Chapter 11.7, Problem 10PSB

a.

To determine

To calculate:Theratio of area of shaded triangle ABCand whole triangle ADE .

Answer to Problem 10PSB

Theratiois $4:81$ .

Explanation of Solution

Given information:

SideAB = 2,

Side AD = 9.

Formula used:

The below property is used:

Area of triangle is proportional to square of its side.

Calculation:

  

Area of triangle is proportional to square of its sides.

  $\begin{array}{l}\frac{A(△ABC)}{A(△ADE)}=\frac{{(AB)}^2}{{(AD)}^2}\\ \frac{A(△ABC)}{A(△ADE)}=\frac{{(2)}^2}{{(9)}^2}\\ \frac{A(△ABC)}{A(△ADE)}=\frac{4}{81}\end{array}$

b.

To determine

To find: The ratio of area of shaded triangle XYZ and whole triangle XWZ .

Answer to Problem 10PSB

The ratio is $2:9$ .

Explanation of Solution

Given information:

Side WY = 7 ,

Side WZ = 9 .

Formula used:

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

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Calculation:

  

The ratio is as follows:

  $\begin{array}{l}\frac{A(△XYZ)}{A(△XWZ)}=\frac{(\frac{1}{2}\times 2h)}{(\frac{1}{2}\times 9h)}\\ \frac{A(△XYZ)}{A(△XWZ)}=\frac{h}{(\frac{9}{2}h)}\\ \frac{A(△XYZ)}{A(△XWZ)}=\frac{2}{9}\end{array}$

c.

To determine

To calculate: The ratio of area of shaded triangle OMN and whole triangle LMN .

Answer to Problem 10PSB

The ratio is $1:3$ .

Explanation of Solution

Given information:

Side OP = 1 ,

Side LP = 3.

Formula used:

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

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Calculation:

  

The ratio is as follows:

  $\begin{array}{l}\frac{A(△OMN)}{A(△LMN)}=\frac{(\frac{1}{2}\times 1b)}{(\frac{1}{2}\times 3b)}\\ \frac{A(△OMN)}{A(△LMN)}=\frac{b}{3b}\\ \frac{A(△OMN)}{A(△LMN)}=\frac{1}{3}\end{array}$

d.

To determine

To find: The ratio of area of shaded triangle OPN and whole triangle KMN .

Answer to Problem 10PSB

The ratio is $4:81$ .

Explanation of Solution

Given information:

Side ON = 15 ,

Side KN = 18.

Formula used:

The below property is used:

Area of triangle is proportional to square of its side.

Calculation:

  

ΔOPM is similar toΔMKN .

Corresponding sides of similar triangles are congruent.

  $MN\cong KN$

Area of triangle is proportional to square of its side.

  $\begin{array}{l}\frac{A(△OPN)}{A(△MKN)}=\frac{{(PN)}^2}{{(MN)}^2}\\ MN\cong KN\\ \frac{A(△OPN)}{A(△MKN)}=\frac{{(12)}^2}{{(18)}^2}\\ \frac{A(△OPN)}{A(△MKN)}=\frac{144}{324}\\ \frac{A(△OPN)}{A(△MKN)}=\frac{4}{9}\end{array}$