Question

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.​

Answer 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.

Answer 2

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$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)

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

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

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

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

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

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

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

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