Question

Plz anyone solve this​

Answer 1

Answer:

The answer is given in the photo..

Step-by-step explanation:

HOPE IT HELPS YOU....

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:a : b \: = \: c : d$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{ {c}^{2} }{ {d}^{2} } = \dfrac{ {a}^{2} + {c}^{2} }{ {b}^{2} + {d}^{2} }$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a : b \: = \: c : d$

Let assume that,

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d} = x$

So,

$\rm :\implies\:a = bx$

and

$\rm :\implies\:c = dx$

Now

Consider,

$\rm :\longmapsto\:\dfrac{ {c}^{2} }{ {d}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {(dx)}^{2} }{ {d}^{2} } \: \: \: \: \: \{ \because \: c = dx \}$

$\rm \:  =  \:  \: \dfrac{ {d}^{2} {x}^{2} }{ {d}^{2} }$

$\rm \:  =  \:  \: {x}^{2}$

$\bf\implies \:\:\dfrac{ {c}^{2} }{ {d}^{2} } = {x}^{2} - - - (1)$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {c}^{2} }{ {b}^{2} + {d}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {(bx)}^{2} + {(dx)}^{2} }{ {b}^{2} + {d}^{2} } \: \: \: \: \: \{ \because \: c = dx \: and \: a = bx\}$

$\rm \:  =  \:  \: \dfrac{ {b}^{2} {x}^{2} + {d}^{2} {x}^{2} }{ {b}^{2} + {d}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {x}^{2} ( {b}^{2} + {d}^{2} )}{ {b}^{2} + {d}^{2} }$

$\rm \:  =  \:  \: {x}^{2}$

$\bf\implies \:\:\dfrac{ {a}^{2} + {c}^{2} }{ {b}^{2} + {d}^{2} } = {x}^{2} - - - (2)$

Thus,

From equation (1) and equation (2), we concluded that

$\rm :\longmapsto\:\dfrac{ {c}^{2} }{ {d}^{2} } = \dfrac{ {a}^{2} + {c}^{2} }{ {b}^{2} + {d}^{2} }$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d}, \: then \:$

$\underbrace{\boxed{ \tt{\dfrac{a}{c} = \dfrac{b}{d} \:is \: called \: alternendo }}}$

$\underbrace{\boxed{ \tt{\dfrac{b}{a} = \dfrac{d}{c} \:is \: called \: invertendo }}}$

$\underbrace{\boxed{ \tt{\dfrac{a + b}{b} = \dfrac{c + d}{d} \:is \: called \: componendo}}}$

$\underbrace{\boxed{ \tt{\dfrac{a - b}{b} = \dfrac{c - d}{d} \:is \: called \: dividendo}}}$

$\underbrace{\boxed{ \tt{\dfrac{a+ b}{a - b} = \dfrac{c + d}{c - d} \:is \: called \:componendo \: and \: dividendo}}}$