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1)We know that the side lengths in traingle ABC are AB = 8, BC = 9, and  CA = 12. Which of the following options is true?
(I). Triangle ABC is acute.
(II). Triangle ABC is obtuse.
(III). Triangle ABC is a right triangle.
(IV). We cannot determine which of the above option is true based on the given side lengths.

 

2)In isosceles triangle APS, we have  angle P is equal to 54 degrees and AP = AS. Which is larger, AP or PS?

 

3)In actute triangle ABC, we know  AB = 7, BC = 8 and that CA is the shortest side. What is the smallest possible integer value of CA?

 

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

 

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

 

6)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?

 

7)We can find an acute triangle with the three altitude lengths 1, 2, and h, if and only if h^2 belongs to interval (p,q). Find (p,q).

Guest Jan 28, 2018
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1)  If this were a right triangle...it would have to be that

 

8^2 + 9^2  = 12^2      but

 

64 + 81  >  144

 

145 > 144

 

Thus....angle ABC isn't quite large enough to be 90°

 

And the other two angles are definitely < 90° each....so this is an acute triangle

 

2) In isosceles triangle APS, we have  angle P is equal to 54 degrees and AP = AS. Which is larger, AP or PS?

                                                       A

 

 

                                           P                   S

 

Angle  APS  = Angle  ASP......So....angle  PAS =  [ 180  - 2*54 ] =    180 - 108  = 72°

And in a triangle.....the greater side is opposite the greater angle......and angle PAS > angle ASP.....so ...... PS > AP

 

 

3)In actute triangle ABC, we know  AB = 7, BC = 8 and that CA is the shortest side. What is the smallest possible integer value of CA?

 

Since this is an acute triangle, we must have that

 

7^2  + CA^2 >  8^2

49 + CA^2  > 64

CA^2 > 15

CA > √15  ⇒    CA > ≈ 3.8

So....the least integer value for CA  is  4

 

 

 

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

 

Since the triangle is obtuse and CA is the longlest side, we must have that

 

AB^2 + BC^2  <  CA^2

7^2  + 8^2  < CA^2

49 + 64 < CA^2

113 < CA^2

√113  < CA

CA > √113 ≈ 10.6

So ......   the smallest possible integer value for CA  is 11

 

 

5)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 4 triangles that have two sides of 3 and 4

And....by the triangular inequality, we have that

 

 

Longest side + Intermediate side > Shortest side

4 + 3  >  Shortest side

7 > Shortest side

Shortest side < 7

 

So......the largest possible value for the shortest side must be < 3

 

 

6)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?

 

Longest possible side length  

8 + 15  >  Third side length

23 > Third side length

Third side length < 23

 

Shortest possible side length

8 + Shortest side length > 15

Shortest side > 7

 

So.....the possible integer side lengths are

8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 19,, 20 21, 22

 

So.....15 integer lengths are possible

 

 

Don't know how to do 7....maybe someone else has a solution......!!!

 

 

cool cool cool                                 

CPhill  Jan 28, 2018

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