Question

if A the am between a and b, prove that (A-a) ^2+(A-b)^2=1/2(a-b) ^2​

Answer 1

Answer:-

Given:

A is the AM (Arithmetic mean) of a and b.

We know that,

AM of a and b = (a + b) / 2

So,

A = (a + b) / 2 -- equation (1).

We have to prove :

(A - a)² + (A - b)² = (1/2) * (a - b)²

$\implies \sf \: \bigg( \dfrac{a + b}{2} - a \bigg) ^{2} + \bigg( \dfrac{a + b}{2} - b \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2}$

[ From equation (1) ]

$\: \implies \sf \: \bigg( \dfrac{a + b - 2a}{2} \bigg) ^{2} + \bigg( \dfrac{a + b - 2b}{2} \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2} \\ \\ \\ \: \implies \sf \:\bigg( \dfrac{ b - a}{2} \bigg) ^{2} + \bigg( \dfrac{a - b}{2} \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2} \: \\ \\ \\ \implies \sf \:\frac{ {(b - a)}^{2} }{4} + \frac{ {(a - b)}^{2} }{4} = \frac{1}{2} \times {(a - b) }^{2} \\ \\ \\ \implies \sf \:\frac{ {(b - a)}^{2} + ( {a - b)}^{2} }{4} = \frac{1}{2} \times {(a - b)}^{2}$

• (a - b)² = a² + b² - 2ab.

$\: \implies \sf \:\frac{ {b}^{2} + {a}^{2} - 2ba + {a}^{2} + {b}^{2} - 2ab}{4} = \frac{1}{2} \times {(a - b)}^{2} \\ \\ \\ \implies \sf \:\frac{2 {b}^{2} + 2 {a}^{2} - 4ab}{4} = \frac{1}{2} \times {(a - b)}^{2} \\ \\\\ \implies \sf \:\frac{2( {b}^{2} + {a}^{2} - 2ab)}{4} = \frac{1}{2} \times {(a - b)}^{2} \\ \\\\ \implies \boxed{\sf \: \frac{(a - b) ^{2} }{2} = \frac{ {(a - b)}^{2} }{2}}$

Hence Proved.

Answer 2

Given :

• if A the am between a and b,

To Prove :

• prove that (A - a)² + (A - b)² = 1 /2(a - b)²

Solution :

$: \implies \sf \: \: \: \: \: \: {(A - a)}^{2} + {(A - b)}^{2} = \frac{1}{2} \times {(a - b)}^{2} \:$

$: \implies \sf \: \: \: \: \: \bigg( \dfrac{a + b}{2} - a \bigg) ^{2} + \bigg( \dfrac{a + b}{2} - b \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2} \\\\\\ : \implies \sf \: \: \: \: \: \bigg( \dfrac{a + b}{2} - a \bigg) ^{2} + \bigg( \dfrac{a + b}{2} - b \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2}$

$: \implies \sf \: \: \: \: \: \: \: \: \: \bigg( \dfrac{a + b - 2a}{2} \bigg) ^{2} + \bigg( \dfrac{a + b - 2b}{2} \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2} \\ \\ \\ \: :\implies \sf \:\: \: \: \: \: \: \: \: \:\bigg( \dfrac{ b - a}{2} \bigg) ^{2} + \bigg( \dfrac{a - b}{2} \bigg) ^{2} = \frac{1}{2} \times {(a - b)}^{2} \: \\ \\ \\ :\implies \sf \: \: \: \: \: \: \: \: \:\:\frac{ {(b - a)}^{2} }{4} + \frac{ {(a - b)}^{2} }{4} = \frac{1}{2} \times {(a - b) }^{2} \\ \\ \\ :\implies \sf \:\: \: \: \: \: \: \: \: \:\frac{ {(b - a)}^{2} + ( {a - b)}^{2} }{4} = \frac{1}{2} \times {(a - b)}^{2} \\ \\ \\ \: :\implies \sf \:\: \: \: \: \: \: \: \: \: \frac{ \cancel{2} \bigg( {(a - b)}^{2} \bigg)}{ \cancel{4} } = \frac{1}{2} \times {(a - b)}^{2} \\ \\ \\ \: :\implies \sf \:\: \: \: \: \: \: \: \: \: \: \frac{1}{2} \times {(a - b)}^{2} = \frac{1}{2} \times {(a - b)}^{2} \:$

Hence proved !!