Chapter 11.7, Problem 22PSC

a.

To determine

To calculate:Thearea of regions $\mathrm{Ⅰ},\mathrm{Ⅱ},\mathrm{Ⅲ}$ and $\mathrm{Ⅳ}$ .

Answer to Problem 22PSC

The area of regions $\mathrm{Ⅰ}=5,\mathrm{Ⅱ}=10,\mathrm{Ⅲ}=20$ and $\mathrm{Ⅳ}=25$ .

Explanation of Solution

Given information:

PQRS is a parallelogram.

T is the midpoint.

  $A(▱PQRS)=60$

Formula used:

The below property is used:

Area of triangle is proportional to square of its side.

Calculation:

  

  $A(\Delta \mathrm{Ⅱ})+A(\Delta \mathrm{Ⅲ})=\frac{A(▱PQRS)}{2}=\frac{60}{2}=30----Equation1$

  $A(\Delta \mathrm{Ⅰ})+A(\Delta \mathrm{Ⅳ})=\frac{A(▱PQRS)}{2}=\frac{60}{2}=30----Equation2$

  $A(\Delta \mathrm{Ⅰ})+A(\Delta \mathrm{Ⅱ})=15----Equation3$

  $A(\Delta \mathrm{Ⅲ})+A(\Delta \mathrm{Ⅳ})=45----Equation4$

  $\Delta \mathrm{Ⅰ}\sim \Delta \mathrm{Ⅲ}$

Area of triangle is proportional to square of its side.

  $\begin{array}{l}\frac{A(\Delta \mathrm{Ⅰ})}{A(\Delta \mathrm{Ⅲ})}=\frac{{(1)}^2}{{(2)}^2}=\frac{1}{4}\\ A(\Delta \mathrm{Ⅲ})=4A(\Delta \mathrm{Ⅰ})\\ A(\Delta \mathrm{Ⅰ})=\frac{A(\Delta \mathrm{Ⅲ})}{4}----Equation4\end{array}$

Substituting Equation 4 in Equation 2, we get

  $\begin{array}{l}A(\Delta \mathrm{Ⅰ})+A(\Delta \mathrm{Ⅱ})=15\\ \frac{A(\Delta \mathrm{Ⅲ})}{4}+A(\Delta \mathrm{Ⅱ})=15\\ A(\Delta \mathrm{Ⅲ})+4A(\Delta \mathrm{Ⅱ})=60----Equation5\end{array}$

Substituting Equation 1 from Equation 5, we get

  $\begin{array}{l}3A(\Delta \mathrm{Ⅱ})=30\\ A(\Delta \mathrm{Ⅱ})=10----Equation6\end{array}$

Substituting Equation 6 in Equation 3, we get

  $A(\Delta \mathrm{Ⅰ})=15-10=5$

Substituting Equation 6 in Equation 1, we get

  $A(\Delta \mathrm{Ⅲ})=30-10=20$

Substituting Equation 6 in Equation 2, we get

  $A(\Delta \mathrm{Ⅳ})=30-5=25$

b.

To determine

To calculate: The ratio of area of regions $\mathrm{Ⅰ}$ and $▱PQRS$ .

Answer to Problem 22PSC

The area ratio is $\frac{y^2}{2x^2+6xy+4y^2}$ .

Explanation of Solution

Given information:

PQRS is a parallelogram.

Tdivides $\overline{QR}$ such that $\frac{QT}{TR}=\frac{x}{y}$

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

Calculation:

  

  $\frac{QT}{TR}=\frac{x}{y}$

  $\begin{array}{l}A(\Delta \mathrm{Ⅰ})=\frac{1}{2}wx\\ A(▱PQRS)=(x+y)(w+z)\end{array}$

  $\Delta \mathrm{Ⅲ}\sim \Delta \mathrm{Ⅰ}$

Corresponding sides of similar triangles are congruent.

  $\frac{w}{z}=\frac{y}{x+y}$ and $\frac{w}{w+z}=\frac{y}{x+2y}$

  $\begin{array}{l}\frac{A(\mathrm{Ⅰ})}{A(▱PQRS)}=\frac{\frac{1}{2}wx}{(x+y)(w+z)}\\ \frac{A(\mathrm{Ⅰ})}{A(▱PQRS)}=\frac{\frac{1}{2}\times y\times y}{(x+y)(x+2y)}\\ \frac{A(\mathrm{Ⅰ})}{A(▱PQRS)}=\frac{\frac{1}{2}{y}^2}{(x^2+3xy+2y^2)}\\ \frac{A(\mathrm{Ⅰ})}{A(▱PQRS)}=\frac{y^2}{(2x^2+6xy+4y^2)}\end{array}$