Chapter 9, Problem 8CR

a

To determine

To find:The measure of side BC

Answer to Problem 8CR

The measure of side $BC=20.25$

Explanation of Solution

Given information:The following information has been given

AB is parallel to DE

  $AD=10$

  $CD=8$

  $CE=9$

  $DE=6$

Calculation:We know that if AB is parallel to DE , then we have

Triangles CDE and CAB are similar.

We know that corresponding sides of similar triangles are in same ratio. Hence, we can say that

  $\frac{AC}{BC}=\frac{DC}{EC}$

Now, substituting the other values, we get

  $BC=20.25$

b

To determine

To find:The measure of side AB

Answer to Problem 8CR

The measure of side $AB=6.75$

Explanation of Solution

Given information:The following information has been given

AB is parallel to DE

  $AD=10$

  $CD=8$

  $CE=9$

  $DE=6$

Calculation:We know that if AB is parallel to DE, then we have

Triangles CDE and CAB are similar.

We know that corresponding sides of similar triangles are in same ratio. Hence, we can say that

  $\frac{DE}{AB}=\frac{DC}{EC}$

Now, substituting the other values, we get

  $AB=6.75$

c

To determine

To find:The type of triangle DEC

Answer to Problem 8CR

Triangle DEC is acute

Explanation of Solution

Given information:The following information has been given

AB is parallel to DE

  $AD=10$

  $CD=8$

  $CE=9$

  $DE=6$

We know that DEC can’t be obtuse. Hence it remains to check from acute and right triangles. Thus, we apply Pythagoras for right triangle, and we get

  $6^2+8^2\ne 9^2$

Thus, we know that it is not a right triangle.

Hence, DEC is an acute triangle