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Hi friends,

 

This one is for Melody and/or Alan and/or CPhill, since they helped me with this before and should recall the original question. 

 

hi guys!...remember the triangle question with the a triangle going into the centre of the previous one?. They stipulate that if the triangles keep on repeating indefnitely to insert into the previous one, calculate the sum of the triangles in terms of p.....sorry to bother you with this, I'm beginning to hate mathematics, it's just not the same anymore...sad

 Feb 17, 2022

Best Answer 

 #1
avatar+117834 
+1

You are refering to this question?

https://web2.0calc.com/questions/triangles_57

Sum of the areas?

 

The sides of the biggest equialateral triangle are    6p

so the height is    3sqrt3*p    unites squared    (30-60-90 triangle)

So the area of the biggest one is   \(0.5*6p*3\sqrt3 p = 9\sqrt3 p^2\)

After that each one is a quarter the size of the first.

 

so it is a GP, we have    \(a = 9\sqrt3 p^2\;\;and\;\;r=0.24\)

 

\(\text{Sum to infinity =}\frac{a}{1-r}\\ S_\infty = \frac{9\sqrt3 p^2}{0.75}=12\sqrt3p^2 \;\;units^2\)

 

 

You can't hate maths Jurriemagic,  that is not allowed   wink

 Feb 17, 2022
 #1
avatar+117834 
+1
Best Answer

You are refering to this question?

https://web2.0calc.com/questions/triangles_57

Sum of the areas?

 

The sides of the biggest equialateral triangle are    6p

so the height is    3sqrt3*p    unites squared    (30-60-90 triangle)

So the area of the biggest one is   \(0.5*6p*3\sqrt3 p = 9\sqrt3 p^2\)

After that each one is a quarter the size of the first.

 

so it is a GP, we have    \(a = 9\sqrt3 p^2\;\;and\;\;r=0.24\)

 

\(\text{Sum to infinity =}\frac{a}{1-r}\\ S_\infty = \frac{9\sqrt3 p^2}{0.75}=12\sqrt3p^2 \;\;units^2\)

 

 

You can't hate maths Jurriemagic,  that is not allowed   wink

Melody Feb 17, 2022

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