Geometry Halp! -- 2 Questions

1) a.  In acute triangle ABC, we know AB = 7, BC = 8, and that Line{CA} is the shortest side. What is the smallest possible integer value of CA?

The angle opposite  the longest side must be < 90°

Therefore

√[ CA^2  + 7^2 ] > 8^2

CA^2  + 49  > 64

CA^2  > 15

CA > 3.8

So....the smallest possible integer value of CA  is  4

b.  In obtuse triangle ABC, we know AB = 7, BC = 8, and that Line{CA} is the longest side. What is the smallest possible integer value of CA?

We must  have  that

AB^2  + BC^2  <  CA^2

7^2  +  8^2  <  CA^2

49  +  64  <  CA^2

113 < CA^2        take the square root of both sides

10.63  < CA   ⇒   CA  > 10.3

So....the shortest  integer side length for CA  is   11

2)  a.Two diagonals of a parallelogram have lengths 6 and 8. What is the largest possible length of the shortest side of the parallelogram?

The diagonals will bisect each other.....so we have a triangle with sides of  3 and 4

The largest possible (integer) length of the shortest side, S, of the parallellogram  must be

3 + S  >  4

S  >  1

So...the largest possible (integer) length of the shortest side  is  2

b.  Two sides of an acute triangle are 8 and 15. How many possible lengths are there for the third side if it is an integer?

Let S be the missing side...so we have....

S + 8 >  15

S >  7  so the shortest integer length is 8

But also...since it's an acute triangle,  the longest possible side is

√ [ 8^2 + 15^2 ]  > S

17 >  S  ⇒   S <  17

So....the greatest possible integer  length of S  is  16

So...the number of possible lengths  is   16  - 8 +  1      =  9