Chapter 11.7, Problem 19PSC

a.

To determine

To calculate:The arearatio of triangleABCandtriangle EBF .

Answer to Problem 19PSC

Theratiois $4:1$ .

Explanation of Solution

Given information:

Triangle ABC is a right triangle.

E and F are midpoints.

D , H and G divide $\overline{AC}$ into four congruent segments.

Calculation:

  

All eight small triangles have the same area so the ratio is

  $\frac{A(ABC)}{A(EBF)}=\frac{8}{2}=4:1$

b.

To determine

To find: The area ratio of triangle ABCandtriangle GFC .

Answer to Problem 19PSC

The ratio is $8:1$ .

Explanation of Solution

Given information:

Triangle ABC is a right triangle.

E and F are midpoints.

D , H and G divide $\overline{AC}$ into four congruent segments.

Calculation:

  

All eight small triangles have the same area so the ratio is

  $\frac{A(ABC)}{A(GFC)}=\frac{8}{1}=8:1$

c.

To determine

To find: The area ratio of triangle ADE andtriangle GFC .

Answer to Problem 19PSC

The ratio is $1:1$ .

Explanation of Solution

Given information:

Triangle ABC is a right triangle.

E and F are midpoints.

D , H and G divide $\overline{AC}$ into four congruent segments.

Calculation:

  

All eight small triangles have the same area so the ratio is

  $\frac{A(ADE)}{A(GFC)}=\frac{1}{1}=1:1$

d.

To determine

To find: The area ratio of rectangle DEFGandtriangle ABC .

Answer to Problem 19PSC

The ratio is $1:2$ .

Explanation of Solution

Given information:

Triangle ABC is a right triangle.

E and F are midpoints.

D , H and G divide $\overline{AC}$ into four congruent segments.

Calculation:

  

All eight small triangles have the same area so the ratio is

  $\frac{A(\square DEFG)}{A(△ABC)}=\frac{4}{8}=1:2$

e.

To determine

To find: The perimeter of rectangleDEFG .

Answer to Problem 19PSC

The perimeter is $30$ .

Explanation of Solution

Given information:

Triangle ABC is a right triangle.

E and F are midpoints.

D , H and G divide $\overline{AC}$ into four congruent segments.

AC = 20.

Calculation:

  

AC = 20.

GD = 10.

  $\overline{DE}$ median to hypotenuse of ΔHEA.

DE = 5.

Perimeter is sum of all sides.

  $\begin{array}{l}P(\square DEFG)=(2\times GD)+(2\times DE)\\ P(\square DEFG)=(2\times 10)+(2\times 5)\\ P(\square DEFG)=(20)+(10)\\ P(\square DEFG)=30\end{array}$