Question

The square of hypotenuse of a right angked triangle is80cm and one side is half the other side find the length of two sides

Answer 1

$<b>$Let in △ABC


=> let AB's length is x cm


=>BC's length is x/2 cm


Applying pythagoras theorem,


(AB)²+(BC)²=(AC)²


Given=> (AC)²=80


${x}^{2} + ( { \frac{x}{2} )}^{2} = 80 \\ \\ {x}^{2} + \frac{ {x}^{2} }{ {4}^{} } = 80 \\ \\ \frac{4 {x}^{2} + {x}^{2} }{4} = 80 \\ \\ \frac{5 {x}^{2} }{4} = 80 \\ \\ {x}^{2} = \frac{80 \times 4}{5} \\ \\ {x}^{2} = 64 \\ \\ x = \sqrt{64} \\ \\ x = \sqrt{8 \times 8}$

x= 8cm


Hence AB = x cm = 8cm 


BC = (x/2)cm = (8/2)cm = 4cm


Thanks


Have a colossal day ahead


$\huge\boxed{\textbf{Be Brainly}}$

Answer 2

Given that square of hypotenuse of a right angled triangle = 80

or h^2 = 80

one side is half the other.

Let the smaller side = x

bigger side = 2x

Using Pythagoras theorem,

(2x)^2 + x^2 = h^2

=> 4x^2 + x^2 = 80

=> 5x^2 = 80

=> x^2 = 80/5 = 16

=> x = √16

=>x = 4 cm

2x = 4 × 2 = 8cm

Hence, the sides are 4cm and 8cm.