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The sides of a triangle are x , x + 1 and 2 x -1. It’s area is x √10 square units. Find the value of x. ​
The sides of a triangle are x , x + 1 and 2 x -1. It’s area is x √10 square units. Find the value of x. ​

```
3 months ago

Anand Kumar Pandey
2937 Points
```							Dear Student,Here is solution:s=(x+x+1+2x-1)/2=4x/2=2xNow Area=√[s(s-x){s-(x+1)}{s-(2x-1)}]=√[2x(2x-x){2x-(x+1)}{2x-(2x-1)}]=√[2x^2(2x-x-1)(2x-2x+1)]=√[2x^2(x-1)×1]=√(2x^3–2x^2)According to question√(2x^3–2x^2)=x√10Or, {√(2x^3–2x^2)}^2={x√10}^2Or, 2x^3–2x^2=10x^2Or, 2x^3–10x^2–2x^2=0Or,2x^3–12x^2=0Or, 2x^2(x-6)=0Either 2x^2=0Or, x=0/OR x-6=0Or,x=6But x can not be 0Therefore x=6Thanks
```
3 months ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions