AB=3.2

BC=8.4

Area=10cm^2

Calculate the perimeter.

Method not answer. Calculator allowed.

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- Jan 8th 2010, 11:53 AMMukilabPerimeter of a triangle
AB=3.2

BC=8.4

Area=10cm^2

Calculate the perimeter.

Method not answer. Calculator allowed. - Jan 8th 2010, 12:01 PMmr fantastic
- Jan 8th 2010, 12:05 PMMukilab
- Jan 8th 2010, 12:07 PMpickslides
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 $\displaystyle a= 8.4 $ and side $\displaystyle c = 3.2$

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

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

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

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

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

Now go ahead a change the world! - Jan 8th 2010, 12:13 PMMukilab
- Jan 8th 2010, 12:22 PMpickslides
- Jan 8th 2010, 12:36 PMMukilab
- Jan 8th 2010, 01:05 PMpickslides
- Jan 8th 2010, 01:15 PMMukilab
sin^{-1}(1/2 x 3.2 x 8.4)

- Jan 8th 2010, 02:15 PMpickslides
- Jan 8th 2010, 03:07 PMArchie Meade
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

$\displaystyle triangle\ area=0.5abSinC$ 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 $\displaystyle area=0.5abSinC$

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

$\displaystyle 0.5abSinC=0.5acSinB=0.5bcSinA$ 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....

Your problem is asking you to work this in reverse.

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.

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

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. - Jan 8th 2010, 03:49 PMSoroban
Hello, Mukilab!

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

Quote:

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

Calculate the perimeter.

Code:`A`

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* \

c = 3.2 * \

* \ b

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B * * * * * * * C

a = 8.4

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

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

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

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

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

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

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

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

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

Got it?

- Jan 8th 2010, 04:15 PMArchie Meade
Or...

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

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

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

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

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

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

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

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

Then $\displaystyle y=8.4-x$

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

$\displaystyle b=\sqrt{y^2+h^2}$ - Jan 8th 2010, 05:16 PMtakis1881
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.