Math, asked by Anonymous, 6 months ago

If the area of a right triangle is 70 cm² and one of the sides containing the right angle is 14 cm. Find the length of other leg.​

Answers

Answered by Anonymous
1

Given :

  • Area of triangle = 70 cm²
  • Length of 1st side = 14 cm

To Find :

  • Length of 2nd side = ?

Solution :

Let, another side = x

We know that,

 \underline{\boxed{\bf{ Area \: of \: triangle = \dfrac{1}{2} \times base \times height }}}

 \underline{\boxed{\bf{ Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}}

We have :

  • Base = x
  • Height = 14 cm
  • Area of triangle = 70 cm²

On substituting the values :

 \sf : \implies 70 \: cm^{2} = \sf\dfrac{1}{2} \times 14 \: cm \times x

 \sf : \implies 70 \: cm^{2} = \sf\dfrac{1}{\cancel{2}} \times \cancel{14}_{7} \: cm \times x

 \sf : \implies 70 \: cm^{2} = 7 \: cm \times x

 \sf : \implies \dfrac{\cancel{70}^{10} \: cm^{2}}{\cancel{7} \: cm} = x

 \sf : \implies 10 \: cm = x

 \underline{\boxed{\bf{x = 10 \: cm}}}

Hence, Length of 2nd side of triangle is 10 cm.

Answered by Anonymous
0

Answer:

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the branch of science concerned with the substances of which matter is composed, the investigation of their properties and reactions, and the use of such reactions to form new substances.

2.

the complex emotional or psychological interaction between people.

"their affair was triggered by intense 

\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}

\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}

\color{yellow} {\Huge {\sf{Solution:}}}

\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}

\tt \color{blue} {\mathrm{Rewrite\:}\sin ^4(x)-\cos ^4(x)\mathrm{\:as\:}(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x))^2-(\cos ^2(x))^2}

\color{fuchsia} {\normalsize {\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}}

\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}

\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}=

\color{fuchsia} {\normalsize \mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}

\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}

\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}=

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=(x+y)(x-y)

(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))

=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))=

\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=(x+y)(x-y)

\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))

(x)=(sin(x)+cos(x))(sin(x)−cos(x))

=(\sin (x)+\cos (x))(\sin (x)-\cos (x))=(sin(x)+cos(x))(sin(x)−cos(x))

\large=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))\ \textless \ br /\ \textgreater \ (x))(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ \large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}=

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=(x+y)(x-y)

\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)

 (x)=(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}=

\mathrm{Cancel\:}\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}:\quad \sin ^2(x)+\cos ^2(x)Cancel \ \textless \ br /\ \textgreater \ (sin(x)+cos(x))(sin(x)−cos(x))

\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)+\cos(x)Cancelthecommonfactor:sin(x)+cos(x)

=\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)-\cos (x))}{\sin (x)-\cos (x)}=

\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cos

\mathrm{Use\:the\:following\:identity}:\quad \cos ^2(x)+\sin

\huge \boxed{\color{red} {\ \huge =1}}

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