Question

Can someone help me with this and explain me how to do it please :) ^

Answer 1

Answer:

x= 4

$perpendicular (p) = \sqrt{x} + 1$

$base(b) = 2 \sqrt{x}$

$hypotenuse(h) = 2 \sqrt{x} + 1$

According to the Pythagoras Theoram,

p²+b²= h²

${( \sqrt{x } + 1) }^{2} + {(2 \sqrt{x} )}^{2} = {(2 \sqrt{x} + 1)}^{2}$

=>x+2√x+1+4x = 4x +4√x +1

=> x = 4x - 4x +4√x- 2√x+ 1- 1

=>x= 2√x

=> x/√x = 2

=> √x = 2

Therefore, x= (√x)² = (2)² = 4

Answer 2

Answer:

4

Step-by-step explanation:

2√x, √x + 1, 2√x + 1 sides of triangle

√x= y

2y, y+1, 2y+1

(2y)²+(y+1)²= (2y+1)²

4y²+y²+2y+1= 4y²+4y+1

y²-2y=0

y=0 (discarded as side can't be zero) or y= 2

so √x=2 ⇒ x= 4