Chapter 7.4, Problem 30WE

(a)

To determine

To Find: the measure of side $VC$ in the figure shown such that $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }V{C}^{\text{'}}=18$

Answer to Problem 30WE

  $42$

Explanation of Solution

Given:

Figure is shown below.

  

  $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }V{C}^{\text{'}}=18$

Calculation:

Since, plane of $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ is similar to the plane of $\Delta ABC$

Since, $ABCD$ is a parallelogram.

  $\begin{array}{l}\angle V=\angle V\therefore \left(\text{Common}\right)\\ \angle VAB=\angle V{A}^{\text{'}}{B}^{\text{'}}\therefore \left(\text{Corresponding angle in two similar plane}\right)\end{array}$

Therefore by AA similarity, $\Delta VAB\sim \Delta V{A}^{\text{'}}{B}^{\text{'}}$

Ratio of corresponding sides in two similar triangles is equal.

  $\begin{array}{l}\frac{VA}{V{A}^{\text{'}}}=\frac{VC}{V{C}^{\text{'}}}\\ \frac{V{A}^{\text{'}}+A{A}^{\text{'}}}{V{A}^{\text{'}}}=\frac{VC}{V{C}^{\text{'}}}\\ \frac{15+20}{15}=\frac{VC}{18}\\ VC=\frac{18\times 35}{15}\\ VC=42\end{array}$

Therefore, the answer is $42$ .

(b)

To determine

To Find: the measure of side $B{B}^{\text{'}}$ in the figure shown such that $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }VB=49$

Answer to Problem 30WE

  $28$

Explanation of Solution

Given:

Figure is shown below.

  

  $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }VB=49$

Calculation:

Since, plane of $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ is similar to the plane of $\Delta ABC$

Since, $ABCD$ is a parallelogram.

  $\begin{array}{l}\angle V=\angle V\therefore \left(\text{Common}\right)\\ \angle VAB=\angle V{A}^{\text{'}}{B}^{\text{'}}\therefore \left(\text{Corresponding angle in two similar plane}\right)\end{array}$

Therefore by AA similarity, $\Delta VAB\sim \Delta V{A}^{\text{'}}{B}^{\text{'}}$

Ratio of corresponding sides in two similar triangles is equal.

  $\begin{array}{l}\frac{VA}{V{A}^{\text{'}}}=\frac{VC}{V{C}^{\text{'}}}\\ \frac{V{A}^{\text{'}}+A{A}^{\text{'}}}{V{A}^{\text{'}}}=\frac{VB}{V{B}^{\text{'}}}\\ \frac{15+20}{15}=\frac{49}{V{B}^{\text{'}}}\\ V{B}^{\text{'}}=\frac{15\times 49}{35}\\ V{B}^{\text{'}}=21\end{array}$

  $\begin{array}{l}VB=B{B}^{\text{'}}+V{B}^{\text{'}}\\ 49=BB+21\\ B{B}^{\text{'}}=28\end{array}$

Therefore, the answer is $28$ .

(c)

To determine

To Find: the measure of side $AB$ in the figure shown such that $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }{A}^{\text{'}}{B}^{\text{'}}=24$

Answer to Problem 30WE

  $56$

Explanation of Solution

Given:

Figure is shown below.

  

  $V{A}^{\text{'}}=15,A^{\text{'}}A=20\text{ and }{A}^{\text{'}}{B}^{\text{'}}=24$

Calculation:

Since, plane of $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ is similar to the plane of $\Delta ABC$

Since, $ABCD$ is a parallelogram.

  $\begin{array}{l}\angle V=\angle V\therefore \left(\text{Common}\right)\\ \angle VAB=\angle V{A}^{\text{'}}{B}^{\text{'}}\therefore \left(\text{Corresponding angle in two similar plane}\right)\end{array}$

Therefore by AA similarity, $\Delta VAB\sim \Delta V{A}^{\text{'}}{B}^{\text{'}}$

Ratio of corresponding sides in two similar triangles is equal.

  $\begin{array}{l}\frac{VA}{V{A}^{\text{'}}}=\frac{VC}{V{C}^{\text{'}}}\\ \frac{V{A}^{\text{'}}+A{A}^{\text{'}}}{V{A}^{\text{'}}}=\frac{AB}{A^{\text{'}}{B}^{\text{'}}}\\ \frac{15+20}{15}=\frac{AB}{24}\\ AB=\frac{24\times 35}{15}\\ AB=56\end{array}$

Therefore, the answer is $56$ .