Chapter 7.4, Problem 31WE

To determine

To Find: the perimeter of $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ in the figure shown such that $V{A}^{\text{'}}=10,VA=25,AB=20\text{,}B{C}^{\text{'}}=14\text{ and }AC=16$

Answer to Problem 31WE

  $23$

Explanation of Solution

Given:

Figure is shown below.

  

  $V{A}^{\text{'}}=10,VA=25,AB=20\text{,}B{C}^{\text{'}}=14\text{ and }AC=16$

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.

  $\begin{array}{l}\frac{VA}{V{A}^{\text{'}}}=\frac{AB}{A^{\text{'}}{B}^{\text{'}}}\\ \frac{25}{10}=\frac{20}{A^{\text{'}}{B}^{\text{'}}}\\ A^{\text{'}}{B}^{\text{'}}=\frac{10\times 20}{25}\\ A^{\text{'}}{B}^{\text{'}}=8\end{array}$

Similarly, $\Delta VAC\sim \Delta V{A}^{\text{'}}{C}^{\text{'}}$ therefore.

  $\begin{array}{l}\frac{VA}{V{A}^{\text{'}}}=\frac{AC}{A^{\text{'}}{C}^{\text{'}}}\\ \frac{25}{10}=\frac{16}{A^{\text{'}}{C}^{\text{'}}}\\ A^{\text{'}}{C}^{\text{'}}=\frac{10\times 16}{20}\\ A^{\text{'}}{C}^{\text{'}}=8\end{array}$

Similarly, $\Delta ABC\sim \Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ therefore.

  $\begin{array}{l}\frac{AC}{A^{\text{'}}{C}^{\text{'}}}=\frac{BC}{B^{\text{'}}{C}^{\text{'}}}\\ \frac{16}{8}=\frac{14}{B^{\text{'}}{C}^{\text{'}}}\\ B^{\text{'}}{C}^{\text{'}}=\frac{14\times 8}{16}\\ B^{\text{'}}{C}^{\text{'}}=7\end{array}$

Perimeter of $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ is.

  $\begin{array}{l}=A^{\text{'}}{B}^{\text{'}}+B^{\text{'}}{C}^{\text{'}}+A^{\text{'}}{C}^{\text{'}}\\ =8+7+8\\ =23\end{array}$

Therefore, the answer is $23$ .