Do you hypotenuse of a triangle is 1 foot more than twice the length of the shorter leg the longer leg is 7 feet longer than the shorter leg find the dimensions of the triangle

We got this equation from reading that the "hypotenuse of a triangle is 1 foot more than twice the length of the shorter leg". I replaced the "hypotenuse of a triangle" with [tex]h[/tex], "is" with [tex]=[/tex], "1 foot more than" with [tex]1+[/tex] and finally "twice the length of the shorter leg" with [tex]2s[/tex].

We also have "longer leg is 7 feet longer than the shorter leg".

I'm going to replace "longer leg" with [tex]L[/tex].

I'm going to replace "is" with [tex]=[/tex].

I'm going to replace "7 feet longer than the shorter leg" with [tex]7+s[/tex].

So we have the equation [tex]L=7+s[/tex].

So we have a right triangle since something there is a side being referred to as the hypotenuse. We can use Pythagorean Theorem to find a relation between all these sides.

So by Pythagorean Theorem, we have: [tex]s^2+L^2=h^2[/tex].

Let's make some substitutions from above:

[tex]s^2+(7+s)^2=(1+2s)^2[/tex]

Let's expand the powers using:

[tex](a+b)^2=a^2+2ab+b^2[/tex]

Applying this now:

[tex]s^2+(49+2(7)s+s^2)=(1+2(1)(2s)+(2s)^2)[/tex]

[tex]s^2+49+14s+s^2=1+4s+4s^2[/tex]

Combine like terms on right hand side:

[tex]2s^2+49+14s=1+4s+4s^2[/tex]

Subtract everything on left hand side to get 0 on that side:

[tex]0=(1-49)+(4s-14s)+(4s^2-2s^2)[/tex]

Simplify:

[tex]0=(-48)+(-10s)+(2s^2)[/tex]

Reorder into standard form for a quadratic:

[tex]0=2s^2-10s-48[/tex]

Every term is even and therefore divisible by 2. I will divide both sides by 2:

[tex]0=s^2-5s-24[/tex]

I'm going to see if this is factoroable.

We need to see if we can come up with two numbers that multiply -24 and add up to be -5.

Those numbers are -8 and 3.

So the factored form is:

[tex]0=(s-8)(s+3)[/tex]

This implies that either [tex]s-8=0[/tex] os [tex]s+3=0[/tex].

The first equation can be solved by adding 8 on both sides: [tex]s=8[/tex].

The second equation can be solved by subtracting 3 on both sides: [tex]s=-3[/tex].

The only solution that makes sense for [tex]s[/tex] is 8 since it can't the shorter length cannot be a negative number.

[tex]s=8[/tex]

[tex]L=7+s=7+8=15[/tex]

[tex]h=1+2s=1+2(8)=1+16=17[/tex]

So the dimensions of the right triangle are:

Shorter leg: 8 units

Longer leg: 15 units

Hypotenuse: 17 units