Chapter 11.5, Problem 25PSC

a.

To determine

To calculate:Thearea of ABCD as sum of triangles CDE , BAE and CEB .

Answer to Problem 25PSC

Thearea of ABCDis $\frac{1}{2}ab+\frac{1}{2}ab+\frac{1}{2}{c}^2$ .

Explanation of Solution

Given information:

Triangle CDEwith sides b , aand c .

Triangle BAE with sides a , b and c .

TriangleCEBwith sidesc and c .

Formula used:

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Calculation:

  

Triangles CDE , BAE and CEB are ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

The area of triangles are as follows:

  $\begin{array}{l}A(△CDE)=\frac{1}{2}\times DE\times CD\\ A(△CDE)=\frac{1}{2}ab\\ A(△BAE)=\frac{1}{2}\times AE\times BA\\ A(△BAE)=\frac{1}{2}ab\\ A(△CEB)=\frac{1}{2}\times EB\times CE\\ A(△CEB)=\frac{1}{2}(c\times c)=\frac{1}{2}{c}^2\\ A(ABCD)=A(△CDE)+A(△BAE)+A(△CEB)\\ A(ABCD)=\frac{1}{2}ab+\frac{1}{2}ab+\frac{1}{2}{c}^2\end{array}$

b.

To determine

To find:The area of ABCD as area of trapezoid with bases $\overline{AB}$ and $\overline{CD}$ .

Answer to Problem 25PSC

The area of ABCD is $\frac{1}{2}(a+b)(a+b)$ .

Explanation of Solution

Given information:

Triangle CDEwith sides b , a and c .

Triangle BAE with sides a , b and c .

TriangleCEB with sides c and c .

Formula used:

  $\begin{array}{l}Areaofatrapezoid=\frac{(a+b)}{2}\times h\\ a=longbase,\\ b=shortbase,\\ h=altitude.\end{array}$

Calculation:

  

Area of Trapezoid $=\frac{(a+b)}{2}\times h$

  $\begin{array}{l}A(ABCD)=\frac{(b+a)}{2}\times (a+b)\\ A(ABCD)=\frac{1}{2}\times (a+b)\times (a+b)\end{array}$

c.

To determine

To simplify:The value of $\frac{1}{2}ab+\frac{1}{2}ab+\frac{1}{2}{c}^2=\frac{1}{2}\times (a+b)\times (a+b)$

Answer to Problem 25PSC

The simplified valueis $c^2=a^2+b^2$ .

Explanation of Solution

Given information:

Triangle CDEwith sides b , a and c .

Triangle BAE with sides a , b and c .

TriangleCEB with sides c and c .

Formula used:

  ${(a+b)}^2=(a^2+2ab+b^2)$

Calculation:

  

It is given that,

  $\begin{array}{l}\frac{1}{2}ab+\frac{1}{2}ab+\frac{1}{2}{c}^2=\frac{1}{2}\times (a+b)\times (a+b)\\ ab+\frac{1}{2}{c}^2=\frac{1}{2}\times (a^2+2ab+b^2)\\ ab+\frac{1}{2}{c}^2=\frac{1}{2}{a}^2+ab+\frac{1}{2}{b}^2\\ \frac{1}{2}{c}^2=\frac{1}{2}{a}^2+\frac{1}{2}{b}^2\\ c^2=a^2+b^2\end{array}$