Question

If a/b=c/d,show that (a+b):(c+d)=√(a^2+b^2):√(c^2+d^2)

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:a\:+\:b\:)\::\:(\:c\:+\:d\:)\:=\:\sqrt{a^2\:+\:b^2}\::\:\sqrt{c^2\:+\:d^2}}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\dfrac{a}{b}\ =\ \dfrac{c}{d}}$

We have to prove that,

$\displaystyle{\sf\:(\:a\:+\:b\:)\::\:(\:c\:+\:d\:) \:=\:\sqrt{a^2\:+\:b^2}\::\:\sqrt{c^2\:+\:d^2}}$

Now,

$\displaystyle{\sf\:\dfrac{a}{b}\ =\ \dfrac{c}{d}}$

Let,

$\displaystyle{\implies\sf\:\dfrac{a}{b}\ =\ \dfrac{c}{d}\:=\:k}$

$\displaystyle{\implies\sf\:a\:=\:bk\:\:\&\:\:c\:=\:dk}$

Now, we have to prove that,

$\displaystyle{\sf\:(\:a\:+\:b\:)\::\:(\:c\:+\:d\:)\:=\:\sqrt{a^2\:+\:b^2}\::\:\sqrt{c^2\:+\:d^2}}$

$\displaystyle{\implies\sf\:\dfrac{(\:a\:+\:b\:)}{(\:c\:+\:d\:)}\:=\:\dfrac{\sqrt{a^2\:+\:b^2}}{\sqrt{c^2\:+\:d^2}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{(\:a\:+\:b\:)}{(\:c\:+\:d\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{bk\:+\:b}{dk\:+\:d}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{b\:\cancel{(\:k\:+\:1\:)}}{d\:\cancel{(\:k\:+\:1\:)}}}$

$\displaystyle{\implies\pink{\sf\:LHS\:=\:\dfrac{b}{d}}}$

Now,

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sqrt{a^2\:+\:b^2}}{\sqrt{c^2\:+\:d^2}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{(\:bk\:)^2\:+\:b^2}{(\:dk\:)^2\:+\:d^2}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{b^2\:k^2\:+\:b^2}{d^2\:k^2\:+\:d^2}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{b^2\:\cancel{(\:k^2\:+\:1\:)}}{d^2\:\cancel{(\:k^2\:+\:1\:)}}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{b^2}{d^2}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{b\:\times\:b}{d\:\times\:d}}}$

$\displaystyle{\implies\pink{\sf\:RHS\:=\:\dfrac{b}{d}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!

Answer 2

$\bf \: Let \: \frac{a}{b} \: = \frac{c}{d} = k \: (say)$

$\bf= > a = bk,c = dk$

$\bf \: L.H.S = \frac{ \: a + b \: }{c + d}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \frac{ \: bk + b \: }{ck + c}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= \frac{ \: (bk + 1) \: }{(dk + 1)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \color{red}= > \frac{ \: b \: }{d}$

$\bf \: R.H.S = \frac{ \: \sqrt{ {a}^{2} + {b}^{2} } }{ \sqrt{ {c}^{2} + {d}^{2} } }$

$\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \frac{ \: \sqrt{ {(bk)}^{2} + {b}^{2} } }{ \sqrt{ {(dk)}^{2} + {b}^{2} \: } }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \frac{ \sqrt{ {b}^{2} {(k}^{2} + 1) } }{ \sqrt{ {d}^{2} {(k}^{2} + 1) } }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \frac{ \sqrt{ \: {b}^{2} } }{ \sqrt{ {d}^{2} } \: }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\bf \color{red}= > \frac{ \: b \: }{ \: d \: }$

$\bf \: Therefore, \color{red} L.H.S=R.H.S$