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

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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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Answer 2

Answer:

Answer in attachment dear..

Chandrasekhara Venkata Raman was an Indian physicist known mainly for his work in the field of light scattering. With his student K. S. Krishnan, he discovered that when light traverses a transparent material, some of the deflected light changes wavelength and amplitude