Question

find the area of the triangle​

Answer 1

Answer:

Given:

b = 12 cm

h = 10 cm

To find:

Area of the triangle

Solution:

Area of the triangle= 1/2* b* h

=1/2*12cm*10 cm

=6cm*10cm

. =60 cm

. . Area of triangle is 60 cm

Answer 2

Answer:

${\pmb{\frak{\underline{\underline{Given...}}}}}$

• $\red\bigstar$ Sides of Triangle = 10cm, 10cm, 12cm

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

${\pmb{\frak{\underline{\underline{To\: Find...}}}}}$

• $\red\bigstar$ Area of Triangle

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

${\pmb{\frak{\underline{\underline{Using \: Formulae ...}}}}}$

$\green\bigstar$ Semi Perimeter of Triangle

$\dag{\underline{\boxed{\sf{\purple{Semi \: Perimeter = \dfrac{a + b + c}{2}}}}}}$

$\green\bigstar$ Area of Triangle by herom Formula

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

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

${\pmb{\frak{\underline{\underline{Solution...}}}}}$

$\ddag{\underline{\frak{\green{Firstly \: finding \: the \: semi \: perimeter \: of \: Triangle..}}}}$

$\quad{ : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{a + b + c}{2}}}}}$

• Substituting the values

$\begin{gathered}\quad{ : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{12 + 10 + 10}{2}}}}} \\ \\ { : \implies{\sf{Semi \: Perimeter = \bf{\dfrac{32}{2}}}}} \\ \\ \qquad{ : \implies{\sf{Semi \: Perimeter = \bf{\cancel{\dfrac{32}{2}}}}}} \\ \\ \qquad{: \implies{\sf{Semi \: Perimeter = \bf{16}}}} \\ \\ \qquad{\dag{\underline{\boxed{\sf{\pink{Semi \: Perimeter = {16}}}}}}}\end{gathered}$

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

$\ddag{\underline{\frak{\green{Now \: finding \: the \: area \: of \: Triangle..}}}}$

$\quad{: \implies{\sf{Area \: of \: Triangle = \bf{\sqrt{s(s - a)(s - b)(s - c)}}}}}$

• Substituting the values

$\begin{gathered}{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{16(16 - 12)(16- 10)(16 - 10)}}}}} \\ \\{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{16(4)(6)(6)}}}}} \\ \\ \quad\qquad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{16 \times 4 \times 6 \times 6}}}}} \\ \\ \quad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{2304}}}}} \\ \\ \qquad\qquad{: \implies{\sf{Area \: of \: \triangle = \bf{\sqrt{48 \times 48}}}}} \\ \\ \qquad{: \implies{\sf{Area \: of \: \triangle = \bf{48}}}} \\ \\ { \dag{\underline{\boxed{\sf{\pink{Area \: of \: Triangle = \bf{48}}}}}}}\end{gathered}$

• Henceforth,The Area of Triangle is 48 cm.

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

${\pmb{\frak{\underline{\underline{Learn \: More ...}}}}}$

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