Question

The lengths of a right triangle are (2x-1) m, 4x m and (4x+1) m where x>0. Find the value of x and area of the triangle

Answer 1

Since, the triangle is right-angled.
So, applying Pythagoras Theorem, which states that -
$Hypotenuse^2 = Base^2 + Height^2$

Here sides are given as,
(2x-1), 4x and (4x+1)m,        where, x>0
=> 4x+1 is the longest side i.e. Hypotenuse.
Assume other sides of triangle to be Base or Height.

Let's apply the Pythagoras Theorem,
$(4x+1)^2 = (4x)^2 + (2x-1)^2$
We get,$16x^2 +8x + 1 = 16x^2 + 4x^2-4x+1$
On Simplifying,
[tex]4x^2-12x = 0 [/tex]
This is the quadratic equation (having highest degree of 2).
So, roots of this equation are either 0 or 3. (x*(x-3) = 0)
As the root can't be 0 as mentioned in the question. 
So only root is 3 i.e. x = 3.

Let's calculate the length of sides,
2x-1 = 2*3 - 1
        = 5m
 
4x = 4*3 
     = 12m

4x+1 = 4*3 + 1
         = 13m

So, sides are 5m, 12m and 13m.

Area of right angled triangle = $\frac{1}{2}* Base*Height$
                                             = $\frac{1}{2}*12*5$
                                             =  6*5
                                             = 30 $m^{2}$


  

Answer 2

Answer:

x = 3

Step-by-step explanation:

did it in rough ...but correct ans