Question

find the area of the triangle whose two of its sides are 13 cm and 14 cm and perimeter is 42 cm​.

Answer 1

third side is equal to 42-(13+14) = 15

a=13,b=14,c=15

Semi-perimeter s=(a+b+c)/2=(13+14+15)/2=21

Area, A = s(s−a)(s−b)(s−c)

A=21(21−13)(21−14)(21−15)

A=84 cm2

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Answer 2

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\red{Given :}}}}}}}\end{gathered}$

• $\leadsto$ Two sides of Triangle = 13 cm and 14 cm.
• $\leadsto$ Perimeter of Triangle = 42 cm

$\begin{gathered} \end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\red{To Find :}}}}}}}\end{gathered}$

• $\leadsto$ Area of Triangle

$\begin{gathered} \end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\red{Using Formula :}}}}}}}\end{gathered}$

$\quad\dag\underline{\boxed{\sf{Perimeter \: of \: \triangle = sum \: of \: all \: sides}}}$

$\quad{\dag{\underline{\boxed{\sf{S = \dfrac{a + b + c}{2}}}}}}$

$\quad{\dag{\underline{\boxed{\sf{Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}}}}}}$

$\begin{gathered} \end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\red{Solution:}}}}}}}\end{gathered}$

$\bigstar{\underline{\underline{\pmb{\frak{\green{Firstly \: finding \: third \: side \: of \: triangle :-}}}}}}$

• Let the the side of triangle be "x" cm.

According to the question

$\quad{\longmapsto{\sf{Perimeter \: of \: \triangle = sum \: of \: all \: sides}}}$

• Substituting the values

$\quad{\longmapsto{\sf{42 \: cm = 13 \: cm + 14 \: cm + x}}}$

$\quad{\longmapsto{\sf{42 = 17 + x}}}$

$\quad{\longmapsto{\sf{42 - 17 = x}}}$

$\quad{\longmapsto{\sf{15 = x}}}$

${\quad{\dag{\underline{\boxed{\sf{\pink{x = 15 \: cm}}}}}}}$

${\therefore{\sf{\underline{\underline{\red{The \: third \: side \: of \: triange\: is \: 15 \: cm..}}}}}}$

$\begin{gathered} \end{gathered}$

$\bigstar{\underline{\underline{\pmb{\frak{\green{Now \: calculating \: the \: semi \: perimeter \: of \: triangle :-}}}}}}$

$\quad{\longmapsto{\sf{S = \dfrac{a + b + c}{2}}}}$

• Substituting the values

$\quad{\longmapsto{\sf{S = \dfrac{13 + 14 + 15}{2}}}}$

$\quad{\longmapsto{\sf{S = \dfrac{42}{2}}}}$

$\quad{\longmapsto{\sf{S = \cancel{\dfrac{42}{2}}}}}$

$\quad{\longmapsto{\sf{S = 21 \: cm}}}$

$\quad\dag{\underline{\boxed{\sf{\pink{Semi \: perimeter = 21 \: cm}}}}}$

${\therefore{\sf{\underline{\underline{\red{The \: semipeter \: of \: triangle\: is \: 21 \: cm..}}}}}}$

$\begin{gathered} \end{gathered}$

$\bigstar{\underline{\underline{\pmb{\frak{\green{Finding \: the \: area \: of \: triangle \: by \: heron's \: Formula :-}}}}}}$

$\quad{\longmapsto{\sf{Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}}}}$

• Substituting the values

$\quad{\longmapsto{\sf{Area \: of \: \triangle = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}}}}$

$\quad{\longmapsto{\sf{Area \: of \: \triangle = \sqrt{21(8)(7)(6)}}}}$

$\quad{\longmapsto{\sf{Area \: of \: \triangle = \sqrt{21 \times 8 \times 7 \times 6}}}}$

$\quad{\longmapsto{\sf{Area \: of \: \triangle = \sqrt{7056 }}}}$

$\quad{\longmapsto{\sf{Area \: of \: \triangle = {84 \: {cm}^{2} }}}}$

$\quad\dag{\underline{\boxed{\sf{\pink{Area \: of \: \triangle = {84 \: {cm}^{2}}}}}}}$

${\therefore{\sf{\underline{\underline{\red{The \: Area \: of \: Triangle \: is \: 84 \: {cm}^{2} ..}}}}}}$

$\begin{gathered} \end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\red{ Learn More :}}}}}}}\end{gathered}$

$\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}$