We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
36
3
avatar+198 

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+104933 
+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+198 
+1

I completely forgot about Heron's Formula! Thanks CPhill

 Nov 8, 2019
 #3
avatar+104933 
+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

37 Online Users

avatar
avatar