Chapter 11.7, Problem 17PSB

To determine

To calculate:Theratio of area of triangle ACDandarea of triangle BCD .

Answer to Problem 17PSB

Theratiois $4:9$ .

Explanation of Solution

Given information:

SideCD = 6 ,

Side DB = 9 .

Formula used:

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

b = base of triangle

h = height of triangle

Calculation:

  

Altitude is mean proportional between segment of hypotenuse.

  $\begin{array}{l}\frac{AD}{6}=\frac{6}{9}\\ 9AD=36\\ AD=\frac{36}{9}=4\end{array}$

The ratio of two triangles is as follows:

  $\begin{array}{l}\frac{A(△ACD)}{A(△BCD)}=\frac{\frac{1}{2}\times AD\times CD}{\frac{1}{2}\times DB\times CD}\\ \frac{A(△ACD)}{A(△BCD)}=\frac{\frac{1}{2}\times 4\times 6}{\frac{1}{2}\times 9\times 6}\\ \frac{A(△ACD)}{A(△BCD)}=\frac{12}{27}\\ \frac{A(△ACD)}{A(△BCD)}=\frac{4}{9}\end{array}$