Perimeter of the triangle

[123]

Lengths of the triangle sides agree to 7 : 8 : 9, while its area is equal to $\displaystyle 540\sqrt{5}$$\displaystyle cm^2$.

Calculate the perimeter of the triangle.

maybe i could write that sides of triangle is 7x, 8x, 9x.
then I have Perimeter=24x
... or maybe i wrong

any solutions

 

[soroban]

Hello, 123!

Your game plan is excellent . . . Keep going!

Lengths of the triangle sides agree to 7 : 8 : 9, while its area is equal to $\displaystyle 540\sqrt{5}$$\displaystyle cm^2$.

Calculate the perimeter of the triangle.

Someone will suggest Heron's Formula . . . I prefer Trigonometry.

            *
           *   *
          *       *  8x
      7x *          *
        *             *
       * @              *
      *   *   *   *   *   *
                9x

$\displaystyle \text{Law of Cosines: }\:\cos\theta \:=\:\frac{(7x)^2 + (9x)^2 - (8x)^2}{2(7x)(9x)} \:=\:\frac{66x^2}{126x^2} \;=\;\frac{11}{21}$

. . $\displaystyle \sin\theta \;=\;\sqrt{1-\cos^2\!\theta} \;=\;\sqrt{1-\left(\tfrac{11}{21}\right)^2} \;=\;\sqrt{\tfrac{320}{441}} \;=\;\tfrac{8\sqrt{5}}{21}$


$\displaystyle \text{Formula: }\;A \;=\;\tfrac{1}{2}ab\sin C$

. . $\displaystyle \text{The area of a triangle is equal to one-half the product of two sides times the sine of the included angle.}$


$\displaystyle \text{Hence, we have: }\;\tfrac{1}{2}(7x)(9x)\left(\tfrac{8\sqrt{5}}{21}\right) \;=\;540\sqrt{5} \quad\Rightarrow\quad 12\sqrt{5}\,x^2 \:=\:540\sqrt{5} \quad\Rightarrow\quad x^2 \:=\:45 \quad\Rightarrow\quad x \:=\:3\sqrt{5}$

\(\displaystyle \text{Therefore: \erimeter}\;=\;24x \;=\;24(3\sqrt{5}) \;=\;72\sqrt{5}\text{ cm.}\)

 

[ashily24dee]

now i know how to solve this kind of problem.. thanks for telling solution step by step..

 

[Deleted member 4993]

Lengths of the triangle sides agree to 7 : 8 : 9, while its area is equal to $\displaystyle 540\sqrt{5}$$\displaystyle cm^2$.

Calculate the perimeter of the triangle.

maybe i could write that sides of triangle is 7x, 8x, 9x.
then I have Perimeter=24x
... or maybe i wrong

any solutions

Since Heron's formula was mentioned:

\(\displaystyle Area \ of \ a \ triangle \ = \ \sqrt{s*(s-a)*(s-b)*(s-c)\)

where a, b & c are the sides of the triangle and:

$\displaystyle s = \frac{a+b+c}{2}$

then

s = 12x

and

$\displaystyle \left [540\sqrt{5}\right ]^2 \ = \ 12*x * 5*x * 4*x * 3*x$

540*540*5 = 720 * x[sup:2jiw9fd7]4[/sup:2jiw9fd7]

x[sup:2jiw9fd7]4[/sup:2jiw9fd7] = 3 * 135 * 5 = 3[sup:2jiw9fd7]4[/sup:2jiw9fd7] * 5[sup:2jiw9fd7]2[/sup:2jiw9fd7]

x = 3 * ?5

Perimeter = 24 * x = 72*?5 cm