Chapter 9.3, Problem 5PSA

a

To determine

To find:The value of a

Answer to Problem 5PSA

The value of $a=9$

Explanation of Solution

Given information:The following equations can be generated by applying Pythagoras in different triangles

  $c^2=27+a^2$ --------(1)

  ${(a+3)}^2=b^2+c^2$ -----------(2)

  $b^2=3^2+27$ ------(3)

Formula used:Linear equations have been solved

Calculation :

Solving the three equations,

  $\begin{array}{l}{b}^2=3^2+27=36\\ b=6\end{array}$

Putting this value in the equation (ii)

  $\begin{array}{l}$ {(a+3)}^2=b^2+c^2$$ {(a+3)}^2=36+c^2$$ \text{using the equation 1, we can write}$$ {\text{(}a+3\text{)}}^2=36+27+a^2$a^2+9+6a=36+27+a^2\\ a=9\end{array}$

b

To determine

To find:The value of ab

Answer to Problem 5PSA

The value of $ab=54$

Explanation of Solution

Given information:The following equations can be generated by applying Pythagoras in different triangles

  $c^2=27+a^2$

  ${(a+3)}^2=b^2+c^2$

  $b^2=3^2+27$

Formula used:Quadratic equations have been solved

Calculation:Hence, solving the three equations, we get

  $a=9$ , and

  $b=6$

Thus, we can say that

  $ab=9\times 6=54$

c

To determine

To find:The value of $a+b+c$

Answer to Problem 5PSA

The value of $a+b+c=15+6\sqrt{3}$

Explanation of Solution

Given information:The following equations can be generated by applying Pythagoras in different triangles

  $c^2=27+a^2$

  ${(a+3)}^2=b^2+c^2$

  $b^2=3^2+27$

Formula used:Quadratic equations have been solved

Calculation:Hence, solving the three equations, we get

  $a=9$

  $b=6$

  $c=6\sqrt{3}$