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

${ \color{aqua} { \underbrace{ \underline{ \color{lime}{ \mathbb{\star SoLuTiOn\star }}}}}}$

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

$altitude \: be \: x - 17$

$In \: \triangle ABC, \angle B =90°$

$By \:Pythagoras \: Theorem$

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

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

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

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

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

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

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

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

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

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

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

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

$x = \frac{34± \sqrt{3844} }{4}$

$x = \frac{34±62}{4}$

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

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

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

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Answer 2

Answer:

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

{ \color{aqua} { \underbrace{ \underline{ \color{lime}{ \mathbb{\star SoLuTiOn\star }}}}}}

⋆SoLuTiOn⋆

let \: the \: base \: be \: 4xletthebasebe4x

altitude \: be \: x - 17altitudebex−17

In \: \triangle ABC, \angle B =90°In△ABC,∠B=90°

By \:Pythagoras \: TheoremByPythagorasTheorem

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

2

=AB

2

+BC

2

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

2

=(x−17)

2

+(x)

2

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

2

+(17)

2

−2(x)(17)+x

2

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

2

+289−34x+x

2

625 = 2 {x}^{2} + 289 - 34x625=2x

2

+289−34x

2 {x}^{2} - 34x + 289 - 625 = 02x

2

−34x+289−625=0

2 {x}^{2} - 34x - 336 = 02x

2

−34x−336=0

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

thisisintheformof

ax

2

+bx+c=0

\begin{lgathered}a = 2 \\ b = - 34 \\ c = - 336\end{lgathered}

a=2

b=−34

c=−336

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

2a

−b±

b

2

−4ac

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

2(2)

−(−34)±

(−34)

2

−4(2)(−366)

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

4

34±

1156+2688

x = \frac{34± \sqrt{3844} }{4}x=

4

34±

3844

x = \frac{34±62}{4}x=

4

34±62

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

4

34+62

orx=

4

34−62

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

4

96

orx=

4

−32

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

x=24orx=−8

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