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# TRIANGLES

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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
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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   Nov 8, 2019
edited by CPhill  Nov 8, 2019
#2
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I completely forgot about Heron's Formula! Thanks CPhill

Nov 8, 2019
#3
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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   CPhill  Nov 8, 2019