# Thread: Finding perimeter and area using polynomials

1. ## Finding perimeter and area using polynomials

One side of an isoceles right triangle is x + 2 and its hypotenuse is x +3. Find the perimeter and the area. p = 2l * 2w

How exactly could I solve this?

I would say:

(x+3)^2 = (x+2)^2 * h is this right just for a start?

If so does it mean (x+3)-(x+2) = b?

I believe the above idea is disappointing to you guys.

2. ## Re: Finding perimeter and area using polynomials

I see $\displaystyle P = (x+2)+(x+2)+(x+3)$ and $\displaystyle A = \frac{1}{2}\times (x+2)\times (x+2)$

3. ## Re: Finding perimeter and area using polynomials

Originally Posted by pickslides
I see $\displaystyle P = (x+2)+(x+2)+(x+3)$ and $\displaystyle A = \frac{1}{2}\times (x+2)\times (x+2)$
So both the length and the width are the same size?

4. ## Re: Finding perimeter and area using polynomials

I would call them base and height, but yes at you are told it is an 'isoceles' right triangle.

Draw one, it'll make more sense.

5. ## Re: Finding perimeter and area using polynomials

Originally Posted by pickslides
I would call them base and height, but yes at you are told it is an 'isoceles' right triangle.

Draw one, it'll make more sense.
Couldn't you also say the perimeter is 2(x+2)2(x+2)? [Sorry for all of the (bad) questions]

P =
Code:
x^3+7x^2+16x+12
a = (1/2)(x^2+4x+4)

6. ## Re: Finding perimeter and area using polynomials

$\displaystyle P = (x+2)+(x+2)+(x+3) = 2(x+2)+(x+3) = 2x+4+x+3$ group like terms.

$\displaystyle A = \frac{1}{2}\times (x+2)\times (x+2)= \frac{1}{2}\times (x+2)^2$

7. ## Re: Finding perimeter and area using polynomials

Originally Posted by vaironxxrd
One side of an isoceles right triangle is x + 2 and its hypotenuse is x +3.
Find the perimeter and the area.
p = 2l * 2w
That triangle is NOT isosceles: isosceles means 2 equal sides, so 2 sides each equal to x +2; then:
hypotenuse = SQRT[(x + 2)^2 + (x + 2)^2] : NOT x + 3

What you have is a right triangle with one leg = x + 2 and the hypotenuse = x + 3; so other leg
= SQRT[(x + 3)^2 - (x + 2)^2]
= SQRT(2x + 5)

So triangle has sides SQRT(2x + 5), x +2 and x + 3

You assumed: p = 2l * 2w
That's perimeter of a rectangle!
You are quite unaware of the basics. Are you attending math classes?

8. ## Re: Finding perimeter and area using polynomials

Originally Posted by Wilmer
That triangle is NOT isosceles: isosceles means 2 equal sides, so 2 sides each equal to x +2; then:
hypotenuse = SQRT[(x + 2)^2 + (x + 2)^2] : NOT x + 3

What you have is a right triangle with one leg = x + 2 and the hypotenuse = x + 3; so other leg
= SQRT[(x + 3)^2 - (x + 2)^2]
= SQRT(2x + 5)

So triangle has sides SQRT(2x + 5), x +2 and x + 3

You assumed: p = 2l * 2w
That's perimeter of a rectangle!
You are quite unaware of the basics. Are you attending math classes?
Yes I am sorry for the errors.

9. ## Re: Finding perimeter and area using polynomials

Originally Posted by vaironxxrd
Yes I am sorry for the errors.
Nobody's asking you to be "sorry"; just be careful, and make sure what you post makes sense...

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# area of a triangle using po

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