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# 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.

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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.

Mar 22, 2015

#1
+94619
+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  !!!

Mar 22, 2015

#1
+94619
+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
#2
+95369
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

Mar 22, 2015
#3
+95369
+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

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

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

$$\\ \mathbf{\mbox{So there are 2 possible exact lengths for the 3rd side}}\\\\ \mathbf{2\sqrt{89}\;cm\;\;\;\; or\;\;\;\;\;2\sqrt{1049}\;cm}$$

Mar 22, 2015
#4
+94619
+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 !!!!

Mar 23, 2015
#5
+95369
0

Yes I had put a plus instead of a minus when I found the product of a conjugal pair.

That was my error. I spent ages looking for that stupid mistake. :)

Mar 23, 2015
#6
+94619
0

My original amswer wasn't entirely correct.....you actually helped me to see MY dumb error...!!

Mar 23, 2015