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A triangle with sides 3a - 1, a2 + 1, and a2 + 2 has a perimeter of 16 units. What is the number of square units in the area of the triangle? I already know the value of is 2 or -3.5

 Nov 8, 2019
edited by mathleteig  Nov 8, 2019
edited by mathleteig  Nov 8, 2019
 #1
avatar+109202 
+2

We have that

 

( 3a - 1) + (a^2  + 1)  + (a^2 + 2)   = 16         simplify

 

2a^2  + 3a  + 2  = 16

 

2a^2  + 3a  -  14  = 0        factor

 

(2a + 7) ( a - 2)  =  0

 

Only the second  linear factor produces a positive for a....

a - 2  = 0

a  = 2

 

So....the sides of the triangle  are

3(2) - 1  =  5

2^2 + 1  =  5

2^2 + 2  =  6

 

The semi-perimeter, s,  =  [ 5 + 5 + 6 ] /2   =  8

 

We can find the area by the use of something known as Heron's Formula :

 

√ [ s ( s - 5) (s - 5) (s - 6)  ]  =

 

√ [ 8 * 3 * 3 * 2 ]  =

 

√ [16 * 9  ]  =

 

√144 =

 

12 units^2

 

 

cool cool cool

 Nov 8, 2019
edited by CPhill  Nov 8, 2019
 #2
avatar+297 
+1

I completely forgot about Heron's Formula! Thanks CPhill

 Nov 8, 2019
 #3
avatar+109202 
+2

THX, Mathleteig  !!!

 

We wouldn't necessarily have to use Heron's Formula......

 

Note that  this is an isosceles triangle  with a base of 6  and sides of  5

 

The altitude  = √ [side ^2  -  (base /2)^2  ]  = √ [ 5^2 - 3^2 ]

 

So....the area is......

 

(1/2) base * height

 

(1/2) 6 *  √[5^2  - 3^2 ]  =

 

3 * √16  =

 

3 * 4  =

 

12 units^2

 

cool cool cool

CPhill  Nov 8, 2019

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