Question

Systems of equations word problem?

the perimeter of a right triangle is 48 inches. It's area is 96 square inches. Find the 3 sides. I know the answer is 16,12,20. I tried both elimination and substitution but I can't get the set-up. I did get a partial answer but could not follow it. I have set up ab/2 = 96 for equation 1 and 48 = a + b + c for equation 2. thanks.

Answers

Hi Kyra Cook

There are 3 sides to the triangle which we need to find

So

we need 3 simultaneous equations to solve

a+b+c = 48....eqn1

1/2 ab = 96....eqn2

a^2+b^2 = c^2....eqn3

from eqn 1

a+b = 48-c

from eqn2

ab = 192

Now

(a+b)^2 = a^2+ b^2 +2ab

(48-c)^2 = c^2 + 384

48^2 -96c + c^2 = c^2 +384

96c = 48^2 - 384

c = 24 - 4 = 20

So

a+b = 48 - 20 = 28 eqn1

and ab = 192...eqn2

Hence

a = 16 and b =12

The 3 sides are

12, 16 and 20

#1

a+b+c=48

ab=192

a^2+b^2=c^2

a=192/b

substitute

192+b^2+cb/b=48

1).192+b^2+cb=48b

substitute

2).36864+b^4/b^2=c^2

now you have your 2 equations and 2 variables

#2

You forgot

a^2+b^2=c^2, it should be

a+b+c=48-----------(1)

ab/2=96--------------(2)

a^2+b^2=c^2-------(3)

Putting (1),(2) in (3), get

(a+b)^2-2ab=c^2=>

(48-c)^2-384=c^2=>

c=20

Then

a+b=48-20=>

a+b=28=>

b=28-a & from (2), get

a(28-a)=192=>

a^2-28a+192=0=>

a=16 or 12

a=16=>b=28-16=12

a=12=>b=28-12=16

So, a=16,b=12,c=20 or a=12, b=16,c=20.

#3

because there are three unknowns, you need three equations to solve it.

1. 48=x+y+z

2. 96=.5xy

3. x^2+y^2=z^2 (its a right triangle)

this actually takes awhile to solve..

first simplify to a system of two equations

z=sqrt(x^2+y^2) from 3.

plug this into 1.

4. 48=x+y+sqrt(x^2+y^2)

5. 96=.5xy

then solve for y in 5. and plug into 4.

y=192/x

48=x+192/x+sqrt(x^2+(192/x)^2)

48=x+192/x+sqrt(x^2+36864/x^2) (multiply everything by x)

48x=x^2+192+sqrt(x^4+36864)

(-x^2+48x-192)^2=x^4+36864

x^4-96x^3+2688x^2-18432x+36864 = x^4+36864

-96x^3+2688x-18432x=0 (divide everything by -96x)

x^2-28x+192=0 (factor)

(x-16)(x-12)=0

x=16,12

plug these into equation 2. from before and you get (16,12) and (12,16)

then plug all of these into equation 3. and you get 20 both times

make it a good day

#4