A triangle has an area of 200cm^2. Two sides of this triangle measure 26 and 40 cm respectively. Find the exact value of the third side.

Guest Mar 22, 2015

#1**+10 **

We'll have to use Heron's Formula to solve this.

Let a = 26, b= 40 and x be the unknown side

s = the semi-perimeter = [x + 26 + 40]/2 = [x + 66]/2

So we have

200 = √[s(s -a)(s-b)(s-x)]

200^2 = [x + 66]/2 * [(x + 66)/2 - 26] * [(x + 66)/2 - 40]* [(x + 66)/2 - x]

200^2 = [x + 66]/2 * [ x + 14]/2 * [x - 14]/2 * [66 - x]/2

200^2 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

40000 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

640000= [x + 66] * [x + 14] * [x - 14] * [66- x]

We could brute force this.....but....I'm going to let WolframAlpha do the heavy lifting...

It returns two solutions

x = 2√89 ≈ 18.87 and 2√1049 ≈ 64.78

Both satisfy the Triangle Inequality.....so.....looking at the Law of Cosines

18.87^2 = 40^2 + 26^2 - 2(40)(26)cosΘ

And the angle between the sides of 26 and 40 could be 22.6°

And using the Law of Sines, we have

sinΘ/40 = sin 22.6/18.87

sinΘ = 40sin22.6/18.87 = 54.55°

And the remaining angle is 180 - 22.6 - 54.55 = 102.85

But this is impossible because it would mean that the greatest angle is opposite the intermediate side

And using the Law of Cosines again, we have...

64.78^2 = 40^2 + 26^2 - 2(40)(26)cosΘ

So, the angle between the sides of 26 and 40 could be 157.38°

And using the Law of Sines again, we have

sinΘ/40 = sin 157.38/64.78 = 13.73°

And the remaining angle is 180 - 157.38 - 13.73 = 8.89°

So, the solution is

Side 64.78, opposite angle = 157.38°

Side 40, oppsite angle 13.73°

Side 26, opposite angle 8.89°

So....the remaining unknown side is 64.78

Here's the approximate triangle

**SEE MELODY'S ANSWER BELOW...THE SECOND SOLUTION IS POSSIBLE !!!**

CPhill Mar 22, 2015

#1**+10 **

Best Answer

We'll have to use Heron's Formula to solve this.

Let a = 26, b= 40 and x be the unknown side

s = the semi-perimeter = [x + 26 + 40]/2 = [x + 66]/2

So we have

200 = √[s(s -a)(s-b)(s-x)]

200^2 = [x + 66]/2 * [(x + 66)/2 - 26] * [(x + 66)/2 - 40]* [(x + 66)/2 - x]

200^2 = [x + 66]/2 * [ x + 14]/2 * [x - 14]/2 * [66 - x]/2

200^2 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

40000 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

640000= [x + 66] * [x + 14] * [x - 14] * [66- x]

We could brute force this.....but....I'm going to let WolframAlpha do the heavy lifting...

It returns two solutions

x = 2√89 ≈ 18.87 and 2√1049 ≈ 64.78

Both satisfy the Triangle Inequality.....so.....looking at the Law of Cosines

18.87^2 = 40^2 + 26^2 - 2(40)(26)cosΘ

And the angle between the sides of 26 and 40 could be 22.6°

And using the Law of Sines, we have

sinΘ/40 = sin 22.6/18.87

sinΘ = 40sin22.6/18.87 = 54.55°

And the remaining angle is 180 - 22.6 - 54.55 = 102.85

But this is impossible because it would mean that the greatest angle is opposite the intermediate side

And using the Law of Cosines again, we have...

64.78^2 = 40^2 + 26^2 - 2(40)(26)cosΘ

So, the angle between the sides of 26 and 40 could be 157.38°

And using the Law of Sines again, we have

sinΘ/40 = sin 157.38/64.78 = 13.73°

And the remaining angle is 180 - 157.38 - 13.73 = 8.89°

So, the solution is

Side 64.78, opposite angle = 157.38°

Side 40, oppsite angle 13.73°

Side 26, opposite angle 8.89°

So....the remaining unknown side is 64.78

Here's the approximate triangle

**SEE MELODY'S ANSWER BELOW...THE SECOND SOLUTION IS POSSIBLE !!!**

CPhill Mar 22, 2015

#2**0 **

I'm impressed Chris :)

check

$${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{26}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{64.78}}\right) = {\frac{{\mathtt{6\,539}}}{{\mathtt{100}}}} = {\mathtt{65.39}}$$

$${\sqrt{{\mathtt{65.39}}{\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{26}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{40}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{64.78}}\right)}} = {\mathtt{199.730\: \!747\: \!341\: \!489\: \!712\: \!9}}$$

yes that is correct

Melody Mar 22, 2015

#3**+5 **

Until I came to this forum I had never heard of Heron's formula so I am going to do it without using Heron's formula.

$$\\A=(1/2)*a*b*sinC\\\\

200=(1/2)*26*40*sinC\\\\

sinC=\frac{400}{26*40}\\\\

sinC=\frac{5}{13}\\\\

cosC=\pm\frac{12}{13} \qquad $draw a triangle and use the pythagorean theorem to help you get this$\\\\

$now use cosine rule$\\\\

c^2=26^2+40^2\pm 2*26*40*\frac{12}{13}\\\\

c^2=26^2+40^2\pm 2*2*40*12\\\\

c^2=356,\;\; or\;\;c^2=4196\\\\

c^2=4*89,\;\; or\;\;c^2=4*1049\\\\

c=2\sqrt{89}\approx 18.8680cm,\;\; or\;\;c=2\sqrt{1049}\approx 64.7765cm$$

-----------------------------------

check - using Heron's formula

check $$2+\sqrt{89}$$ first

$$\\$s=the semi perimeter $\\\\

s= \frac{26+40+2\sqrt{89}}{2}=33+\sqrt{89}\\\\

A=\sqrt{s(s-a)(s-b)(s-c)}\\\\

A^2=s(s-a)(s-b)(s-c)\\\\

A^2=(33+\sqrt{89})(33+\sqrt{89}-2\sqrt89)(33+\sqrt{89}-26)(33+\sqrt{89}-40)\\\\

A^2=(33+\sqrt{89})(33-\sqrt{89})(7+\sqrt{89})(-7+\sqrt{89})\\\\

A^2=(33^2-89)(89-49)\\\\

A^2=1000*40\\\\

A^2=40000\\\\

A=200$$

**OR**

now check the other one using the approximation.

semiperimeter= $${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{64.776\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{26}}\right) = {\mathtt{65.388\: \!25}}$$

Area

$${\sqrt{{\mathtt{65.388\: \!25}}{\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{40}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{26}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{64.776\: \!5}}\right)}} = {\mathtt{200.003\: \!028\: \!623\: \!087\: \!389\: \!9}}$$

------------------------------------------------

$$\\

\mathbf{\mbox{So there are 2 possible exact lengths for the 3rd side}}\\\\

\mathbf{2\sqrt{89}\;cm\;\;\;\; or\;\;\;\;\;2\sqrt{1049}\;cm}$$

Melody Mar 22, 2015

#4**+5 **

Thanks, Melody......when I checked again, that second solution IS possible, too!!!

Here is the other triangle that Melody found....

**GOOD JOB, MELODY !!!! A POX ON ME !!!!**

CPhill Mar 23, 2015