Chapter 12.8, Problem 27E

(a)

To determine

To find: The coordinate proof of the given statement.

Answer to Problem 27E

The work done by a force in moving an object from $A$ and B is $13$ units.

Explanation of Solution

Given information:

The midpoint of the hypotenuse of a right angle is the same distance from each vertex of the triangle.

  

The above graph is a right angle triangle. From the graph it can be seen that the point $P\left(2,1\cdot 5\right)$ on the hypotenuse.

Now calculate the distance between the points $AP,BP,CP$ to prove that the hypotenuse of a right angle triangle is equidistant from each of its vertex.

The distance between two points can be calculated by the formula,

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

The length of $AP$ can be calculated by Substituting the value in the given formula,

  $\begin{array}{c}{d}_{BP}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{{\left(4-2\right)}^2+{\left(0-1\cdot 5\right)}^2}\end{array}$

    $\begin{array}{c}=2\cdot 5\end{array}$

  $\begin{array}{c}{d}_{AP}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{{\left(2-0\right)}^2+{\left(1\cdot 5-0\right)}^2}\end{array}$

    $\begin{array}{c}=2\cdot 5\end{array}$

Similarly, the length of $BP$ and $CP$ can be calculated as,

  $\begin{array}{c}{d}_{BP}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{{\left(4-2\right)}^2+{\left(0-1\cdot 5\right)}^2}\end{array}$

    $\begin{array}{c}=2\cdot 5\end{array}$

And,

  $\begin{array}{c}{d}_{CP}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{{\left(2-0\right)}^2+{\left(1\cdot 5-3\right)}^2}\end{array}$

    $\begin{array}{c}=2\cdot 5\end{array}$

From the above conclusion it can be clearly seen that the values of $AP,BP,CP=2\cdot 5$ .

Therefore, the midpoint of a hypotenuse of right angle triangle is equidistant from each of its vertex.

(b)

To determine

To find:The coordinate proof of the statement.

Answer to Problem 27E

The triangle $ACD$ is a right angle triangle with length equal to $2\sqrt{2}\text{units}$ and the hypotenuse is equal to $4\text{units}$ .

Explanation of Solution

Given information:

Any two congruent right isosceles triangles can be combined to form a single isosceles triangle.

  

In triangle $ABD$ and $BDC$ ,

  $\begin{array}{l}<ABD\cong <CBD\text{}\text{}\left(\therefore {90}^{\circ }\right)\\ <ADB\cong <CDB\left(\therefore \text{corresponding}\text{angles}\right)\\ BD=BD\left(\text{common}\right)\end{array}$

Hence, the triangles are congruent.

The length of both the triangles is $2\text{units}$ and hypotenuse is equal to $2\sqrt{2}\text{units}$ .

Thus, both these combine to form a single isosceles triangle.

Therefore, the triangle $ACD$ is a right angle triangle with length equal to $2\sqrt{2}\text{units}$ and the hypotenuse is equal to $4\text{units}$ .