# Length of a third side of triangle

#### Joystar1977

##### Active member
Math Problem: Find the length of the third side of a triangle if the area of the triangle is 18 and two of its sides have lengths of 5 and 10.

Which one of these are correct when I am working them out? If none of these are correct, then can somebody please help me solve this math problem step-by-step?

First way I worked out the problem:

A=18=0.5*5*10*sin(x)

x = 46 degrees

====

c^2 = 10^2+5^2-2*50*cos(46) = 55,53

third side:

c = 7,45

Second way I worked out the problem:

I know A = 18 units², a = 5 units, and b = 10 units

given A = (absin(C))/2

=> 18 = 25sin(C)

=> 18/25 = sin(C)

=> C = sin⁻¹(18/25)

given c² = a² + b² -2abcos(C)

=> c² = 25 + 100 -100cos(sin⁻¹(18/25))

=> c = √(125 -100cos(sin⁻¹(18/25)))

=> c = √(125 - 4√301) units

=> c ≈ 7.4567 ( 4 dp)

Third way I worked out the problem:

Use Heron's formula for triangle:

Suppose the third side is x, and the others are 5 and 10, so by Heron formula, we get:

Area = sqrt(s(s-a)(s-b)(s-c))

where s=semi perimeter, a,b,c are sides of triangle, a=x, b=5, c=10

so, s=1/2.(x+5+10)

=1/2(x+15)

s-a= 1/2x +15/2 -x= 15/2-x/2

s-b=x/2+15/2-10/2=x/2+5/2

s-c=x/2+15/2-20/2=x/2-5/2

Area= sqrt(x/2+15/2)(15/2-x/2)(x/2+5/2)(x/2-5/…

324=(15/2+x/2)(15/2-x/2)(x/2+5/2)(x/2-…

324=(225/4 -x^2/4)(x^2/4-25/4)..multiply by 4 to get

1296=(225-x^2)(x^2-25)

225x^2-225*25-x^4+25x^2-1296=0

-x^4+250x^2-6921=0

-(x^4-250x^2+6921)=0

-((x^2-125)-8704)=0

(x^2-16sqrt34-125)(x^2+16sqrt34-125)=0

x^2=16sqrt34+125

x=sqrt(16sqrt 34+125)

=14.775

or

x^2=125-16sqrt34

x=sqrt(125-16sqrt34)

=5.63

So, the length of the third side is 14.775 or 5.63

I am really lost and confused on this problem.

#### MarkFL

Staff member
I would look at the two case: In both cases, we have:

$$\displaystyle A=\frac{1}{2}bh$$

$$\displaystyle 18=25\sin(\theta)$$

Case 1:

$$\displaystyle \theta=\pi-\sin^{-1}\left(\frac{18}{25} \right)$$

Using the law of cosines, we may write:

$$\displaystyle x=\sqrt{10^2+5^2-2\cdot5\cdot10\cos\left(\pi-\sin^{-1}\left(\frac{18}{25} \right) \right)}$$

Case 2:

$$\displaystyle \theta=\sin^{-1}\left(\frac{18}{25} \right)$$

Using the law of cosines, we may write:

$$\displaystyle x=\sqrt{10^2+5^2-2\cdot5\cdot10\cos\left(\sin^{-1}\left(\frac{18}{25} \right) \right)}$$

You should be able to obtain an exact value for $x$ in both cases (you have already found the exact value for the acute case), and these do agree with the two positive roots that Heron's formula gives.

What do you find?

MHB Math Helper