Question

34
The sides of the triangle are x cm, x + 1 cm, and 2x-1 cm. Its area is x root of 10 square cm. Find the value
of x

Answer 1

Solution :

We are given with the lengths of all sides of a triangle and the area.

Therefore we need to solve this problem using heron's formula

By Heron's Formula

Area of the triangle A = √[ s(s - a)(s - b)(s - c) ]

Where a, b, c are lengths of the sides and s is the semi-perimeter and s = (a + b + c)/2

Now, let't start solving

Lengths of sides of the triangle :

• a = x cm
• b = (x + 1) cm
• c = (2x - 1) cm

$\text{Semi-perimeter of the triangle s } = \dfrac{a + b + c}{2}$

$\implies s = \dfrac{x + (x + 1) + (2x - 1)}{2}$

$\implies s = \dfrac{x + x + 1+ 2x - 1}{2}$

$\implies s = \dfrac{4x}{2}$

$\implies \boxed{ s = 2x}$

$\text{Area of the triangle A } = \sqrt{s(s - a)(s - b)(s - c)}$

Substituting the values

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

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

$\implies A = \sqrt{2x^{2} (x - 1)( 1)}$

$\implies A = \sqrt{2x^{3} - 2 {x}^{2} }$

Given : A = x√10 cm²

$\implies \sqrt{2x^{3} - 2 {x}^{2} } = x \sqrt{10}$

Squaring on both sides

$\implies 2x^{3} - 2 {x}^{2} = 10 {x}^{2}$

$\implies 2x^{2}(x -1) = 10 {x}^{2}$

$\implies x -1 = \dfrac{10 {x}^{2} }{2 {x}^{2} }$

$\implies x -1 = 5$

$\implies x = 5 + 1$

$\implies \boxed{x = 6}$

Hence, the value of x is 6 cm.

Answer 2

$\bold{Given}\begin{cases}\underline{\footnotesize\sf{\star\:\:Sides\:of\:the\:Triangle}}\\\sf{side_1=x\:cm}\\\sf{side_2=(x+1)\:cm}\\\sf{Side_3=(2x-1)\:cm}\\\sf{Area=x\sqrt{10}\:cm^{2}}\\ \sf{\footnotesize\sf{Find\:the\:value\:of\:x}}\end{cases}$

Formula used :-----

• By heron's Formula Area of ∆ = √s(s-a)(s-b)(s-c) where s is semi-perimeter of ∆ ..

Solution :-----

First we need to Find semi-perimeter of the ∆ .

semi-perimter of ∆ with sides as a, b and c is =

$s = \frac{a + b + c}{2}$

Putting values we get,

$\star \: s = \frac{x + x + 1 + 2x - 1}{2} \\ \\ \star \: s = \frac{4x}{2} \\ \\ \star \: s = 2x$

Now, putting this in above formula we get,

(s-a) = (2x-x) = x

(s-b) = [2x-(x+1)] = x-1

(s-c) = [2x-(2x-1)] = 1

So, Area of ∆ will be =

$\star \: area = \sqrt{2x \times x \times (x - 1) \times 1} \\ \\ \star \: area = \sqrt{2x^{2}(x - 1) }$

Now,

Given Area of ∆ is = x√10

so,

$\sqrt{2x^{2}(x - 1) } = x \sqrt{10} \\ \\ squaring \: both \: sides \: we \: get \\ \\ \implies \: \cancel{2 {x}^{2}} (x - 1) = \cancel{10 {x}^{2}} \\ \\ \implies \: x - 1 = 5 \\ \\ \implies \: x = 5 + 1 = 6$

Hence, Value of x is $\red{\bold{6}}$

(Hope it Helps you)