Find the area of the triangle.

[Juri]

This question was in my worksheet and I was wondering if I solved it right.

Question: Find the area of ABC where the length of 3 sides is AB = √5, BC = 2, CA = 3.



I used Heron’s formula. In order to apply it, I need 1/2 of the perimeter of the triangle (the value of S).

Formula: S = (a + b + c) / 2
•S = (√5 + 2 + 3) / 2
•S = (√5 + 5) / 2
•S = 3.62 (decimal form)

Heron’s formula: A = √S(S - a)(S - b)(S - c)

A = √3.62(1.38)(1.62)(0.61)
A = √4.94
A = 2.22

Therefore, the area of the triangle for this problem is 2.22. Please correct me if I am wrong.

 

[Steven G]

No, the answer is not correct since (sqrt(5) + 5)/2 does not equal 3.62. Besides where does that side of sqrt(5) come from? Your diagram says that the sides are 2, 3 and 5

On 2nd thought, you can't have a triangle with sides 2, 3 and 5

 

[Juri]

Thanking you for pointing out the error, I fixed the diagram. Since √5+5 / 2 isn’t 3.62, I will assume it is √10 / 2.


Formula: S = (a + b + c) / 2
•S = (√5 + 2 + 3) / 2
•S = (√5 + 5) / 2
•S = √10 / 2
•S = 1.58 (decimal form)

Heron’s formula: A = √S(S - a)(S - b)(S - c)

A = √1.58(0.66)(-0.42))(-1.42)
A = √0.61
A = 0.78

 

[Steven G]

I have something for you to think about.

Isn't 0 + 5 = 5??
Isn't sqrt(5) >0?
Isn't 3=sqrt(9) < sqrt(10) < sqrt(16) = 4

How can it be that sqrt(5) + 5, which is more than 5, possible equal sqrt(10) which is between 3 and 4?


Using this logic we get results like 4+7 = sqrt(16) + 7 = sqrt(23) which is between 4 and 5. That is 4+7 is between 4 and 5 and not 11????

 

[Juri]

Your explanation made me understand that I should not have added √5 + 5 because it would result to √10. I remember now that numbers inside radicals should not be added with whole numbers in the first place.

I calculated √5 + 5 / 2 in the calculator and got 3.62 since I rounded it up.


Please explain to me why it is not equivalent to 3.62. I am guessing it is because the format is wrong.

 

[Dr.Peterson]

You've made some mistakes along the way due to notation (neglecting parentheses) and order of operations (the radical has to be done before the addition); but what you have in the OP (as well as your intermediate result in post #5) is correct apart from some rounding errors.

I did the calculation without rounding, by storing s in the calculator (3.6180339887498948482045868343656) and using that (recalling it from memory) rather than typing in a rounded value. The result is that the final radicand is exactly 5, not 4.94! Try it yourself.

(In principle you could do the exact calculation using radicals, and hopefully find that exact result, but I haven't tried it. It might be a good exercise for you.)

 

[Cubist]

There's a shortcut available if you spot that angle ABC is a right angle, using Pythagoras (√5)² + 2²= 3²

 

[Juri]

You've made some mistakes along the way due to notation (neglecting parentheses) and order of operations (the radical has to be done before the addition); but what you have in the OP (as well as your intermediate result in post #5) is correct apart from some rounding errors.

I did the calculation without rounding, by storing s in the calculator (3.6180339887498948482045868343656) and using that (recalling it from memory) rather than typing in a rounded value. The result is that the final radicand is exactly 5, not 4.94! Try it yourself.

(In principle you could do the exact calculation using radicals, and hopefully find that exact result, but I haven't tried it. It might be a good exercise for you.)

Thank you for the in-depth explanation. It really helped me a lot, I recalculated everything and just like how you stated, I got the same answer of 5 as well, which means that

A = √5

Now everything makes sense, thank you so much for your help!!

 

[Steven G]

(1/4)*sqrt {(2+3+sqrt(5)) * (2+3-sqrt(5)) * (2-3+sqrt(5)) * (-2+3+sqrt(5))} = (1/4)*sqrt{ (5+sqrt(5))*(5-sqrt(5))*(-1+sqrt(5))*(1+sqrt(5))} = (1/4) sqrt{(25-5)(-1+5)} = (1/4) sqrt(20*4) = (1/4)4sqrt(5) = sqrt(5)

No round off errors, no complicated multiplications requiring a calculator. The area equals sqrt(5)

 

[Steven G]

Your explanation made me understand that I should not have added √5 + 5 because it would result to √10. I remember now that numbers inside radicals should not be added with whole numbers in the first place.

I calculated √5 + 5 / 2 in the calculator and got 3.62 since I rounded it up.
View attachment 17955

Please explain to me why it is not equivalent to 3.62. I am guessing it is because the format is wrong.

(sqrt(5) + 5) is not 3.62 because as you wrote, it is approximately 3.61803

 

[Juri]

(1/4)*sqrt {(2+3+sqrt(5)) * (2+3-sqrt(5)) * (2-3+sqrt(5)) * (-2+3+sqrt(5))} = (1/4)*sqrt{ (5+sqrt(5))*(5-sqrt(5))*(-1+sqrt(5))*(1+sqrt(5))} = (1/4) sqrt{(25-5)(-1+5)} = (1/4) sqrt(20*4) = (1/4)4sqrt(5) = sqrt(5)

No round off errors, no complicated multiplications requiring a calculator. The area equals sqrt(5)

Whoops, there goes my careless mistake! Thank you.

 

[Dr.Peterson]

Thank you for the in-depth explanation. It really helped me a lot, I recalculated everything and just like how you stated, I got the same answer of 5 as well, which means that

A = √5

Now everything makes sense, thank you so much for your help!!

I don't see that you've commented specifically on Cubist's much quicker method in post #7.

Since it's a right triangle, you can take the base as √5 and height as 2, and the area is 2*√5/2 = √5.

What Jomo did (post #9) is what I said you could do but I hadn't bothered. It is much more impressive.