# Thread: Perimeter of a triangle

1. ## Perimeter of a triangle

AB=3.2
BC=8.4
Area=10cm^2

Calculate the perimeter.

2. Originally Posted by Mukilab
AB=3.2
BC=8.4
Area=10cm^2

Calculate the perimeter.

What have you tried? Where are you stuck?

3. Originally Posted by mr fantastic
What have you tried? Where are you stuck?
I don't know the formula at all for this. I cannot even attempt to try.

4. Hi there,

Let me first say there is probably a few ways to do this other than the way I am going to suggest. Something with Herron's formula may be easier.

My solution takes a few steps so be patient.

First of all, in triangle with sides a,b,c and with angles A,B,C the perimeter needs all sides to be found and summed.

You have information about the area and sides $a= 8.4$ and side $c = 3.2$

Now use $Area = \frac{1}{2}\times a \times c \times \sin(B)$ to solve for $B$

Here is the sum, use inverse $\sin$ to complete.

$10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)$

After you have your angle value for $B$ you should employ the $\cos$ rule to find $b$

It is $b = \sqrt{a^2+c^2-2ac\times \cos(B)}$

Now go ahead a change the world!

5. Originally Posted by pickslides
Hi there,

Let me first say there is probably a few ways to do this other than the way I am going to suggest. Something with Herron's formula may be easier.

My solution takes a few steps so be patient.

First of all, in triangle with sides a,b,c and with angles A,B,C the perimeter needs all sides to be found and summed.

You have information about the area and sides $a= 8.4$ and side $c = 3.2$

Now use $Area = \frac{1}{2}\times a \times c \times \sin(B)$ to solve for $B$

Here is the sum, use inverse $\sin$ to complete.

$10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)$

After you have your angle value for $B$ you should employ the $\cos$ rule to find $b$

It is $b = \sqrt{a^2+c^2-2ac\times \cos(B)}$

Now go ahead a change the world!
I don't know how to use reverse sin for lines, only for angles and google is not helping me. Although thank you for trying

6. Originally Posted by Mukilab
I don't know how to use reverse sin for lines,

use $\sin^{-1}$ on the side length, it will work.

7. Originally Posted by pickslides
use $\sin^{-1}$ on the side length, it will work.
What do you mean side length? I know how to use sin^{-1} with angles. I tried getting the whole thing in (apart from the area) into that and there was a math error as spoken by my calculator.

8. Originally Posted by Mukilab
What do you mean side length? I know how to use sin^{-1} with angles. I tried getting the whole thing in (apart from the area) into that and there was a math error as spoken by my calculator.

Show me your workings I will try to correct them.

9. sin^{-1}(1/2 x 3.2 x 8.4)

10. Originally Posted by Mukilab
sin^{-1}(1/2 x 3.2 x 8.4)
This is how it goes

$10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)$

$\frac{10}{\frac{1}{2}\times 8.4 \times 3.2} = \sin(B)$

$\sin(B) = \frac{10}{\frac{1}{2}\times 8.4 \times 3.2}$

Now take the inverse.

11. Hi Mukilab,

The side measurements you are given are

3.2cm and 8.4cm.

If you knew the 3rd side length, you could add all 3 lengths together
to calculate the perimeter.

The clue you are given to find the 3rd side is.. triangle area = 10 sq cm.

You must understand how the area of a triangle is calculated in order to
go in reverse.
Inverse Sine is not confusing when you have mastered Sine.
Inverse addition is subtraction, inverse multiplication is division,
inverse square is square root and so on.

Triangle area is half the base times the perpendicular height.
When you are not given the perpendicular height directly, then the

$triangle\ area=0.5abSinC$ gives the same result.

You should study why this is so....
if "a" is the base, then bSinC is the perpendicular height.

I want to add something here to make this more complete.

The formula $area=0.5abSinC$

means that if you have the 3 dimensions of a corner then you can calculate the area.

$0.5abSinC=0.5acSinB=0.5bcSinA$ all give triangle area.

Remember, side a is opposite angle A, b is opposite B, c is opposite C.

You need to study and practice with that.

If that's very clear, you can continue....

You are given the area and two sides.

What dimensions calculate the area?

two sides and the angle that those two sides make! ok?

So 0.5(3.2)(8.4)SinX=10, where X is the angle between those sides.

$SinX=\frac{10}{(0.5)(3.2)(8.4)}$

Therefore X is the inverse Sine of that fraction.

That's putting it simply, but unfortunately, there are 2 possible answers.
Your calculator gives you the one that is less than 90 degrees.
The other one is 180 degrees minus the one your calculator gave you.

Just work with the one your calculator gave you.

there are more steps after you've got the angle.
You should just go step by step.

So try to calculate the angle between the two lines first.

12. Hello, Mukilab!

We don't need inverse trig functions . . . just Pythagorus.

$AB\,=\,3.2,\;\;BC\,=\,8.4,\;\;\text{Area}\:=\:10\t ext{ cm}^2$

Calculate the perimeter.
Code:
                  A
o
*  \
c = 3.2  *     \
*        \ b
*           \
* θ            \
B *  *  *  *  *  *  * C
a = 8.4

$\text{Area: }\:A \:=\:\tfrac{1}{2}ac\,\sin B$

We have: . $\text{Area} \:=\:\tfrac{1}{2}(3.2)(8.4)\sin B \:=\:10$

. . $\sin B \:=\:\frac{10}{13.44} \quad \Rightarrow\quad \sin B \:=\:\frac{125}{168}$ .[1]

Since $\sin^2\!B + \cos^2\!B \:=\:1\text{, then: }\:\cos B \:=\:\sqrt{1-\sin^2\!B}$

Substitute [1]: . $\cos B \:=\:\sqrt{1 - \left(\frac{125}{168}\right)^2} \quad\Rightarrow\quad \cos B \:=\:\frac{\sqrt{12599}}{168}$

Law of Cosines: . $b^2 \;=\;a^2+c^2-2ac\cos B$

We have: . $b^2 \;=\;8,4^2 + 3.2^2 - 2(8.4)(3.2)\left(\frac{\sqrt{12599}}{168}\right)$

. . . . . . . . . $=\;70.56 + 10.24 - 35.91848549 \;=\;44.88151451$

Hence: . $b \;=\;\sqrt{44.88151451} \;=\;6.699366724 \;\approx\;6.7\text{ cm}$

Got it?

13. Or...

$10=0.5(3.2)(8.4)SinB=3.2(4.2)SinB$

$SinB=\frac{(\frac{10}{4.2})}{3.2}$

which is $\frac{opposite}{hypotenuse}$

for the right-angled triangle on the left with perpendicular height $\frac{10}{4.2}$

Then split the base "a" into 2 parts x and y.

$x^2+(\frac{10}{4.2})^2=3.2^2$

$x^2=3.2^2-(\frac{10}{4.2})^2$

$x=\sqrt{3.2^2-(\frac{10}{4.2})^2}$

Then $y=8.4-x$

Finally $y^2+h^2=b^2$

$b=\sqrt{y^2+h^2}$

14. You can try Heron's formula for the area of a triangle :
E=sqrt[t(t-a)(t-b)(t-c)], sqrt:square root, t=1/2 of perimeter or (a+b+c)/2.
I tried the method, but i found out that it leads to a 4th degree equation(if you take y=b^2 the equation becomes quadratic) with no solutions!!! Check the facts and figures of your problem again. Maybe i had made a mistake during my calculations.