Chapter 11, Problem 15RP

a.

To determine

To find: The area ratio of triangle LMN and triangle XYZ .

Answer to Problem 15RP

The area ratio is $5:8$ .

Explanation of Solution

Given information:

In ΔLMN ,

Side LO = 3 ,

Side MN = 5 .

In ΔXYZ ,

Side XK = 2 ,

Side YZ = 12 .

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:

  $\begin{array}{l}\frac{A(△LMN)}{A(△XYZ)}=\frac{\left(\frac{1}{2}\times MN\times LO\right)}{\left(\frac{1}{2}\times YZ\times XK\right)}\\ \frac{A(△LMN)}{A(△XYZ)}=\frac{\left(\frac{1}{2}\times 5\times 3\right)}{\left(\frac{1}{2}\times 12\times 2\right)}\\ \frac{A(△LMN)}{A(△XYZ)}=\frac{7.5}{12}=\frac{5}{8}\end{array}$

b.

To determine

To calculate: The area ratio of triangle ABC and triangle DEF .

Answer to Problem 15RP

The area ratio is $9:16$ .

Explanation of Solution

Given information:

In ΔABC ,

Side AB = 18 ,

Side BC = 12 .

In ΔDEF ,

Side DE = 24 ,

Side DF = 14 .

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.

  $\begin{array}{l}\frac{A(△ABC)}{A(△DEF)}=\frac{{(AB)}^2}{{(DE)}^2}\\ \frac{A(△ABC)}{A(△DEF)}=\frac{{(18)}^2}{{(24)}^2}\\ \frac{A(△ABC)}{A(△DEF)}={\left(\frac{3}{4}\right)}^2\\ \frac{A(△ABC)}{A(△DEF)}=\frac{9}{16}\end{array}$