# 1) One leg of a right triangle has length 20 and the radius of the inscribed circle is 6. Compute the length of the hypotenuse of the right triangle. Answer: 29 2) Let a be a positive number...

1) One leg of a right triangle has length 20 and the radius of the inscribed circle is 6. Compute the length of the hypotenuse of the right triangle.

Answer: 29

2) Let a be a positive number such that a^2009-2a + 1 = 0: Find all possible values of the sum S = 1 + a + a^2 +… + a^2008:

Answer: 2 & 2009 (I can find 2009 but I can’t find out how to get 2)

3) Suppose (a + bi)^2 = 4(b + ai); where a > 0; b > 0; and i =rad -1; If a + bi =r(cosx + isin x), what is the ordered pair (r,x)? ( x should be in radians)

Answer:( 4, pi/6)

Only one question can be answered either 1 or 2 or 3.

### 1 Answer | Add Yours

1.the length of tangents from exterior point to circle are equal in length.

So leg of right angle triangle legth 20 can be written as 6+14 ( radius of inscibed cirle is 6). another leg can be written as x+6 and hypotenuse can be wriiten as 14+6 ,where x is length to be determined. By Pythagoras theorem

(14+x)^2=20^2+(x+6)^2

196+x^2+28x=400+x^2+12x+36

28x-12x=436-196

16x=240

x=15

length of hypotenuse= 14+15=29

2.It is GP ,first term is 1,common ration a and number of terms 2009

SUM= 1.(a^2009-1)/(a-1) ,provided a >1 (i)

given a^2009-2a+1=0

a^2009=2a-1 (ii)

substituting a^2009 from (ii) in (i)

SUM= (2a-1-1)/(a-1)=2

3. (a+ib)^2=4(b+ai)

a^2-b^2+2iab=4b+4ai

compare real and imaginary parts we have

a^2-b^2=4b (i)

2ab=4a (ii)

from (i) and (ii) we get

b=2 and a=sqrt(12)

now sqrt(12)+2i= rcos(x)+irsin(x)

comparing real and imaginary parts we have

r cos(x)= sqrt(12) (iii)

r sin(x)= 2 (iv)

sqraring adn adding (iii) and (iv) ,we have

r^2cos^2(x)+r^2sin^2(x)=12+4 (cos^(x)+sin^2(x)=1)

r^2=16

r=4

divide (iv) by (iii),we have

tan(x)=2/sqrt(12)=sqrt(1/3)=tan(pi/6)

x=pi/6

(r,x)=(4,pi/6)