If x , y , z are in continued proportion, prove that: ( x + y ) 2 ( y + z ) 2 = x z .

Given that, x , y , z are in continued proportion. Then, x y = y z y 2 = z x . . . ( i ) Now, by componendo, we get x + y y = y + z z and, by alternedo, we get x + y y + z = y z Now, squaring on both sides, we get ( x + y ) 2 ( y + z ) 2 = y 2 z 2 from equation ( i ) , ( x + y ) 2 ( y + z ) 2 = z x z 2 Hence, ( x + y ) 2 ( y + z ) 2 = x z

If x , y , z are in continued proportion, prove that: ( x + y ) 2 ( y + z ) 2 = x z .

Given that, x , y , z are in continued proportion. Then, x y = y z y 2 = z x . . . ( i ) Now, by componendo, we get x + y y = y + z z and, by alternedo, we get x + y y + z = y z Now, squaring on both sides, we get ( x + y ) 2 ( y + z ) 2 = y 2 z 2 from equation ( i ) , ( x + y ) 2 ( y + z ) 2 = z x z 2 Hence, ( x + y ) 2 ( y + z ) 2 = x z

Given x = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 - b 2 , use componendo and dividendo to prove that: b 2 = 2 a 2 x x 2 + 1 .

Given that, x = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 - b 2 Now, by componendo and dividendo, we get x + 1 x - 1 = a 2 + b 2 + a 2 - b 2 + a 2 + b 2 - a 2 - b 2 a 2 + b 2 + a 2 - b 2 - a 2 + b 2 - a 2 - b 2 x + 1 x - 1 = 2 a 2 + b 2 2 a 2 - b 2 Then, squaring on both sides, we get x 2 + 2 x + 1 x 2 - 2 x + 1 = a 2 + b 2 a 2 - b 2 Again, applying componendo and dividendo, we get x 2 + 2 x + 1 + x 2 - 2 x + 1 x 2 + 2 x + 1 - x 2 + 2 x - 1 = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 + b 2 2 ( x 2 - 1 ) 4 x = 2 a 2 2 b 2 x 2 - 1 2 x = a 2 b 2 Hence, b 2 = 2 a 2 x x 2 + 1

If x = a + 1 + a - 1 a + 1 - a - 1 , using properties of proportion show that: x 2 - 2 a x + 1 = 0 .

Given that, x = a + 1 + a - 1 a + 1 - a - 1 Now, applying componendo and dividendo, we get x + 1 x - 1 = a + 1 + a - 1 + a + 1 + a - 1 a + 1 + a - 1 - a + 1 - a - 1 x + 1 x - 1 = 2 a + 1 2 a - 1 x + 1 x - 1 = a + 1 a - 1 Then, squaring on both sides, we get x + 1 x - 1 2 = a + 1 a - 1 ( x + 1 ) 2 ( a - 1 ) = ( x - 1 ) 2 ( a + 1 ) ( x 2 + 2 x + 1 ) ( a - 1 ) = ( x 2 - 2 x + 1 ) ( a + 1 ) a x 2 + 2 a x + a - ( x 2 + 2 x + 1 ) = a x 2 - 2 a x + a + ( x 2 - 2 x + 1 ) 4 a x = 2 x 2 + 1 2 a x = x 2 + 1 x 2 - 2 a x + 1 = 0

If x a = y b = z c show that: x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c .

Given that, x a = y b = z c = m Let, x = a m , y = b m , z = c m L . H . S . = x 3 a 3 + y 3 b 3 + z 3 c 3 = ( a m ) 3 a 3 + ( b m ) 3 b 3 + ( c m ) 3 c 3 = a 3 m 3 a 3 + b 3 m 3 b 3 + c 3 m 3 c 3 = m 3 + m 3 + m 3 = 3 m 3 Now, R . H . S . = 3 x y z a b c = 3 ( a m ) ( b m ) ( c m ) a b c = 3 m 3 Hence, L . H . S . = R . H . S .

If b is the mean proportion between a and c , show that a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 .

Given that, b is mean proportional between a and c i.e., a b = b c . Let, a b = b c = x ⇒ b = c x , a = b x = ( c x ) x = c x 2 L . H . S . = a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = ( c x 2 ) 4 + ( c x 2 ) 2 ( c x ) 2 + ( c x ) 4 ( c x ) 4 + ( c x ) 2 c 2 + c 4 = c 4 x 8 + c 4 x 6 + c 4 x 4 c 4 x 4 + c 4 x 2 + c 4 = c 4 x 4 ( x 4 + x 2 + 1 ) c 4 ( x 4 + x 2 + 1 ) = x 4 Now, R . H . S . = a 2 c 2 = ( c x 2 ) 2 c 2 = c 2 x 4 c 2 = x 4 Thus, L . H . S . = R . H . S . Hence, a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2

E M B I B E

Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 18

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IMPORTANT

Earn 100

If $x, y, z$ are in continued proportion, prove that: $\frac{(x + y)^{2}}{(y + z)^{2}} = \frac{x}{z} .$

Important Questions on Ratio and Proportion (Including Properties and Uses)

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Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 19

Given $x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}},$ use componendo and dividendo to prove that: $b^{2} = \frac{2 a^{2} x}{x^{2} + 1} .$

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Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 20

If $\frac{x^{2} + y^{2}}{x^{2} - y^{2}} = 2 \frac{1}{8},$ find $\frac{x}{y}$ .

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Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 20

If $\frac{x^{2} + y^{2}}{x^{2} - y^{2}} = 2 \frac{1}{8},$ find $\frac{x^{3} + y^{3}}{x^{3} - y^{3}}$ .

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IMPORTANT

Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 21

Using componendo and dividendo, find the value of $x$ : $\frac{\sqrt{3 x + 4} + \sqrt{3 x + 5}}{\sqrt{3 x + 4} - \sqrt{3 x - 5}} = 9 .$

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Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 22

If $x = \frac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}$ , using properties of proportion show that: $x^{2} - 2 a x + 1 = 0$ .

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IMPORTANT

Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 23

Given

\frac{x^{3} + 12 x}{6 x^{2} + 8} = \frac{y^{3} + 27 y}{9 y^{2} + 27}

. Using componendo and dividendo, find

x : y .

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IMPORTANT

Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 24

If $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ show that:

$\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 x y z}{a b c} .$

HARD

10th ICSE

IMPORTANT

Concise Mathematics>Chapter 7 - Ratio and Proportion (Including Properties and Uses)>EXERCISE>Q 25

If $b$ is the mean proportion between $a$ and $c$ , show that $\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}} .$