Chapter 2, Problem 35RP

a.

To determine

The area of $\Delta \text{BDE}$

Answer to Problem 35RP

The area of $\Delta \text{BDE = 28 sq units}$

Explanation of Solution

Given:

Line AB is parallel to ED.

AB = 8 units, ED = 8 units, BD = 4 + 3 = 7 units

Concept Used:

  $Areaofrightangletriangle=\frac{1}{2}\times base\times height$

Calculation:

?BDE is a right angle triangle.

Base BD = 7 units

Height ED = 8 units

Area of right-angle triangle

= $\frac{1}{2}\times base\times height$

  $=\frac{1}{2}\times 7\times 8=28squnits$

b.

To determine

To compare the area of $\Delta \text{ABC and }\Delta \text{EDC}$

Answer to Problem 35RP

Area of $\Delta \text{ABC}$ = Area of $\Delta \text{EDC}$

Explanation of Solution

Given:

Line AB is parallel to ED.

AB = ED

Concept Used:

If two triangles are congruent then their area are also equal.

Calculation:

In $\Delta \text{ABC and }\Delta \text{EDC}$

Line AB is parallel to ED.

  $\begin{array}{l}So,\angle BAC=\angle EDC(Alternateinteriorangle)\\ AB=ED(Given)\\ \angle ABC=\angle DEC(Alternateinteriorangle)\\ \\ \therefore \Delta ABC\cong \Delta EDC(ASAcriteria)\\ Areaof\Delta ABC=\Delta EDC\end{array}$