Chapter 9.4, Problem 24PSC

To determine

To calculate: The perimeter of a triangleDCB .

Answer to Problem 24PSC

The perimeter of a triangleDCB is 31.41.

Explanation of Solution

Given information:

Side $AD=20,$

Side $AB=10,$

Side $CD=12.$

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,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Side CA and DB can be calculated by using Pythagoras theorem.

In right angle triangle DCA , we get

  ${\left(DA\right)}^2={\left(CD\right)}^2+{\left(CA\right)}^2$

  $\begin{array}{l}{\left(20\right)}^2={\left(12\right)}^2+{\left(CA\right)}^2\\ 400=144+{\left(CA\right)}^2\\ {\left(CA\right)}^2=400-144\\ {\left(CA\right)}^2=256\\ \therefore CA=16\end{array}$

  $\begin{array}{l}CA-BA=CB\\ CB=16-10=6\end{array}$

In right angle triangle DCB , we get

  $\begin{array}{l}{\left(DB\right)}^2={\left(CD\right)}^2+{\left(CB\right)}^2\\ {\left(DB\right)}^2={\left(12\right)}^2+{\left(6\right)}^2\\ {\left(DB\right)}^2=144+36\\ {\left(DB\right)}^2=180\\ \therefore DB=\sqrt{180}\\ \therefore DB=13.41\end{array}$

  $\begin{array}{l}P=CD+DB+CB\\ P=12+13.41+6\\ \therefore P=31.41\end{array}$