Question

the altitude of a right triangle is 7cm less than its base. if the hypotenuse is 13cm, find the other two sides

Answer 1

Let the base =   x cm

Given that The altitude of a right triangle is 7 cm less than its base

Altitude is =   x - 7 cm

Given that hypotenuse = 13cm


Applying Pythagoras theorem,   

 base2+ altitude2 = hypotenuse2

plug the values  we get  

     x2+ (  x – 7)2 = 132

    x2+   x2+ 49 – 14  x = 169

    2  x2 – 14  x + 48 – 169 = 0

    2  x2 – 14  x  – 120 = 0

Divide   by 2 to both side to simplify it

    x2 – 7  x  – 60 = 0

      x2 – 12  x + 5  x – 60 = 0

      x (  x – 12) + 5 (  x – 12) = 0

    (  x – 12)(  x + 5) = 0

    x – 12 = 0 or   x + 5 = 0

    x = 12 or   x = –5

length can not negative so that x can not equal to – 5

base     x = 12cm

 altitude = 12 – 7 = 5cm

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assume the base be (m) and height be (m - 7)

Hypotenuse = 13 cm

Situation :-

√m² + (m - 7)² = 13

m² + (m - 7)² = √13

m² + (m² - 14m + 49) = 169

2m² - 14m - 120 = 0

${\boxed{\sf\:{Take\;2\;common\;in\;LHS}}}$

m² - 7m - 60 = 0

★Factorise the middle term in LHS we get :-

m² - 12m + 5m - 60 = 0

m(m - 12) + 5(m - 12) = 0

(m - 12)(m + 5) = 0

m - 12 = 0

m = 12

m + 5 = 0

m = -5

★Here we have (-5) but sides of triangle can't be in negative.

★Hence :-

${\boxed{\sf\:{Base=12cm,Length\;of\;height=12 - 7 = 5cm}}}$

$\boxed{\begin{minipage}{14 cm} Additional Information \\ \\ \ A\; Quadratic\; Equation\;has\;three\;equal\;roots \\ \\ 1)Real\;and\;Distinct \\ \\ 2)Real\;and\;Coincident \\ \\ 3) Imaginary \\ \\ Note:-Third\; Imaginary\;is\;not\;taken\;in\;class\;10th \\ \\ If\;p(x)\;is\;a\; quadratic\; polynomial\;then\;p(x)=0\;is\;called\; Quadratic\; Polynomial \\ \\ General\;Formula=ax^2+bx+c=0 \\ \\ A polynomial\;whose\;degree\;will\;be\;2\;is\; considered\;as\; Quadratic\; Polynomial \\ \\ Rules\;for\; solving\; Quadratic\; Equations:- \\ \\ Put\;all\;the\;terms\;into\;RHS\;and\;make\it\;zero \\ \\ Substitute\;all\; factors\;equal\;to\;Zero\;Get\;a\;equal\; solution \end{minipage}}$