Chapter 11.7, Problem 22WE

(a)

To determine

To Find: The area of $\Delta ABE$ .

Answer to Problem 22WE

  $ar(\Delta ABE)=24c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Observe the figure.

  $\begin{array}{l}ar(ABCD)=base\times height\\ ar(ABCD)=AN\times DC\\ ar(ABCD)=AN\times (DE+EC)\\ ar(ABCD)=AN\times (2EC+EC)\\ ar(ABCD)=AN\times (3EC)\\ ar(ABCD)=3AN\times EC\\ 48=3AN\times EC\\ AN\times EC=16\end{array}$

Now

Area of triangle $\Delta ABE$

  $\begin{array}{l}=\frac{1}{2}\times AN\times AB\\ =\frac{1}{2}\times AN\times DC\\ =\frac{1}{2}\times AN\times 3EC\\ =\frac{3}{2}\times 16\\ =24c{m}^2\end{array}$

Hence,

  $ar(\Delta ABE)=24c{m}^2$ .

(b)

To determine

To Find: The area of $\Delta BEC$ .

Answer to Problem 22WE

  $ar(\Delta BEC)=8c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Observe the figure.

  $\begin{array}{l}ar(ABCD)=base\times height\\ ar(ABCD)=AN\times DC\\ ar(ABCD)=AN\times (DE+EC)\\ ar(ABCD)=AN\times (2EC+EC)\\ ar(ABCD)=AN\times (3EC)\\ ar(ABCD)=3AN\times EC\\ 48=3AN\times EC\\ AN\times EC=16\end{array}$

Now

Area of triangle $\Delta BEC$

  $\begin{array}{l}=\frac{1}{2}\times AN\times EC\\ =\frac{1}{2}\times 16\\ =8c{m}^2\end{array}$

Hence,

  $ar(\Delta BEC)=8c{m}^2$ .

(c)

To determine

To Find: The area of $\Delta ADE$ .

Answer to Problem 22WE

  $ar(\Delta ADE)=16c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Observe the figure.

  $\begin{array}{l}ar(ABCD)=base\times height\\ ar(ABCD)=AN\times DC\\ ar(ABCD)=AN\times (DE+EC)\\ ar(ABCD)=AN\times (2EC+EC)\\ ar(ABCD)=AN\times (3EC)\\ ar(ABCD)=3AN\times EC\\ 48=3AN\times EC\\ AN\times EC=16\end{array}$

Now

Area of triangle $\Delta ADE$

  $\begin{array}{l}=\frac{1}{2}\times AN\times DE\\ =\frac{1}{2}\times AN\times 2EC\\ =AN\times EC\\ =16c{m}^2\end{array}$

Hence,

  $ar(\Delta ADE)=16c{m}^2$ .

(d)

To determine

To Find: The area of $\Delta CEF$ .

Answer to Problem 22WE

  $ar(\Delta CEF)=4c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Observe the figure.

  $\begin{array}{l}ar(ABCD)=base\times height\\ ar(ABCD)=AN\times DC\\ ar(ABCD)=AN\times (DE+EC)\\ ar(ABCD)=AN\times (2EC+EC)\\ ar(ABCD)=AN\times (3EC)\\ ar(ABCD)=3AN\times EC\\ 48=3AN\times EC\\ AN\times EC=16\end{array}$

Triangles $\Delta CEFand\Delta ADE$ are similar thus

  $\frac{DE}{EC}=\frac{AN}{FM}=2$

Now

Area of triangle $\Delta CEF$

  $\begin{array}{l}=\frac{1}{2}\times EC\times FM\\ =\frac{1}{2}\times EC\times \frac{AN}{2}\\ =\frac{16}{4}\\ =4c{m}^2\end{array}$

Hence,

  $ar(\Delta CEF)=4c{m}^2$ .

(e)

To determine

To Find: The area of $\Delta DEF$ .

Answer to Problem 22WE

  $ar(\Delta DEF)=8c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Area of triangle $\Delta DEF$

  $\begin{array}{l}=\frac{1}{2}\times DE\times FM\\ =\frac{1}{2}\times 2EC\times \frac{AN}{2}\\ =\frac{1}{2}\times EC\times AN\\ =\frac{1}{2}\times 16\\ =8c{m}^2\end{array}$

Hence,

  $ar(\Delta DEF)=8c{m}^2$ .

(f)

To determine

To Find: The area of $\Delta BEF$ .

Answer to Problem 22WE

  $ar(\Delta BEF)=12c{m}^2$

Explanation of Solution

Given: Area of parallelogram is $48c{m}^2$ and $DE=2EC$ .

  

Area of triangle $\Delta BEF$

  $\begin{array}{l}=\frac{1}{2}\times (EN\times AC+EC+FM)\\ =\frac{1}{2}\times (16+8)\\ =12c{m}^2\end{array}$

Hence,

  $ar(\Delta BEF)=12c{m}^2$ .