URGENT help needed before 4 PM

Question

Hi guys, anyone can help me with these questions asap? I have my exam at 5 today. Thanks

1) What is the greatest possible area of a triangle with one vertex at the center of a circle with r of 1 and the other two vertices on the circle?

a. Square root of 3/4
b. 1/2
c. Pi/4
d. 1
e. square root of 2

Answer is 1/2

Also see attached 2 more with correct answers squared

walker · Answer

S=2*1/2*r*cos(a/2)*r*sin(a/2)=r^2*sin(a/2)*cos(a/2)=1/2*r^2*sin(a)
a e{0..180} - center angle.

max(sin(2a)) when a=90,
so,
S=1/2*1^2*1=1/2

GOOD LUCK !!!!!!!!

walker · Answer

for 35/37

1..1000 0.5 ==> 5$
1000..3000 1.0 ==>2000 ==> 20$
3000..6000 1.5 ==>3000 ==> 45$
6000..10000 2.0 ==> 4000 ==> 80$

5+20+45+80 =150$

walker · Answer

for 17/37

y - Lee in May
x - others

pm=y/(x+y)=0.6 - May
pj=1.2y/(x+1/2y) - June

y/(x+y)=0.6 ==> 0.4y=0.6x ==> x=2/3y

pj=1.2y(2/3y+1.2y)=3.6/5.6~0.64 ==> 64%

GMATBLACKBELT · Answer

Walker has got a great explanation, but you shouldnt have to use sin and cos on the test.

Anyway u look at this, its going to be an iscosles triangle.

So we have 1, 1, sqrt2 as our sides.

Our height is going to equal (sqrt2/2)^2+B^2=1^2

2/4+B^2=1--> B^2=1/2 so B= sqrt2/2

BaseH/2=Area

Base=sqrt2 Height=sqrt2/2 --> (sqrt2*sqrt2/2)/2 --> 1/2 (b/c 2/2=1)

sry i think is too late for ur test. GL let us know how u did!

kevincan · Answer

Suppose the vertices have coordinates A(0,0) and B(1,0) and C(x,sqrt(1-x^2))
Area =sqrt(1-x^2)/2
Max area (when x=0)=1/2

tarek99 · Answer

gmatblackbelt, how did you get the 1, 1, sqrt2 as our sides? obviously the 1's are the 2 equal sides since they are also the radius of the cirle. but how did you get to the sqrt2?

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