Question

in a right-angled triangle ABC, right angled at B, AB = x/2 ,BC = x + 2 and AC = x + 3. Find
value of x.​

Answer 1

Answer:

Step-by-step explanation:

(x/2)² + (x + 2)² = (x + 3)²

x²/4 + x² + 4 + 4x = x² + 9 + 6x

x²/4 + x² - x² + 4 - 9 + 4x - 6x = 0

x²/4 - 5 - 2x = 0

(x² - 20 - 8x)/4 = 0

x² - 8x - 20 = 0

x² - 10x + 2x - 20 = 0

x(x - 10) + 2(x - 10) = 0

(x + 2)(x - 10) = 0

x + 2 = 0 ⠀⠀⠀⠀⠀or ⠀⠀⠀⠀⠀⠀⠀x - 10 = 0

x = -2 ⠀⠀⠀⠀⠀⠀⠀or ⠀⠀⠀⠀⠀⠀⠀x = 10

Hope it helps u...

Answer 2

Given: In a right-angled triangle ABC, right angled at B, AB = x/2 ,BC = x + 2 and AC = x + 3.

To find: The value of x.

Solution:  

The values of the sides of the triangle are given in terms of x. According to the question, AB forms the perpendicular, BC forms the base and AC forms the hypotenuse of the right-angled triangle. Now, using the Pythagoras theorem, the following equation is formed.

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

$(x+3)^{2} = (\frac{x}{2} )^{2} + (x+2)^{2}$

$x^{2} + 9 + 6x = \frac{x^{2} }{4} + x^{2} + 4 + 4x$

On rearranging, a quadratic equation is obtained as follows.

$x^{2} -8x -20 = 0$

On solving the quadratic equation, the value of x is found to be,

$x = 10$ or $x = -2$

Therefore, the value of x is 10 or -2.