Chapter 9.7, Problem 25PSC

a

To determine

To calculate: The Length of DC.

Answer to Problem 25PSC

Length of DC is $(2+2\sqrt{3})$ units.

Explanation of Solution

Given information:Here, we have given a trapezoid ABCD in which $AB\parallel CD$ , $AB=AD=4$ , $\angle C=45°$ and $\angle C=45°$

Formula used: Sum of co-interior angles is $180°$ .

Calculation: Draw a perpendicular from D to AB such that $DE\perp AB$ and a perpendicular from B to DC such that $BF\perp DC$ .

Now, as DEBF is a rectangle,

  $\angle FDE=\angle FBE=90°$

Now, $\angle A+\angle D=180°$ and $\angle C+\angle B=180°$ (Co-interior angles)

  $\Rightarrow 60°+\angle D=180°$ and $45°+\angle B=180°$

  $\Rightarrow \angle D=120°$ and $\angle B=135°$

Now, $\angle ADE=120°-90°=30°$

Similarly, $\angle BFC=135°-90°=45°$

Now, in $\Delta DAE$ ,

Angles are $90°,60°,30°$ ,

Hypotenuse (DA) = $2a$

  $\begin{array}{l}\Rightarrow 4=2a\\ \Rightarrow a=2\end{array}$

Base (AE) = $2$ (As perpendicular bisects the base)

Perpendicular (DE) = $a\sqrt{3}=2\sqrt{3}$

Now, $DE=BF$ (Opposite sides of rectangle are equal)

  $\Rightarrow BF=2\sqrt{3}$

Similarly, in $\Delta BFC$ ,

Angles are $90°,45°,45°$ ,

  $\Rightarrow$ Perpendicular (BE) = Base (EC) = $2\sqrt{3}$

DC = BE (Opposite sides of rectangle)

  $\Rightarrow$ DE = $2$

  $\therefore$ DC = DE + EC

  $=2+2\sqrt{3}$

b

To determine

To calculate:The Length of BC

Answer to Problem 25PSC

Length of BC is $2\sqrt{6}$ units.

Explanation of Solution

Given information:Here, we have given a trapezoid ABCD in which $AB\parallel CD$ , $AB=AD=4$ , $\angle C=45°$ and $\angle C=45°$

Formula used: Hypotenuse of triangles having angles $90°,45°,45°$ is $x\sqrt{2}$ .

Calculation: In $\Delta BEC$ ,

Angles are $90°,45°,45°$ ,

Hypotenuse will be $x\sqrt{2}$ and other two sides will be x.

Here, Base = $2\sqrt{3}$

  $\Rightarrow$ Hypotenuse (BC) = $2\sqrt{3}(\sqrt{2})=2\sqrt{6}$