Question

The perimeter of a triangle is 5x^2+3x-1 and two of its sides are 2x-5+7x^2 and 3x^2+4-2x^3+x.find the third side of the rectangle

Answer 1

Third side of the triangle is $\mathbf{2x^{3}-5x^{2}}$

Step-by-step explanation:

1. In question you have write find out third side of rectangle but it should be third side of triangle.

2. Let triangle is ABC. Let

   $\mathbf{Perimeter \ (P)= 5x^{2}+3x-1}$

   $\mathbf{AB=2x-5+7x^{2}}$

   CA = unknown

   $\mathbf{BC=3x^{2}+4-2x^{3}+x}$

3. We know that perimeter of triangle is

   P = AB + BC + CA

   Putting respective value in  above equation

  $\mathbf{5x^{2}+3x-1=+2x-5+7x^{2}+3x^{2}+4-2x^{3}+x+CA}$

  ⇒$\mathbf{5x^{2}+3x-1=-2x^{3}+10x^{2}+3x-1+CA}$

  So

  $\mathbf{CA=5x^{2}+3x-1+2x^{3}-10x^{2}-3x+1}$

  $\mathbf{CA=2x^{3}-5x^{2}}$ = This is third side of triangle.

 

 

Answer 2

Step-by-step explanation:

Third side of the triangle is \mathbf{2x^{3}-5x^{2}}2x

3

−5x

2

Step-by-step explanation:

1. In question you have write find out third side of rectangle but it should be third side of triangle.

2. Let triangle is ABC. Let

\mathbf{Perimeter \ (P)= 5x^{2}+3x-1}Perimeter (P)=5x

2

+3x−1

\mathbf{AB=2x-5+7x^{2}}AB=2x−5+7x

2

CA = unknown

\mathbf{BC=3x^{2}+4-2x^{3}+x}BC=3x

2

+4−2x

3

+x

3. We know that perimeter of triangle is

P = AB + BC + CA

Putting respective value in above equation

{5x^{2}+3x-1=+2x-5+7x^{2}+3x^{2}+4-2x^{3}+x+CA}5x

2

+3x−1=+2x−5+7x

2

+3x

2

+4−2x

3

+x+CA

⇒\{5x^{2}+3x-1=-2x^{3}+10x^{2}+3x-1+CA}5x

2

+3x−1=−2x

3

+10x

2

+3x−1+CA

So

{CA=5x^{2}+3x-1+2x^{3}-10x^{2}-3x+1}CA=5x

2

+3x−1+2x

3

−10x

2

−3x+1

CA=2x^{3}-5x^{2}}CA=2x

3

−5x

2

= This is third side of triangle.