Question

How many different obtuse triangles with integer side lengths and a perimeter of 20 can be drawn that no two of them are congruent?

Answer 1

Given :- How many different obtuse triangles with integer side lengths and a perimeter of 20 can be drawn that no two of them are congruent ?

Solution :-

→ Perimeter of ∆ = 20

→ Semi perimeter = 20/2 = 10m .

so, longest side must be less than 10 m .

then, Possible values of sides as a, b and c where c is the largest side will be :-

when c is 9 :-

• a = 9 , b = 2
• a = 8 , b = 3
• a = 7, b = 4
• a = 5 , b = 6
• given that, no two of them are congruent . so no need to take next values .

when c is 8 :-

• a = 8, b = 4
• a = 7 , b = 5
• a = 6, b = 6

when c is 7 :-

• a = 7 , b = 6 .

now we know that, for an obtuse angle ∆,

• c² > a² + b² .

checking now, we get,

• 9² > 9² + 2² ≠ Not possible .
• 9² > 8² + 3² => 81 > 64 + 9 = Possible .
• 9² > 7² + 4² => 81 > 49 + 16 = Possible .
• 9² > 5² + 6² => 81 > 25 + 36 = Possible .
• 8² > 8² + 4² ≠ Not possible .
• 8² > 7² + 5² => 64 > 49 + 25 ≠ not Possible .
• 8² > 6² + 6² => 64 > 36 + 36 ≠ not Possible .
• 7² > 7² + 6² => 49 > 49 + 36 ≠ not Possible .

therefore, total number of different obtuse triangles with integer side lengths and a perimeter of 20 can be drawn that no two of them are congruent will be only 3 and sides of these 3 ∆'s will be (3,8,9) , (4,7,9) and (5,6,9) .

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Answer 2

Given :  Perimeter of triangle 20 cm  

To Find :  How many different obtuse triangles with integer side lengths and a perimeter of 20 can be drawn that no two of them are congruent

Solution:

Sum of two side of a triangle is greater than third  side

Hence each side is less than semi perimeter

Let say sides a , b , c  

then a + b + c  =  20   and  a≠b≠c

a , b , c  < 20/2

=> a , b , c < 10

Let say

a =  9  then  b , c  

( 8 , 3)  , ( 7 , 4)  , ( 6 , 5)  

a =  8   then  b , c  

( 7 , 5)

So 4 possible triangles

( 9 , 8 , 3) , (9 , 7 , 4) , ( 9 , 6 , 5) , ( 8 , 7 , 5 )  

now for obtuse triangle

square of largest side >  sum of square of other two sides

Hence   a² > b² + c²  as we have considered a as largest side

9²  > 8² + 3²    

9²  > 7² + 4²  

9²  > 6² + 5²

but 8² < 7² + 5²   hence 8 , 7 , 5 is not possible

so possible triangles are  ( 9 , 8 , 3) , ( 9 , 7 , 4) , ( 9 , 6 , 5)

Hence  three triangles are possible.

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