram ABCD=80, on the other hand S_ABCD=AD*h=b*h (here h is height of parallelogram ABCD, h=CF, F belongs to AD )
S_1 is the area of small parallelogram MNOL(there MN=OL=(3/4)*a, ML=NO=(3/4)*b))
S_1=ML*x=(3/4)*x, x-is the height of parallelogram MNOL, x=Official Guide, G belongs to MD.
Consider triangle CDF and triangle ODG, these triangles are similar,
so x/h=((3/4)*a)/a and x=(3/4)*h (and h=80/b)
at last S_1=ML*x=(3/4)*b*x=(3/4)*b*(3/4)*(80/b)=45
S_0(the area of the cutout piece)=80-45=35
Best wishes!
Katya