Question

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that
(i) b² = h² + a² + x² -2ax
(ii) b² = a² +c² - 2ax

Answer 1

SOLUTION :  

(i) In ∆ADC,  

AC² = AD² + DC²

[By using pythagoras theorem]

b² = h² + (a -x)²

b² = h² + a² + x² -2ax

[(a-b)² = a² + b² -2ab]

(ii) In ∆ ADB,

AB² = BD² + AD²

[By using pythagoras theorem]

c² = x² + h² ….. . (1)

b² = h² + a² + x² -2ax

[From part (i)]

b² = a² + (h² + x²) -2ax

b² = a² + c² -2ax

[From eq 1]

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Answer 2

(i)In ∆ADC, using Pythagoras theorem

${AC}^{2} = {AD}^{2} + {CD}^{2} \\ {b}^{2} = {h}^{2} + {(a - x)}^{2} \\\boxed{\bold{ {b}^{2} = {h}^{2} + {a}^{2} + {x }^{2} - 2ax}} \: \: .............(1)$

(ii)In ∆ADB, using Pythagoras theorem
${AB}^{2} = {BD}^{2} + {AD}^{2} \\ {c}^{2} = {x}^{2} + {h}^{2} ............(2)$
from eq (1) and (2)
$\boxed{\bold{{b}^{2} = {a}^{2} + {c}^{2} - 2ax}}$