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.
Probability = 66/112 * 46/111 ... Please simplfy the fractions
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
Preparing for the BIG FIGHT
Just did it !! (Nike
#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?
Within my grasp!
- Rep Power
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
Yes you guys are right. Though I do not have the answers, the approach looks right. Thanks.
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