Question

The three sides of a triangle are consecutive integrals. Let these sides be$(x-1)cm,xcmand(x+1)cm.$ If the area of this triangle is $30\sqrt{2}c{m}^2,$ then the value of $(x^2+96)(x^2-100)$ is

• A.
  0
• B.
  1
• C.
  $\sqrt{2}$
• D.
  $\sqrt{3}$

Answer 1

The correct option is A

0
The sides of the triangle are given as $(x-1)cm,xcmand(x+1)cm.\therefore Perimeter=(x–1)+x+(x+1)=3xcm\Rightarrow Semi-perimeter,\left(s\right)=\frac{3x}{2}cm\therefore \text{Area of the triangle}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{3x}{2}\times [\frac{3x}{2}-(x-1\left)\right][\frac{3x}{2}-x][\frac{3x}{2}-(x+1\left)\right]}=\sqrt{\frac{3x}{2}\times (\frac{x}{2}+1)\times \left(\frac{x}{2}\right)\times (\frac{x}{2}-1)}=\frac{\sqrt{3}}{2}x\times \sqrt{\frac{x+2}{2}\times \frac{x-2}{2}}=\frac{\sqrt{3}}{4}x\times \sqrt{s^2-4}c{m}^2\Rightarrow 30\sqrt{2}=\frac{\sqrt{3}}{4}\times x^2\times \sqrt{s^2-4}(\because \text{Area of triangle =}30\sqrt{2}c{m}^2)$

Squaring both sides, we get

$\Rightarrow 1800=\frac{3}{16}{x}^2(x^2-4)\Rightarrow x^2(x^2–4)=9600\Rightarrow x^4–4x^2–9600=0\Rightarrow x^4+96x^2–100x^2–9600=0\Rightarrow x^2(x^2+96)–100(x^2+96)=0\Rightarrow (x^2+96)(x^2–100)=0$

Hence, the correct answer is option (a).