Chapter 8.1, Problem 41WE

a.

To determine

To fill: The value of $a^2$ and $b^2$ in the given figure.

Answer to Problem 41WE

  $a^2=c^2-b^2$ , $b^2=c^2-a^2$

Explanation of Solution

Given:

Figure

  

Concept used:

Pythagorean Theorem

In a right angled triangle, with perpendicular $a$ , base $b$ and hypotenuse $h$

  $a^2+b^2=h^2$

Calculation:

In the given triangle,

  $a^2=c^2-b^2$

And $b^2=c^2-a^2$

Conclusion:

Therefore, $a^2=c^2-b^2$ and $b^2=c^2-a^2$ .

b.

To determine

To add: The equations in part (a).

Answer to Problem 41WE

  $a^2+b^2=c^2$

Explanation of Solution

Given:

Figure

  

Formula used:

Pythagorean Theorem

In a right angled triangle, with perpendicular $a$ , base $b$ and hypotenuse $h$

  $a^2+b^2=h^2$

Calculation:

Since,

  $a^2=c^2-b^2$ and $b^2=c^2-a^2$ .

Adding both the equations,

  $\Rightarrow a^2+b^2=\left(c^2-b^2\right)+\left(c^2-a^2\right)$

  $=c^2-b^2+c^2-a^2$

  $=c^2+c^2-b^2-a^2$

[Rearranging the terms]

  $\begin{array}{l}=2c^2-\left(b^2+a^2\right)\\ =2c^2-\left(a^2+b^2\right)\end{array}$

  $\begin{array}{l}\Rightarrow a^2+b^2=2c^2-\left(a^2+b^2\right)\\ \Rightarrow a^2+b^2+\left(a^2+b^2\right)=2c^2\\ \Rightarrow 2\left(a^2+b^2\right)=2c^2\end{array}$

  $\Rightarrow a^2+b^2=c^2$

[Dividing by $2$ on both sides]

Conclusion:

Therefore, $a^2+b^2=c^2$ .