Question

the perimeter of a triangle is 15/2cm.if two of its sides measure 7/3 cm and 19/6 cm, find the third side. ​

Answer 1

Given

• The perimeter of a triangle = 15/2cm.
• First side (AB) = 7/3 cm
• Second side (BC) = 19/6 cm

To find

• The third side

Solution

• Let the triangle be ABC.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

★ We know that

$\underline{\boxed{\tt{Sum\: of\: all\: sides = Perimeter\: of\: a\: triangle}}}$

$\tt:\implies\: \: \: \: \: \: \: \: {AB + BC + CA = \dfrac{15}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{7}{3} + \dfrac{19}{6} + CA = \dfrac{15}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{14 + 19}{6} + CA = \dfrac{15}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\cancel{\dfrac{33}{6}} + CA = \dfrac{15}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{11}{2} + CA = \dfrac{15}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {CA = \dfrac{15}{2} - \dfrac{11}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {CA = \dfrac{15 - 11}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {CA = \cancel{\dfrac{4}{2}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {CA = 2}$

⠀

$\sf\pink{⟶}$ Therefore, Third side i.e.,

• CA = 2 cm

━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\large\bold{\underline{\underline{{Given:-}}}}$

${\bullet\:}$ Perimeter of the triangle = $\rm{\dfrac{15}{2}\:cm}$

${\bullet\:}$ First side = $\rm{\dfrac{7}{3}\:cm}$

${\bullet\:}$ Second Side = $\rm{\dfrac{19}{6}\:cm}$

$\large\bold{\underline{\underline{{To\:Find:-}}}}$

${\bullet\:}$ Third side = ?

$\large\bold{\underline{\underline{{Solution:-}}}}$

${\bullet\:}$ Let the triangle be ABC.

[ Refer to the attachment ↑]

Now,

${\bullet\:}$ We have to find the third side,

So,

$\red{\longrightarrow\:\:}$ $\rm{AB+BC+CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{7}{3} +\dfrac{19}{6} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{(\dfrac{7}{3}  \times\dfrac{2}{2}) +( \dfrac{19}{6} \times \dfrac{1}{1}  )+CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{14}{6} +\dfrac{19}{6} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{14+19}{6} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{33}{6} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{33 \div 3}{6\div 3} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{\dfrac{11}{2} +CA=\dfrac{15}{2} }$

$\red{\longrightarrow\:\:}$ $\rm{CA=\dfrac{15}{2}-\dfrac{11}{2}  }$

$\red{\longrightarrow\:\:}$ $\rm{CA=\dfrac{15-11}{2}  }$

$\red{\longrightarrow\:\:}$ $\rm{CA=\dfrac{4}{2}  }$

$\red{\longrightarrow\:\:}$ $\rm{CA=\dfrac{4 \div2}{2\div 2}  }$

$\red{\longrightarrow\:\:}$ $\rm{CA=\dfrac{2}{1}  }$

$\red{\longrightarrow\:\:}$ $\rm{CA=2 }$

$\large\boxed{\underline{\green{\rm \therefore\:The \:third\:side = 2\:cm  }}}$

________________________________________