Question

The hypotenuse of an isosceles right triangle is 6cm longer than either of its legs. Note that an Isosceles right triangle is a right triangle whose legs are the same length, find the exact length of its legs and it’s hypotenuse

Answer 1

The length of its legs = 6 cm

The hypotenuse = 12 cm

What is the exact length of its legs and it’s hypotenuse?

Let

Each leg of the isolation right triangle = x

Hypotenuse = x + 6

Hypotenuse² = Opposite² + Adjacent²

(x + 6)² = x² + x²

(x + 6)² = 2x²

Find the square root of both sides

x + 6 = √2x²

x + 6 = 2x

Subtract x from both sides

6 = 2x - x

6 = x

So,

x = 6 cm

Therefore,

Each leg of the isosceles right triangle is 6 cm

Hypotenuse = x + 6

= 6 + 6

= 12 cm

Answer 2

We know by the pythagorean theorem that

We know that the length of the hypotenuse squared will be equal to the sum of the legs squared. The problem says that the legs have the exact same length and the hypotenuse is 6cm longer, so we can write

Where "a" is the leg length, see that we can apply the pythagorean theorem here, and it will be

`a^2+a^2=(a+6)^2`

See that now c = a + 6, and b = a.

We can simplify that expression

`2a^2=(a+6)^2`

We know that

`(a+6)^2=a^2+12a+36`

Therefore our equation will be

`2a^2=a^2+12a+36`

Now we pass all the terms for one side and we will have a quadratic equation

`-a^2+12a+36=0`

We can use the formula for the quadratic equation and find out the solutions

`\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Using it

`\frac{-12\pm\sqrt[]{12^2-4\cdot(-1)\cdot36}}{2\cdot(-1)_{}}`

Now we can just do all the calculus

`\frac{-12\pm\sqrt[]{144^{}+144}}{-2_{}}=\frac{12\pm\sqrt[]{2\cdot12^2}}{2}=\frac{12\pm12\sqrt[]{2}}{2}`

Then the solution are

`\begin{cases}a_1=6+6\sqrt[]{2} \\ a_2=6-6\sqrt[]{6}\end{cases}`

Even though we have two solution, see that the second one is negative, and we can't have negative length! Then the length of its legs will be

`a=6+6\sqrt[]{6}`

And the hypotenuse will be a + 6, then

`h=6+6+6\sqrt[]{6}=12+\sqrt[]{6}`

Therefore, the legs and the hypotenuse length is

`\begin{gathered} l=6+\sqrt[]{6} \\ h=12+6\sqrt[]{6} \end{gathered}`

We can write it approximately as

`\begin{gathered} l=14.485\text{ cm} \\ h=20.485\text{ cm} \end{gathered}`

If we want a more rough approximation we can say it's

`\begin{gathered} l=14.5\text{ cm} \\ h=20.5\text{ cm} \end{gathered}`