S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6C

  Exercise 6C

Question 1

Ans:

Given,

$\begin{aligned}\frac{x}{a} &=\frac{y}{b}=\frac{z}{c} \\\text { let } \frac{x}{a}=\frac{y}{b} &=\frac{z}{c}=k .\end{aligned}$

$\therefore x=a k, y=b k$ and $z=c k$

(i) $\frac{x^{3}}{a^{3}}-\frac{y^{3}}{b^{3}}+\frac{z^{3}}{c^{3}}=\frac{x y z}{a b c}$

$\frac{(a k)^{3}}{a^{3}}$ - $\frac{(b k)^{3}}{b^{3}}$$+\frac{(c k)^{3}}{c^{3}}$

$\frac{a k\times b k\times ck}{a b c}$

$\frac{a^{3} k^{3}}{a^{2}}-\frac{b^{3} k^{3}}{ b^{3}}+\frac{c^{3} k^{3}}{c^{3}}= k^{3}$

$k^{3}-k^{2}+ k^{3}=k^{3}$

$k^{3}=k^{3}$

Hence, proved: L.H.S=R.H.S 

(ii)  $\left(\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}$

$\left(\frac{a^{2} \times a^{2} x^{2}+b^{2} \times b^{2} k^{2}+c^{2} \times c^{2} k^{2}}{a^{3} \times a k+b^{3} \times b k+c^{3}+c k}\right)^{\frac{3}{2}}=\sqrt{\frac{a k \times b k \times c k}{a b c}}$

$\left(\frac{a^{4} k^{2}+b^{4} k^{2}+c^{4} k^{2}}{a^{4} k+b^{4} k+c^{4} k}\right)^{\frac{3}{2}}=$ $\sqrt{\frac{a b c k^{3}}{a b c}}$

$\left(\frac{k^{2}\left(a^{4}+b^{4}+c^{4}\right)}{k\left(a^{4}+b^{4}+c^{4}\right)}\right)^{\frac{3}{2}}$  $=\sqrt{K^{3}}$

$(K)^{\frac{3}{2}}=\left(K^{3}\right)^{\frac{1}{2}}$

$K^{\frac{3}{2}}=K^{\frac{3}{2}} .$ 

Hence prove : L.H.S =R.H.S

(iii)

$\begin{aligned}&\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c} \\&\frac{a k \times c k+a . c}{a k \times c k-a c}=\frac{b k \times c k+b c}{b k \times c k-b c} \\&\frac{a c k^{2}+a c}{a c k^{2}-a c}=\frac{b c k^{2}+b c}{b c k^{2}-b c}\end{aligned}$

$\frac{a c\left(k^{2}+1\right)}{a c\left(k^{2}-1\right)}=\frac{b c\left(k^{2}+1\right)}{b c\left(k^{2}-1\right)}$

$\frac{k^{2}+1}{k^{2}-1}=\frac{k^{2}+1}{k^{2}-1}$

L.H.S =R.H.S 

Hence proved

(iv) $a b\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{2+c}{c}\right)^{3}=27(x+a)(y+b)(2+c)$ $a b\left(\left(\frac{a k+a}{a}+\frac{b k+b}{b}+\frac{c k+c}{c}\right)^{3}=27(a k+a)(b k+b)(c k+c)\right.$

$a b c\left(\frac{a(k+1)}{a}+\frac{b(k+1)}{b}+\frac{c(k+1)}{c}\right)^{3}=$ $27 a(k+1) \cdot b(k+1)-c(k+1)$

$a b c(k+1+k+1+k+1)^{3}$ =$27 a b c(k+1)^{3}$

$a b c(3 k+3)^{3}=27 a b c(k+1)^{3}$

$a b c \cdot(3)^{3}(k+1)^{3}=27 a b c(k+1)^{3}$

$27 a b c(k+1)^{3}=27 a b c(k+1)^{3}$

L.H.S = R.H.S 

Hence proved 

(iv)$\left(\frac{3 x^{3}+5 y^{3}+7 z^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{3 \times a^{3} k^{3}+5 \times  b^{3} \times  k^{3}+7 \times c^{3} k^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{3 a^{3} k^{3}+5 b^{3} k^{3}+7 c^{3} k^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{k^{3}\left(3 a^{3}+5 b^{3}+7 b^{3}\right)}{\left(3 a^{3}+5 b^{3}+7 c^{3}\right)}\right)^{\frac{1}{3}}$

=$\left(k^{3}\right)^{\frac{1}{3}}$

=$k^{\left(3 \times \frac{1}{3}\right)}$

Hence proved 

Question 2

Ans:

Given,

$\text { let } \begin{aligned}\frac{a}{b}=\frac{c}{d} &=\frac{e}{f} \\\frac{a}{b}=\frac{c}{d} &=\frac{e}{f}=k\end{aligned}$

$\therefore a=b k, c=d k$ and $e=f k$

(i) $\frac{p a^{3}+q c^{3}+r c^{3}}{p b^{3}+q d^{3}+r f^{3}}=\frac{a c c}{b d f}$

$\frac{p b^{3} k^{3}+q d^{3} k^{3}+r j^{3} k^{3}}{p b^{3}+q d^{3}+rf^{3}}=$

$\frac{k^{3}\left(p b^{3}+q d^{3}+r f^{3}\right)}{\left(p b^{3}+q d^{3}+rf^{3}\right)}$

$k^{3}=k^{3} \text {. }$

Hence, prove.

(ii)$\sqrt{\frac{\dot{a}^{4}+c^{4}}{b^{4}+d^{4}}}=\frac{p a^{2}+q c^{2}}{p b^{2}+q d^{2}}$

$\sqrt{\frac{b^{4} k^{4}+d^{4} k^{4}}{b^{4}+d^{4}}}=\frac{p b^{2} k^{2}+q d^{2} k^{2}}{p b^{2}+q d^{2}}$

$\sqrt{\frac{k^{4}\left(b^{4}+d ^{4}\right)}{b^{4}+d^{4}}}=\frac{k^{2}\left(p b^{2}+q d^{2}\right)}{p b^{2}+q d^{2}}$

$\begin{aligned}&\sqrt{k^{4}}=k^{2} \\&\sqrt{k^{2} \times  k^{2}}=k^{2}

\\&k^{2}=k^{2}\end{aligned}$

Hence, proved

(iii) $\frac{2 a^{4} b^{2}+3 a^{2} c^{2}-5 c^{4}f}{2 b^{6}+3 b^{2} f^{2}-5f^{5}}$ $=\frac{a^{4}}{b^{4}}$

$\frac{2 b^{4} k^{4} b^{2}+3 b^{2} k^{2} f^{2} k^{2}-5 f^{4} k^{4} f}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=$ $\frac{b^{4} k^{4}}{b^{4}}$

$\frac{2 b^{6} k^{4}+3 b^{2} f^{2} k^{4}-5 f^{5} k^{4}}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=k^{4}$

$k^{4} \frac{\left(2 b^{6}+3 b^{2} f^{2}-5 f^{5}\right)}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=k^{4}$

$k^{4}=k^{4} .$

Hence, proved.

Question 3

Ans:

Given 

a, b, c are in continued proportion 

$\therefore \quad a: b:: b: c .$

$\frac{a}{b}=\frac{b}{c} .$

Let $\frac{a}{b}=\frac{b}{c}=k$

$b=c k, a=b k=c k-k=c k^{2}$

(i) $(a+b+c)(a-b+c)=a^{2}+b^{2}+c^{2} .$

$\left(c k^{2}+c k+c\right)\left(c k^{2}-ck+c)=\left(ck^{2}\right)^{2}+(c k)^{2}+c^{2}\right.$

$c\left(k^{2}+k+1\right) c\left(k^{2}-k+1\right)=c^{2} k^{4}+c^{2} k^{2}+c^{2}$

$c^{2}\left(k^{4}+k^{2}+1\right)=c^{2}\left(k^{4}+k^{2}+1\right)$

Hence proved 

(ii) $\frac{a^{2}+b^{2}}{b^{2}+c^{2}}=\frac{a}{c}$

$\frac{\left(c k^{2}\right)^{2}+(c k)^{2}}{(c k)^{2}+c^{2}}$ =$\frac{ ck^{2}}{c}$

$\frac{c^{2} k^{4}+c^{2} k^{2}}{c^{2} k^{2}+c^{2}}=\frac{c k^{2}}{c}$

$k^{2}=k^{2}$

Hence proved

(iii)  $\frac{a^{3}+b^{3}+c^{3}}{a^{2} b^{2} c^{2}}=\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$

$\frac{\left(c k^{2}\right)^{3}+(c k)^{3}+c^{3}}{\left(c k^{2}\right)^{2} \cdot(c k)^{2} \cdot c^{2}}=\frac{b^{3} \cdot c^{3}+a^{3} \cdot c^{3}+a^{3}-b^{3}}{a^{3} b^{3} c^{3}}$

$\frac{c^{3} k^{6}+c^{3} k^{3}+c^{3}}{c^{2} k^{4} \cdot c^{2} k^{2} \cdot c^{2}}=\frac{(c k)^{3} \cdot c^{3}+\left(c k^{2}\right)^{3}-c^{3}+\left(c k^{2}\right)^{3}-(c k)^{3}}{\left(c k^{2}\right)^{3}-(c k)^{3}-c^{3}}$

$\frac{c^{3}\left(x^{6}+k^{3}+1\right)}{c^{2} k^{6} \cdot c^{2} k^{2}-c^{2}}=\frac{c^{3} k^{3} \cdot c^{3}+c^{3} k^{6} \cdot c^{3}+c^{3} k^{6} \cdot c^{3} k^{3}}{c^{3} k^{6} \cdot c^{3} k^{3} \cdot c^{3}}$

$\frac{c^{3}\left(k^{6}+k^{3}+1\right)}{c^{6}.k^{6}}=\frac{c^{6}-k^{3}+c^{6} \cdot k^{6}+c^{6} k^{9}}{c^{3} k^{6} \cdot c^{3} k^{3} \cdot c^{3}}$

$\frac{k^{6}+k^{3}+1}{c^{3} k^{6}}$ = $\frac{1+k^{3}+k^{6}}{c^{3} k^{6}}$

$\frac{1+k^{3}+k^{6}}{c^{3} k^{6}}=\frac{1+k^{3}+1k^{6}}{c^{3} k^{6}}$

Hence proved 

(iv) $\left(4 a^{2}+7 a b+9 b^{2}\right):\left(4 b^{2}+7 b c+9 c^{2}\right)=a: c .$

$\frac{4 a^{2}+7 a b+9 b^{2}}{4 b^{2}+7 b c+9 c^{2}}=\frac{a}{c} .$

$\frac{4\left(c k^{2}\right)^{2}+7 c k^{2} \cdot c k+9(c k)^{2}}{4 \cdot(c k)^{2}+7 c k \cdot c+9 \cdot c^{2}}=$ $\frac{ck^{2}}{c}$

$\frac{4 c^{2} k^{4}+7 c^{2} k^{3}+9 c^{2} k^{2}}{4 c^{2} k^{2}+7 c^{2} k+9 c^{2}}=k^{2}$

$k^{2}=k^{2}$ 

Hence proved 

Question 4

Ans: Given,

$a, b, c, d$ are in continued proportion

$\therefore \quad \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$

let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$.

$c=d k, b=c k-d k \cdot k=d k^{2} a=b k=d k^{2} k=d k^{3}$

a, b ,c , d are in continued proportion 

$\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k$ 

$c=d k, b=c k \Rightarrow b=d k^{2} .$

$a=d k,=d k^{2} k=d k^{3}$

$(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-{d})^{2}$

(image to be added)

(ii) $\sqrt{\frac{a^{5}+b^{2} c^{2}+a^{3} c^{2}}{b^{4} c+d^{4}+b^{2} c d^{2}}}$ = $\frac{a}{d}$

$\sqrt{\frac{\left(d k^{3}\right)^{5}+\left(d k^{2}\right)^{2}(d k)^{2}+\left(d k^{3}\right)^{5}-(d k)^{2}}{\left(d k^{2}\right)^{4} \cdot d k+d^{4}+\left(d k^{2}\right)^{2} \cdot d k \cdot d^{2}}}=\frac{d k^{3}}{d}$

$\sqrt{\frac{d^{5} k^{15}+d^{2} k^{4} \cdot d^{2} k^{2}+d^{3} k^{9} \cdot d^{2} k^{2}}{d^{4} k^{8} \cdot d k+d^{4}+d^{2} \cdot k^{4} \cdot d k \cdot d^{2}}}=\frac{d k^{3}}{d}$

$\sqrt{\frac{d^{5} k^{15}+d^{4} \cdot k^{6}+d^{5} k^{11}}{d^{5} k^{9}+d^{4}+d^{5} k^{5}}}=k^{3}$

$\sqrt{k^{6}}=k^{3}$

$\sqrt{k^{3} \times k^{3}}=k^{3}$

$k^{3}=k^{3}$

Hence proved 

(iii) $\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}$

$\sqrt{d k^{3} \cdot d k^{2}}-\sqrt{d k^{2} \cdot d k}+\sqrt{d k \cdot d}=\sqrt{\left(d k^{3}-d k^{2}+d k\right)\left(d k^{2}-d k+d\right)}$

$\sqrt{d^{2} k^{5}}-\sqrt{d^{2} k^{3}}+\sqrt{d^{2} k}=\sqrt{d k\left(k^{2}-k+1\right) d\left(k^{2}-k+1\right)}$

$d k^{2} \sqrt{k}-d k \sqrt{k}+d \sqrt{k}=\sqrt{d^{2} k\left(k^{2}-k+1\right)^{2}}$

$d \sqrt{k}\left(k^{2}-k+1\right)=d \sqrt{k}\left(k^{2}-k+1\right) .$

Hence proved 

(iv) $\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^{3}+5 b^{3}}{5 a^{3}+7 b^{3}}$

$\frac{3 d k^{3}+5 d}{5 d k^{3}+7 d}=\frac{3\left(d k^{3}\right)^{3}+5\left(d k^{2}\right)^{3}}{5\left(d k^{3}\right)^{3}+7\left(d k^{2}\right)^{3}}$

$\frac{d\left(3 k^{3}+5\right)}{d\left(5 k^{3}+7\right)}=\frac{3 d^{3} k^{9}+5 d^{3} k^{6}}{5 d^{3} k^{9}+7 d^{3} k^{6}}$

$\frac{3 k^{3}+5}{5 k^{3}+7}=\frac{3 k^{3}+5}{5 k^{3}+7}$

Hence proved 

(v) $\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+d^{3}}=\frac{a}{d}$

$\frac{\left(d k^{3}\right)^{3}+\left(d k^{2}\right)^{3}+(d k)^{3}}{\left(d k^{2}\right)^{3}+(d k)^{3}+d^{3}}=$ $\frac{d k^{3}}{d}$

$\frac{d^{3} k^{9}+d^{3} k^{6}+d^{3} k^{3}}{d^{3} k^{6}+d^{3} k^{3}+d^{3}}=k^{3}$

$\frac{d^{3} k^{3}\left(k^{6}+k^{3}+1\right)}{d^{2}\left(k^{6}+k^{3}+1\right)}=k^{3}$

$k^{3}=k^{3}$

Hence, proved

(vi) (to be added)

Question 5

Ans: Given,

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

$\operatorname{let} \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k .$

$\therefore x=a k, y=b k$ and $z=c k .$

$\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{c(+a)(z-x)}=3$

$\frac{a-a k-b \cdot b k}{(a+b)(a k-b k)}+\frac{b b k-c \cdot c k}{(b+c)(b k-c k)}+\frac{c(c k-a \cdot a k}{(c+a)(c k-a k)}=3$

$\frac{a^{2} k-b^{2} k}{(a+b)(a k-b k)}+\frac{b^{2} k-c^{2} k}{(b+c)(b k-c k)}+\frac{c^{2} k-a^{2} k}{(c+a)(c k-9 k)}=3$

$\frac{k^{2}\left(a^{2}-b^{2}\right)}{(a+b) k(a-b)}$+ $\frac{K\left(b^{2}-c^{2}\right)}{(b+c)) k(b-c)}$

$\frac{(a+b)(a-b)}{(a+b)(a-b)}+\left(\frac{(b+b)(b-c)}{(b+x) \cdot(b-5)}+\frac{(c+a)(c-a)}{(c+a)(-c-a)}=3\right.$

$1+1+1=3$

3=3 Hence proved

Question 6

Ans: Given,

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{Z}{a+b-c}$

$=\frac{x+y+z}{b+c-a+c+a-b+a+b-c}$

$=\frac{x+y+z}{a+b+c}$

Hence, proved

Question 7

Ans: Given,

$\frac{a}{b+c}=\frac{b}{c+a} = \frac{c}{a+b}$

Let $\frac{a}{b+c}=\frac{b}{c+c}=\frac{c}{a+b}=k$

$a=k(b+c), b=k(c+a), c=k(a+b)$

$\therefore a(b-c)+b(c-a)+c(a-b)=0$

$k\left(b^{2}-c^{2}\right)+k\left(c^{2}-a^{2}\right)+k\left(a^{2}-b^{2}\right)=0$

$K \times 0=0$

0=0 

Hence proved 

Question 8

Ans: Given,

$\begin{aligned} a x &=b y=c z \\ \text { Leb } a x &=b y=c z=k . \\ x &=\frac{1}{4}, y=\frac{k}{b} \text {. and } z=\frac{k}{c .} \\ \therefore \frac{x^{2}}{y z} &+\frac{y^{2}}{2 x}+\frac{z^{2}}{x y}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}} \end{aligned}$

$\frac{\left(\frac{k}{a}\right)^{2}}{\frac{k}{3} \times \frac{k}{c}}+\frac{\left(\frac{k}{b}\right)^{2}}{\frac{k}{c} \times \frac{k}{a}}+\frac{\left(\frac{k}{c}\right)^{2}}{\frac{k}{a} \times \frac{k}{b}}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

$\frac{\frac{k^{2}}{a^{2}}}{\frac{k^{2}}{b c}}+\frac{\frac{k^{2}}{b^{2}}}{\frac{k^{2}}{ca^{2}}}+\frac{\frac{k^{2}}{c^{2}}}{\frac{k^{2}}{a b}}=\frac{b^{2}}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

$\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

HENCE PROVED