Question

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

Answer 1

Let base = x cm
     height= (x - 7) cm 



By Pythagoras Theorem,

$x^2 + (x - 7)^2 = 13^2$

$x^2 + x^2 + 49 - 14x = 169 \\ \\ 2x^2 + 49 - 169 - 14x = 0 \\ \\ 2x^2 - 14x -120 = 0 \\ \\ 2(x^2 - 7x - 60) = 0 \\ \\ x^2 - 7x - 60 = 0 \\ \\ x^2 - (12 - 5)x - 60 = 0 \\ \\ x^2 - 12x + 5x - 60 =0 \\ \\ x(x - 12) + 5(x - 12) = 0 \\ \\ (x - 12)(x + 5) = 0$


By Zero product rule,

x - 12 = 0   and  x + 5 = 0 

x =12   and x  = - 5 

Taking positive value, x =12  cm


Base = x = 12 cm 

height = (x - 7) = 12 - 7 = 5cm



i hope this will help you


(-:

Answer 2

☜☆☞hey. dear friend!!!!! ☜☆☞


here is your answer...☞
_______

given that the altitude the right angle 7cm 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
2x² – 14x + 48 – 169 = 0
2x² – 14x – 120 = 0
Divide by 2 to both side to simplify it

=> x² – 7x – 60 = 0
=> x²– 12x + 5x – 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_____answer

altitude = 12 – 7 = 5cm_____answer


hope it will help you

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