Chapter 6, Problem 12MPS

To determine

To Solve:

The given problem.

Answer to Problem 12MPS

15 units.

Explanation of Solution

Given:

Given statement is :

The lengts of the sides of a right triangle are consecutive multiples of 3.Find the length of the hypotenuse.

Calculation:

Let

a = length of one side

a + 3 = length of the other side

a + 6 = length of the hypotenuse [ since the hypotenuse is the longest side]

The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the squares of the sides of the right angles triangle.

So,

  $\begin{array}{l}{a}^2+{\left(a+3\right)}^2={\left(a+6\right)}^2\\ \Rightarrow a^2+a^2+2\cdot a\cdot 3+9=a^2+2\cdot a\cdot 6+\mathrm{36......}[{\left(x+y\right)}^2=x^2+2xy+y^2]\\ \Rightarrow 2a^2+6a+9=a^2+12a+36\\ \Rightarrow 2a^2-a^2+6a-12a+9-36=\mathrm{0.........}[\text{subtract (}{a}^2+12a+36)\text{ from each side]}\\ \Rightarrow a^2-6a-27=0\\ \Rightarrow a^2-9a+3a-27=\mathrm{0........}\left[\begin{array}{l}factor\text{ the middle term such that the sum of the factor is equal to }\\ \text{the middle term and product of the factors is equal to the product of the}\\ \text{first and the last term}\text{.}\end{array}\right]\\ \Rightarrow a(a-9)+3(a-9)=0\\ \Rightarrow (a+3)(a-9)=0\\ \Rightarrow a+3=0\text{ or }a-9=\mathrm{0.........}[ab=0\iff a=0orb=0]\\ \Rightarrow a=-3\text{ or }a=9\\ \mathrm{Sin}ce\text{ a is the length of the triangle , it cannot be negative}\text{.}\end{array}$

So,

a = 9 units

a + 3 =12 units

a + 6 = 15 units

So, the hypotenuse is 15 units.