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