Chapter 9.4, Problem 42E

a.

To determine

T0 draw the next right triangle with legs of $1$ unit long using the hypotenuse of the previous triangle as one leg.

Answer to Problem 42E

The diagram of next right triangles with legs of $1$ unit long using the hypotenuse of the previous triangle as one leg.

  

Explanation of Solution

Given: Right triangle with legs of $1$ unit long using the hypotenuse of the previous triangle as one leg.

Calculation:

Draw the right angles based on the previous right triangle with the help of formula:

  $\begin{array}{l}$ {\text{Hypotenuse}}^2={\text{ (side1)}}^2+{(side2)}^2$\Rightarrow \text{Hypotenuse}=\text{}\sqrt{{\text{1}}^2+{\text{ 1}}^2}=\sqrt{2}\end{array}$

Similarly, the next hypotenuse values could be found.

  

b.

To determine

To label the length of each hypotenuse in a right triangle.

Answer to Problem 42E

The length of each hypotenuse in a right triangle: $\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},\sqrt{9},\sqrt{10}$

  

Explanation of Solution

Given: Right triangle with legs of $1$ unit long using the hypotenuse of the previous triangle as one leg.

Calculation: Label the length of each hypotenuse in a right triangle by calculate the distance of hypotenuse equal to the addition of square of side1 and side2.

  $\begin{array}{l}$ {\text{Hypotenuse}}^2={\text{ (side1)}}^2+{(side2)}^2$\Rightarrow \text{Hypotenuse 1}=\text{}\sqrt{{\text{1}}^2+{\text{ 1}}^2}=\sqrt{2}\\ \Rightarrow \text{Hypotenuse 2}=\text{}\sqrt{{\text{1}}^2+{\sqrt{2}}^2}=\sqrt{3}\\ \Rightarrow \text{Hypotenuse 3}=\text{}\sqrt{{\text{1}}^2+{\sqrt{3}}^2}=\sqrt{4}\\ \Rightarrow \text{Hypotenuse 4}=\text{}\sqrt{{\text{1}}^2+{\sqrt{4}}^2}=\sqrt{5}\\ \Rightarrow \text{Hypotenuse 5}=\text{}\sqrt{{\text{1}}^2+{\sqrt{5}}^2}=\sqrt{6}\\ \Rightarrow \text{Hypotenuse 6}=\text{}\sqrt{{\text{1}}^2+{\sqrt{6}}^2}=\sqrt{7}\\ \Rightarrow \text{Hypotenuse 7}=\text{}\sqrt{{\text{1}}^2+{\sqrt{7}}^2}=\sqrt{8}\\ \Rightarrow \text{Hypotenuse 8}=\text{}\sqrt{{\text{1}}^2+{\sqrt{8}}^2}=\sqrt{9}\\ \Rightarrow \text{Hypotenuse 9}=\text{}\sqrt{{\text{1}}^2+{\sqrt{9}}^2}=\sqrt{10}\end{array}$

c.

To determine

To make a list of the hypotenuse lenghs in a triangle.

Answer to Problem 42E

The list of the hypotenuse lenghs in a triangle are: $\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},\sqrt{9},\sqrt{10}etc.$

Explanation of Solution

Given: Right triangle with legs of $1$ unit long using the hypotenuse of the previous triangle as one leg.

Calculation:

  $\begin{array}{l}$ {\text{Hypotenuse}}^2={\text{ (side1)}}^2+{(side2)}^2$\Rightarrow \text{Hypotenuse}=\text{}\sqrt{{\text{1}}^2+{\text{ 1}}^2}=\sqrt{2}\end{array}$

Similarly, the next hypotenuse values could be found.

The list of the hypotenuse lenghs in a triangle are: $\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},\sqrt{9},\sqrt{10}etc.$