Chapter 11.7, Problem 23WE

(a)

To determine

To Find: $\frac{Areaof\Delta DEF}{Areaof\Delta ABF}$

Answer to Problem 23WE

  $\frac{1}{9}$

Explanation of Solution

Given: Area of parallelogram.

  

In triangle $\Delta DEFand\Delta ABF$

  $\begin{array}{l}\angle AFB=\angle DFE\\ \angle FAB=\angle FDE(\text{corresponding angles)}\\ \text{thus }\\ \Delta DEF\sim \Delta AFB\end{array}$

Now

  $\begin{array}{l}\frac{Areaof\Delta DEF}{Areaof\Delta ABF}=\frac{\frac{1}{2}\cdot b_1\cdot h_1}{\frac{1}{2}\cdot b_2\cdot h_2}\\ \frac{b_1}{b_2}=\frac{h_1}{h_2}=\frac{4}{12}=\frac{1}{3}\\ i.e.\\ \frac{Areaof\Delta DEF}{Areaof\Delta ABF}=\frac{1}{3}\cdot \frac{1}{3}=\frac{1}{9}\\ \frac{Areaof\Delta DEF}{Areaof\Delta ABF}=\frac{1}{9}\end{array}$

Hence,

  $\frac{Areaof\Delta DEF}{Areaof\Delta ABF}=\frac{1}{9}$

(b)

To determine

To Find: $\frac{Areaof\Delta DEF}{Areaof\Delta CEB}$

Answer to Problem 23WE

  $\frac{1}{4}$

Explanation of Solution

Given: Area of parallelogram.

  

In triangle $\Delta DEFand\Delta CEB$

  $\begin{array}{l}\angle FDE=\angle BEC\\ \angle DFE=\angle CBE(alternating\text{angles)}\\ \text{thus }\\ \Delta DEF\sim \Delta CEB\end{array}$

Now

  $\begin{array}{l}\frac{Areaof\Delta DEF}{Areaof\Delta CBE}=\frac{\frac{1}{2}\cdot b_1\cdot h_1}{\frac{1}{2}\cdot b_2\cdot h_2}\\ \frac{b_1}{b_2}=\frac{h_1}{h_2}=\frac{4}{8}=\frac{1}{2}\\ i.e.\\ \frac{Areaof\Delta DEF}{Areaof\Delta CBE}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}\\ \frac{Areaof\Delta DEF}{Areaof\Delta CBE}=\frac{1}{4}\end{array}$

Hence,

  $\frac{Areaof\Delta DEF}{Areaof\Delta CBE}=\frac{1}{4}$

(c)

To determine

To Find: $\frac{Areaof\Delta DEF}{AreaofDEBA}$

Answer to Problem 23WE

  $\frac{1}{8}$

Explanation of Solution

Given: Area of parallelogram.

  

In triangle $\Delta DEFandtrap.DEBA$

  $\frac{Areaof\Delta DEF}{Areaof\Delta ABF}=\frac{1}{9}$

  $\begin{array}{l}ar(ABF)=9\{ar(DEF)\}\\ ar(DEBA)=8\cdot \{ar(DEF)\}\end{array}$

Now

  $\begin{array}{l}\frac{Areaof\Delta DEF}{8·AreaofDEF}=\frac{1}{8}\\ Thus\\ \frac{Areaof\Delta DEF}{AreaofDEBA}=\frac{1}{8}\end{array}$

Hence,

  $\frac{Areaof\Delta DEF}{AreaofDEBA}=\frac{1}{8}$