Question

In triangle ABC with AB = 7 cm, BC = 5 cm and AC = 6 cm. The height ADfrom A to BC is 6 cm. Find the area of AABC. What will be the heightfrom B to AC i.e., BE?​

Answer 1

${\huge{\boxed{\sf{\green{Question}}}}}$

In triangle ABC with AB = 7 cm, BC = 5 cm and AC = 6 cm. The height ADfrom A to BC is 6 cm. Find the area of AABC. What will be the heightfrom B to AC i.e., BE?​

${\huge{\underline{\bf{\pink{Answer}}}}}$

$\begin{gathered} \\ \Large{\bf{\green{\underline{GiVeN\::}}}} \\ \end{gathered}$

ABC is a right angle triangle

∠B is right angled, i.e. 90°.

AB (Height) = 24 cm

BC (Base) = 7 cm

$\begin{gathered} \\ \Large{\bf{\pink{\underline{To\:FiNd\::}}}} \\ \end{gathered}$

Sin A , Cos A

Sin C , Cos C

$\begin{gathered} \\ \Large{\bf{\purple{\underline{CaLcULaTioN\::}}}} \\ \end{gathered}$

$\begin{gathered}\bf\orange{According\:to\:Pythagoras\:theorem,} \\ \end{gathered}$

$\begin{gathered}\red\bigstar\:\:{\underline{\blue{\boxed{\bf{\green{(Hypotenuse)^2\:=\:(Height)^2\:+\:(Base)^2\:}}}}}} \\ \end{gathered} ★$

$\begin{gathered}:\implies\:\:\bf{(Hypotenuse)^2\:=\:(24)^2\:+\:7^2\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{(Hypotenuse)^2\:=\:576\:+\:49\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{(Hypotenuse)^2\:=\:625\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{Hypotenuse\:=\:\sqrt{625}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\blue{Hypotenuse\:(AC)\:=\:25\:cm} \\ \\ \end{gathered}$

$\begin{gathered}\bf\pink{We\:know\:that,} \\ \end{gathered}$

$\begin{gathered}\orange\checkmark\:\:{\underline{\boxed{\bf{\purple{Sin\:\theta\:=\:\dfrac{Perpendicular}{Hypotenuse}}}}}} \\ \end{gathered}$

Perpendicular is also the opposite side.

$\begin{gathered}\green\checkmark\:\:{\underline{\boxed{\bf{\blue{Cos\:\theta\:=\:\dfrac{Base}{Hypotenuse}}}}}} \\ \end{gathered}$

Base is also the adjacent side.

$\begin{gathered}:\implies\:\:\bf{Sin\:A\:=\:\dfrac{BC}{AC}\:} \\ \end{gathered} :⟹SinA= ACBC$

$\begin{gathered}:\implies\:\:\bf\red{Sin\:A\:=\:\dfrac{7}{25}\:} \\ \end{gathered}$

$\begin{gathered}\Large\bf\purple{And,} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{Cos\:A\:=\:\dfrac{AB}{AC}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\pink{Cos\:A\:=\:\dfrac{24}{25}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{Sin\:C\:=\:\dfrac{AB}{AC}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\orange{Sin\:C\:=\:\dfrac{24}{25}\:} \\ \end{gathered}$

$\begin{gathered}\Large\bf\blue{And,} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{Cos\:C\:=\:\dfrac{BC}{AC}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\green{Cos\:C\:=\:\dfrac{7}{25}\:} \\ \end{gathered}$