Chapter 9.4, Problem 30PSC

a.

To determine

To show:The square of the length of the second leg is equal to sum of the length of the first leg and the hypotenuse.

Explanation of Solution

Given information:

  $\begin{array}{l}oneleg=x,\hfill \\ hypotenuse=x+1,\hfill \\ secondleg=y\hfill \end{array}$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Proof:

  

In the above right angle triangle,

  ${\left(x+1\right)}^2={\left(x\right)}^2+{\left(y\right)}^2$

  $x^2+2x+1={\left(x\right)}^2+{\left(y\right)}^2$

  $\therefore y^2=2x+1$ .

  $x^2+2x+1={\left(x\right)}^2+{\left(y\right)}^2$ and $y^2=2x+1$ .

b.

To determine

To show:The statement“If a square of the length of the one of the legs of a right triangle is equal to the sum of the lengths of the other leg and the hypotenuse, then the lengths of the second leg and hypotenuse are consecutive integers” is not necessarily true.

Explanation of Solution

Given information:

A right angle triangle with legs $4,7\frac{1}{2},$ and hypotenuse $8\frac{1}{2}$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

Proof:

  

In above right angle triangle, we get

  $4^2+{(7\frac{1}{2})}^2={(8\frac{1}{2})}^2$

  $7\frac{1}{2}$ and $8\frac{1}{2}$ are not integers.

The statement is not necessarily true as sum can be also formed with the help of fractions.