Chapter 15, Problem 7CR

a.

To determine

To find: the length of DE of the given triangle.

Answer to Problem 7CR

  $DE=3$ .

Explanation of Solution

Given:

  

In the given triangle,

  $\begin{array}{l}BE\parallel CD,\\ AB=8,BC=4\\ AE=6,BE=9\end{array}$ .

Calculation:

According to the question, the dimensions of the triangles are given,

Now, apply Thale’s theorem

  $\begin{array}{l}\frac{AB}{BC}=\frac{AE}{DE}\\ \frac{8}{4}=\frac{6}{DE}\\ DE=\frac{6\times 4}{8}\\ DE=3\end{array}$

Therefore, the length of DE of the given triangle is $DE=3$ .

b.

To determine

To find: the length of CD of the given triangle.

Answer to Problem 7CR

  $CD=13.5$ .

Explanation of Solution

Given:

  

In the given triangle,

  $\begin{array}{l}BE\parallel CD,\\ AB=8,BC=4\\ AE=6,BE=9\end{array}$ .

Calculation:

According to the question, the dimensions of the triangles are given,

Now, apply Thale’s theorem

  $\begin{array}{l}\frac{AB}{AC}=\frac{AE}{AD}=\frac{BE}{CD}\\ \frac{8}{4+8}=\frac{9}{CD}\\ \frac{8}{12}=\frac{9}{CD}\\ CD=\frac{9\times 12}{8}\\ CD=13.5\end{array}$

Therefore, the length of DE of the given triangle is $CD=13.5$ .

c.

To determine

To check: whether the triangle ABE is a right triangle.

Answer to Problem 7CR

No.

Explanation of Solution

Given:

  

In the given triangle,

  $\begin{array}{l}BE\parallel CD,\\ AB=8,BC=4\\ AE=6,BE=9\end{array}$ .

Calculation:

According to converse of Pythagoras theorem, $\Delta ABE$ is a right triangle, if and only if when $A{B}^2+A{E}^2=B{E}^2$ . Therefore, Now, substitutes the given points in the above equation.

  $\begin{array}{l}A{B}^2+A{E}^2=B{E}^2\left\{\text{given}\right\}\\ 8^2+6^2=9^2\{\text{subtutes these value}}\\ 64+36=81\left\{\text{simplify}\right\}\\ 100\ne 81\end{array}$

Hence, the triangle ABE is not the right triangle.