Question

The sides of a triangle are x, x+1, 2x-1 and it's area is 2x root10. Find x

Answer 1

Answer:

Given that,

Sides of traingle=x,x+1,2x-1

Area=2x√10

S=a+b+c÷2

=(x+x+1+2x-1)÷2

=4x÷2

=2x

To find:

Value of X

Area of traingle=$= \sqrt{s(s - a)(s - b)(s - c)}$

$2x \sqrt{10} = \sqrt{2x(2x - x)(2x - x + 1)(2x - (2x - 1)}$

$2x \sqrt{10} = \sqrt{2x(x)(x + 1)(2x - 2x + 1}$

$2x \sqrt{10} = \sqrt{2x(x)(x + 1)(1)}$

$2x \sqrt{10} = \sqrt{ {2x}^{2} (x + 1)}$

$2x \sqrt{10} = x\sqrt{2(x + 1)}$

On squaring both sides we get

2x²×10=x²×2(x+1)

2x²×10=x²×2(x+1)

2×10=2(x+1)

20=2(x+1)

10=x+1

10-1=x

$\huge{\underline{\underline{\tt{\blue{Hence\ X=9}}}}}$

Hence, x=9