Question

`Q) The perimeter of a triangle is 48 cm. the length of the sides are in the ratio 3:4:5. Find the length of each side of triangle.`​

Answer 1

Given :

• Perimeter of triangle = 48 cm
• Ratio of sides = 3:4:5

To Find :

• Measure of the sides = ?

SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Perimeter{\small_{(Triangle)}} = a + b + c }}}}}$

Where :

• ➻ a = Side 1
• ➻ b = Side 2
• ➻ c = Side 3

$\\ \qquad{\rule{150pt}{1pt}}$

~ Let the Ratios :

• ➻ Side 1 = 3y
• ➻ Side 2 = 4y
• ➻ Side 3 = 5y

$\\ \qquad{\rule{150pt}{1pt}}$

~ Calculating the Value of y :

${\longmapsto{\qquad{\sf{ Perimeter = a + b + c }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 48 = 3y + 4y + 5y }}}} \\ \\ \\ \ \ {\longmapsto{\qquad{\sf{ 48 = 12y }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ \dfrac{48}{12} = y }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ \cancel\dfrac{48}{12} = y }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\pink{\frak{ y = 4 }}}}}}}}$

$\\ \qquad{\rule{150pt}{1pt}}$

~ Calculating the Sides :

• ➳ Side 1 = 3y = 3(4) = 12 cm
• ➳ Side 2 = 4y = 4(4) = 16 cm
• ➳ Side 3 = 5y = 5(4) = 20 cm

$\\ \qquad{\rule{150pt}{1pt}}$

~ Therefore :

❛❛ Three sides of the Triangle are 12 cm , 16 cm and 20 cm . ❜❜

$\\ {\underline{\rule{300pt}{9pt}}}$

Answer 2

To find : length of each side of triangle.

Given :  The perimeter of a triangle is $48$ cm . Length of the sides are in the ratio $3:4:5$ .

Solution :

• Let , $3x ,4x ,5x$ be the sides of triangle in the ratio .
• As per given data we know that perimeter of a triangle is $48$ cm .
• If , $a,b,c$ are sides of triangle then perimeter of triangle is given by ,
• Perimeter of triangle = $a + b+ c$

       $48 = 3x + 4x + 5x \\12x= 48\\x = 4$

• Now , substituting the value of $x$ in the sides of triangle .

      $3x = 3\times 4 =12\ cm \\4x = 4\times 4 =16\ cm \\5x = 5\times 4 = 20\ cm$

Hence , sides of triangle are , $12\ cm , 16\ cm , 20\ cm$ .