Chapter 11.2, Problem 22PSB

a.

To determine

To find the area of a trapezoid AXDF by dividing it into triangle AZX, triangle FYD and rectangle AZFY.

Answer to Problem 22PSB

The area of a trapezoid AXDF is $39\sqrt{3}$ .

Explanation of Solution

Given:

In Trapezoid AXDF,

  $\begin{array}{l}\angle A=120^{\circ }\\ AX=FD=6,\\ AF=10.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

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

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

Calculation:

  

In an isosceles trapezoid any upper base angle is supplementary to lower base angle.

  $\angle X\cong \angle D={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Draw altitude $\overline{FY}$ .

  $\Delta FDYisa30^{o}60^{o}90^o\Delta .$

  $\begin{array}{l}YD=\frac{1}{2}(FD)\\ YD=\frac{1}{2}(6)=3\end{array}$

In right angled $\Delta FYD$ , we get

  ${\left(FD\right)}^2={\left(FY\right)}^2+{\left(YD\right)}^2$

  $\begin{array}{l}{\left(6\right)}^2={\left(FY\right)}^2+{\left(3\right)}^2\\ {\left(FY\right)}^2=36-9\\ FY=\sqrt{27}\\ FY=3\sqrt{3}\end{array}$

  $\Delta AXZisa30^{o}60^{o}90^o\Delta .$

Similarlyin right angled $\Delta AZX$ , we get

  $AZ=3\sqrt{3}$

  $\begin{array}{l}A(\Delta AZX)=\frac{1}{2}*XZ*AZ\\ =\frac{1}{2}*3*3\sqrt{3}\\ =\frac{9\sqrt{3}}{2}\\ A(\Delta FYD)=\frac{1}{2}*YZ*FY\\ =\frac{1}{2}*3*3\sqrt{3}\\ =\frac{9\sqrt{3}}{2}\end{array}$

  $\begin{array}{l}FY=(YD)\sqrt{3}\\ FY=3\sqrt{3}\\ A(\square AZYF)=ZY*FY=10*3\sqrt{3}=30\sqrt{3}\end{array}$

  $\begin{array}{l}Areaoftrapezoid=A(\Delta AZX)+A(\Delta FYD)+A(\square AZYF)\\ =\frac{9\sqrt{3}}{2}+\frac{9\sqrt{3}}{2}+30\sqrt{3}=39\sqrt{3}\end{array}$

b.

To determine

To find the area of a trapezoid ABCD by dividing it into triangle AEB, triangle DFC and rectangle EBCFwith side AD as 14 and side BC as 8.

Answer to Problem 22PSB

The area of a trapezoid ABCD is $33$ .

Explanation of Solution

Given:

In Trapezoid ABCD,

  $\begin{array}{l}\angle A={45}^{\circ }\\ AD=14,\\ BC=8.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

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

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

Calculation:

  

Draw altitude $\overline{BE}$ and $\overline{CF}$ .

  $\Delta ABE\cong \Delta DCF$

  $\Delta ABEisa45^o{45}^{o}90^o\Delta .$

  $\begin{array}{l}AE+FD=14-8\\ AE+FD=6\end{array}$

If corresponding angles are congruent, then corresponding sides are also congruent.

  $\begin{array}{l}\overline{AE}\cong \overline{EB}\\ AE=3\end{array}$

  $\begin{array}{l}A(\Delta AEB)=\frac{1}{2}*AE*EB\\ =\frac{1}{2}*3*3\\ =\frac{9}{2}\\ =4.5\end{array}$

  $\Delta DCFisa45^o{45}^{o}90^o\Delta .$

  $\begin{array}{l}A(\Delta CFD)=\frac{1}{2}*FD*CF\\ =\frac{1}{2}*3*3\\ =\frac{9}{2}\\ =4.5\end{array}$

  $\begin{array}{l}Areaofrectangle=b*h\\ A(\square EBCF)=BC*FC=8*3=24\end{array}$

  $\begin{array}{l}Areaoftrapezoid=A(\Delta AEB)+A(\Delta CFD)+A(\square EBCF)\\ =4.5+4.5+24=33\end{array}$

c.

To determine

To compute the area of a trapezoidBADC by dividing it into triangle BXA, triangle CYD and rectangle BXYCwith angle A as ${45}^{\circ }$ andangle B as $35^{\circ }$ .

Answer to Problem 22PSB

The area of a trapezoid BADC is $183.42$ .

Explanation of Solution

Given:

In Trapezoid BADC,

  $\begin{array}{l}\angle A={45}^{\circ }\\ \angle D=35^{\circ }\\ CD=16,\\ BC=12.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

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

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

Calculation:

  

Draw altitude $\overline{BX}$ and $\overline{CY}$ .

  $\Delta CDYisa30^o{60}^{o}90^o\Delta .$

  $\begin{array}{l}CY=\frac{1}{2}(CD)\\ CY=\frac{1}{2}(16)=8\end{array}$

  $\begin{array}{l}YD=(CY)(\sqrt{3})\\ A(\Delta CYD)=\frac{1}{2}*YD*CY\\ =\frac{1}{2}*8\sqrt{3}*8\\ =32\sqrt{3}\end{array}$

  $\Delta BAXisa45^o{45}^{o}90^o\Delta .$

If corresponding angles are congruent, then corresponding sides are also congruent.

  $\begin{array}{l}\overline{BX}\cong \overline{CY,}\\ \overline{BX}=8,\\ \overline{BX}\cong \overline{AX}\end{array}$

  $\begin{array}{l}A(\Delta BXA)=\frac{1}{2}*AX*BX\\ =\frac{1}{2}*8*8\\ =32\end{array}$

  $A(\square BXYC)=XY*CY=12*8=96$

  $\begin{array}{l}Areaoftrapezoid=A(\Delta BXA)+A(\Delta CYD)+A(\square BXYC)\\ =32+32\sqrt{3}+96=128+32\sqrt{3}=183.42\end{array}$