Chapter 4, Problem 15PST

(a)

To determine

To graph: The type of transformation and show that it is a congruence transformation.

Explanation of Solution

Given information: The vertices of $\Delta ABC$ are $A\left(0,0\right),B\left(2a,b\right)\text{and}C\left(4a,0\right)$ .

Graph: The given vertices are plotted on coordinate grid.

  

(b)

To determine

To find: The expression for distance $AB$ .

Answer to Problem 15PST

The required expression is $AB=\sqrt{4a^2+b^2}$ .

Explanation of Solution

Given information: The vertices of $\Delta ABC$ are $A\left(0,0\right),B\left(2a,b\right)\text{and}C\left(4a,0\right)$ .

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: Take both points $A\left(0,0\right)\text{and}B\left(2a,b\right)$ ,

Applying distance formula to get expression for distance $AB$ ,

  $\begin{array}{l}AB=\sqrt{{\left(2a-0\right)}^2+{\left(b-0\right)}^2}\\ AB=\sqrt{4a^2+b^2}\end{array}$

(c)

To determine

To find: The expression for distance $BC$ .

Answer to Problem 15PST

The required expression is $BC=\sqrt{4a^2+b^2}$ .

Explanation of Solution

Given information: The vertices of $\Delta ABC$ are $A\left(0,0\right),B\left(2a,b\right)\text{and}C\left(4a,0\right)$ .

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: Take both points $B\left(2a,b\right)\text{and}C\left(4a,0\right)$ ,

Applying distance formula to get expression for distance $BC$ ,

  $\begin{array}{l}BC=\sqrt{{\left(4a-2a\right)}^2+{\left(0-b\right)}^2}\\ BC=\sqrt{4a^2+b^2}\end{array}$

(d)

To determine

To find: The conclusion about $\Delta ABC$ .

Answer to Problem 15PST

The $\Delta ABC$ is an equilateral triangle.

Explanation of Solution

Given information: The vertices of $\Delta ABC$ are $A\left(0,0\right),B\left(2a,b\right)\text{and}C\left(4a,0\right)$ .

Calculation: From part (b) and part (c), we get,

  $AB=\sqrt{4a^2+b^2}$ and $BC=\sqrt{4a^2+b^2}$

Hence, $AB=BC$

Therefore, the triangle $\Delta ABC$ is an equilateral triangle.