Question

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The altitude of a right angle triangle is 17cm less than its base. If the hupotenuese is 25cm. Find the other two sides..................

Answer 1

$let \: the \: base \: be \: 4x$

$altitude \: be \: x - 17$

$In \: \triangle ABC, \angle B =90°$

$By \:Pythagoras \: Theorem$

${AC}^{2}={AB}^{2}+{BC}^{2}$

${25}^{2} = {(x - 17)}^{2} + {(x)}^{2}$

$625 = {x}^{2} + {(17)}^{2} - 2(x)(17) +{x}^{2}$

$625 = {x}^{2} + 289 - 34x + {x}^{2}$

$625 = 2 {x}^{2} + 289 - 34x$

$2 {x}^{2} - 34x + 289 - 625 = 0$

$2 {x}^{2} - 34x - 336 = 0$

$this \: is \: in \: the \: form \: of \\ a {x}^{2} + bx + c = 0$

$a = 2 \\ b = - 34 \\ c = - 336$

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{ -( - 34)± \sqrt{ {( - 34)}^{2} - 4(2) (- 366)} }{2(2)}$

$x = \frac{34± \sqrt{1156 + 2688} }{4}$

$x = \frac{34± \sqrt{3844} }{4}$

$x = \frac{34±62}{4}$

$x = \frac{34 + 62}{4} \: \: or \: \: x = \frac{34 - 62}{4}$

$x = \frac{96}{4} \: \: or \: \: x = \frac{ - 32}{4}$

${\boxed {\boxed {x = 24 \: \: or \: \: x = - 8}}}$

Answer 2

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The altitude of a right angle triangle is 17cm less than its base. If the hupotenuese is 25cm. Find the other two sides

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$let \: the \: base \: be \: 4x$

$altitude \: be \: x - 17$

$In \: \triangle ABC, \angle B =90°$

$By \:Pythagoras \: Theorem$

${AC}^{2}={AB}^{2}+{BC}^{2}$

${25}^{2} = {(x - 17)}^{2} + {(x)}^{2}$

$625 = {x}^{2} + {(17)}^{2} - 2(x)(17) +{x}^{2}$

$625 = {x}^{2} + 289 - 34x + {x}^{2}$

$625 = 2 {x}^{2} + 289 - 34x$

$2 {x}^{2} - 34x + 289 - 625 = 0$

$2 {x}^{2} - 34x - 336 = 0$

$this \: is \: in \: the \: form \: of \\ a {x}^{2} + bx + c = 0$

$a = 2 \\ b = - 34 \\ c = - 336$

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{ -( - 34)± \sqrt{ {( - 34)}^{2} - 4(2) (- 366)} }{2(2)}$

$x = \frac{34± \sqrt{1156 + 2688} }{4}$

$x = \frac{34± \sqrt{3844} }{4}$

$x = \frac{34±62}{4}$

$x = \frac{34 + 62}{4} \: \: or \: \: x = \frac{34 - 62}{4}$

$x = \frac{96}{4} \: \: or \: \: x = \frac{ - 32}{4}$

${\boxed {\boxed {x = 24 \: \: or \: \: x = - 8}}}$

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