I don't know how to answer number 9 and 10. Can someone please help me?

AB = AC = 32

SR = TR = 7

Solution:

Question 9:

Tangent theorem of a circle:

Lengths of tangents drawn from an same external point to a circle are equal.

In figure AB and AC are tangents drawn from same an external point A.

By tangent theorem of a circle,

⇒ AB = AC

Divide both side of the equation by x.

Divide by 2 on both side of the equation.

Substitute x = 4 in AB and AC.

Question 10:

By tangent theorem of a circle,

⇒ SR = TR

Multiply by 7 on both side of the equation, we get

Divide by y on both side of the equation, we get

Substitute y = 7 in SR and TR

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!