Chapter 11.7, Problem 18PSC

a.

To determine

To calculate:The arearatio of triangleWYZandtriangle XYZ .

Answer to Problem 18PSC

Theratiois $1:1$ .

Explanation of Solution

Given information:

In trapezoidWXYZ ,

Side WX = 12,

Side ZY = 16.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

Triangle WYZ andtriangle XYZ have same base and height.

The area ratio is as follows:

  $\begin{array}{l}\frac{A(WYZ)}{A(XYZ)}=\frac{\left(\frac{1}{2}\times WZ\times 16\right)}{\left(\frac{1}{2}\times XY\times 16\right)}\\ XY=WZ\\ \frac{A(WYZ)}{A(XYZ)}=\frac{1}{1}=1\end{array}$

b.

To determine

To find: The area ratio of triangle WXZ andtriangle WXY .

Answer to Problem 18PSC

The ratio is $1:1$ .

Explanation of Solution

Given information:

In trapezoidWXYZ ,

Side WX = 12,

Side ZY = 16.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

Triangle WXZ andtriangle WXY have same base and height.

The area ratio is as follows:

  $\begin{array}{l}\frac{A(WXZ)}{A(WXY)}=\frac{\left(\frac{1}{2}\times WZ\times 12\right)}{\left(\frac{1}{2}\times XY\times 12\right)}\\ XY=WZ\\ \frac{A(WXZ)}{A(WXY)}=\frac{1}{1}=1\end{array}$

c.

To determine

To find: The area ratio of triangle WPZ andtriangle XPY .

Answer to Problem 18PSC

The ratio is $1:1$ .

Explanation of Solution

Given information:

In trapezoidWXYZ ,

Side WX = 12,

Side ZY = 16.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

  $\begin{array}{l}\Delta WPZ=\Delta WXZ-\Delta WPX\hfill \\ \Delta XPY=\Delta WXY-\Delta WPX\hfill \end{array}$

Since ΔWXY and ΔWXZ are a ratio of $1:1$ , the triangles left after subtracting ΔWPX from each are also in a ratio of $1:1$ .

d.

To determine

To find: The area ratio of triangle WPX andtriangle ZPY .

Answer to Problem 18PSC

The ratio is $9:16$ .

Explanation of Solution

Given information:

In trapezoidWXYZ ,

Side WX = 12,

Side ZY = 16.

Formula used:

The below property is used:

Area of triangle is proportional to square of its side.

Calculation:

  

Triangle WPX andtriangle ZPY are similar by AA similarity rule.

The area ratio is as follows:

  $\begin{array}{l}\frac{A(WPX)}{A(ZPY)}=\frac{{(WX)}^2}{{(ZY)}^2}\\ \frac{A(WPX)}{A(ZPY)}=\frac{{(12)}^2}{{(16)}^2}\\ \frac{A(WPX)}{A(ZPY)}=\frac{144}{256}\\ \frac{A(WPX)}{A(ZPY)}=\frac{9}{16}\end{array}$

e.

To determine

To find: The area ratio of triangle WPX andtriangle XPY .

Answer to Problem 18PSC

The ratio is $3:4$ .

Explanation of Solution

Given information:

In trapezoidWXYZ ,

Side WX = 12,

Side ZY = 16.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

Triangle WPX andtriangle ZPY are similar by AA similarity rule.

  $\begin{array}{l}\frac{A(WPX)}{A(ZPY)}=\frac{{(WX)}^2}{{(ZY)}^2}\\ \frac{A(WPX)}{A(ZPY)}=\frac{{(12)}^2}{{(16)}^2}\\ \frac{A(WPX)}{A(ZPY)}=\frac{144}{256}\\ \frac{A(WPX)}{A(ZPY)}=\frac{9}{16}\\ \frac{WP}{PY}=\frac{3}{4}\end{array}$

Triangle WPX andtriangle XPY have same heights.

The area ratio is as follows:

  $\begin{array}{l}\frac{A(△WPX)}{A(△XPY)}=\frac{\frac{1}{2}\times 3\times h}{\frac{1}{2}\times 4\times h}\\ \frac{A(△WPX)}{A(△XPY)}=\frac{3}{4}\end{array}$