Question

Two equal sides of a triangle are each 4m less than three times the third side. Find the dimensions of the triangle, if the perimeter is 55m.

Answer 1

Answer:

We don't know what the length of the 3rd side is, so we'll call this length "x"

We do know the other 2 side lengths. They are each 3x - 4.

 

If all three side lengths gives us a perimeter of 55 then  

all we have to do is add up all three sides. This gives  

us an equation.....

 

3x - 4  +  3x - 4  +  x = 55   Now just solve...

 

7x - 16 = 55

7x -16 + 16 = 55 + 16

7x = 63

7x/7 = 63/7

x = 9

 

This value of "x" is the length of the third side. If we  

substitute it into the expression for the other two sides, 3x - 4,  

we get....

 

3x - 4

3(9) - 4

27 - 4

23

 

And to check....

9 + 23 + 23 = 55

55 = 55  YEAH!!  You did it!

 

Smile! Have a great day!

Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

Two equal sides of a triangle are such 4 m less than three times the third side. If the perimeter is 55 m.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The dimensions of the triangle.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the third side of Δ be r m

$\bf{We\:have}\begin{cases}\sf{Two\:equal\:side\:of\:\triangle=(3r-4)m}\\ \sf{Perimeter\:of\:\triangle\:=55m}\end{cases}}$

We know that formula of the perimeter of triangle :

$\bf{\boxed{\bf{Perimeter\:of\:\triangle=Side+Side+Side}}}}}$

• 1st side = (3r-4) m
• 2nd side = (3r-4) m
• 3rd side = r m

A/q

$\longrightarrow\sf{(3r-4)+(3r-4)+r=55}\\\\\longrightarrow\sf{3r-4+3r-4+r=55}\\\\\longrightarrow\sf{7r-8=55}\\\\\longrightarrow\sf{7r=55+8}\\\\\longrightarrow\sf{7r=63}\\\\\longrightarrow\sf{r=\cancel{\dfrac{63}{7} }}\\\\\longrightarrow\sf{\pink{r=9\:m}}$

Thus;

$\bf{\underline{\underline{\bf{The\:dimension's\:of\:the\:\triangle\::}}}}}$

$\bullet\sf{1^{st}\:side=(3r-4)m=[3(9)-4]=(27-4)=\red{23\:m}}\\\bullet\sf{2^{nd}\:side=(3r-4)m=[3(9)-4]=(27-4)=\red{23\:m}}\\\bullet\sf{3^{rd}\:side=(r)m=\red{9\:m}}$