Chapter 11.2, Problem 11PSA

a.

To determine

To find the area of a trapezoid ABCD by dividing it into triangle ABE and rectangle AECD.

Answer to Problem 11PSA

The area of a trapezoid ABCD is $84$ .

Explanation of Solution

Given:

  $\Delta AEB$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}DC=7,\\ AD=10,\\ BC=14.\end{array}$

Formula Used:

  $\begin{array}{l}Areaofrectangle=b\ast h\\ b=Base,\\ h=Height\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

  $\begin{array}{l}AD=EC=10\\ BE=BC-EC=14-10=4\\ AE=DC=7\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta AEB)=\frac{1}{2}*BE*AE\\ =\frac{1}{2}*4*7\\ =14\end{array}$

  $\begin{array}{l}Areaofrectangle=b\ast h\\ A(\square AEDC)=CE*DC=10*7=70\end{array}$

  $\begin{array}{l}TotalArea=A(\Delta AEB)+A(\square AEDC)\\ =14+70=84\end{array}$

b.

To determine

To find the area of a trapezoid ABCD by dividing it into triangle AEB, triangle DFC and rectangle AEFD.

Answer to Problem 11PSA

The area of a trapezoid ABCD is $128.$

Explanation of Solution

Given:

  $\Delta AEB$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\Delta DFC$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}BC=19,\\ AD=13,\\ AE=DF=8.\end{array}$

Formula Used:

  $\begin{array}{l}Areaofrectangle=b\ast h\\ b=Base,\\ h=Height\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

  $\begin{array}{l}AD=EF=13\\ BE+CF=19-13=6\\ BE=CF=3\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta AEB)=\frac{1}{2}*BE*AE\\ =\frac{1}{2}*3*8\\ =12\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta DFC)=\frac{1}{2}*FC*DF\\ =\frac{1}{2}*3*8\\ =12\end{array}$

  $\begin{array}{l}\text{Area of rectangle}=b\ast h\\ A(\square AEFD)=EF*DF=13*8=104\end{array}$

  $\begin{array}{l}TotalArea=A(\Delta AEB)+A(\Delta DFC)+A(\square AEDC)\\ =12+12+104=128\end{array}$