Chapter 11.1, Problem 3ACYP

a)

To determine

To find perimeterof each triangle.

Answer to Problem 3ACYP

  $\begin{array}{l}\text{Perimeter of }\mathrm{\Delta }\text{ABD = 100}\text{.36 in}\text{.}\\ \text{Perimeter of }\mathrm{\Delta }\text{ACD = 117 in}\text{.}\\ \text{Perimeter of }\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}\text{ = 97}\text{.36 in}\text{.}\end{array}$

Explanation of Solution

Given information :

There are three triangles in below figure

  $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{D},\mathrm{\Delta }\mathrm{A}\mathrm{C}\mathrm{D}$ and $\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}$

  

Concept used:

Perimeter of the triangle is the length of the outline

Pythagoras Theorem:

In a right-angle triangle: the square of the hypotenuse is equal to sum of the square of the base and altitude(opposite)

Calculation:

First, find side $\mathrm{B}\mathrm{D}$ by Pythagoras Theorem

  $\begin{array}{l}\mathrm{B}{\mathrm{D}}^2=\mathrm{B}{\mathrm{C}}^2+\mathrm{C}{\mathrm{D}}^2\\ \mathrm{B}{\mathrm{D}}^2={27}^2+{30}^2\\ \mathrm{B}{\mathrm{D}}^2=729+900\\ \mathrm{B}\mathrm{D}=\sqrt{1629}\\ \mathrm{B}\mathrm{D}\approx 40.36\end{array}$

Perimeter of $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{D}$

  $\begin{array}{l}\mathrm{P}=\mathrm{A}\mathrm{B}+\mathrm{B}\mathrm{D}+\mathrm{A}\mathrm{D}\\ =19\text{in}\text{.}+40.36\text{ in}\text{.}+41\text{ in}\text{.}\\ =100.36\text{}\text{in}\text{.}\end{array}$

Perimeter of $\mathrm{\Delta }\mathrm{A}\mathrm{C}\mathrm{D}$

  $\begin{array}{l}\mathrm{P}=\mathrm{A}\mathrm{C}+\mathrm{C}\mathrm{D}+\mathrm{A}\mathrm{D}\\ =\left(\mathrm{A}\mathrm{B}+\mathrm{B}\mathrm{C}\right)\text{in}\text{.}+30\text{ in}\text{.}+41\text{ in}\text{.}\\ =\left(19+27\right)\text{in}\text{.}+30\text{ in}\text{.}+41\text{ in}\\ =117\text{ in}\text{.}\end{array}$

Perimeter of $\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}$

  $\begin{array}{l}\mathrm{P}=\mathrm{B}\mathrm{C}+\mathrm{C}\mathrm{D}+\mathrm{B}\mathrm{D}\\ =27\text{in}\text{.}+30\text{ in}\text{.}+40.36\text{ in}\text{.}\\ =97.36\text{}\text{in}\text{.}\end{array}$

b)

To determine

To find areaof each triangle.

Answer to Problem 3ACYP

  $\begin{array}{l}\text{Area of }\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{D}{\text{ = 285 in}}^2\\ \text{Area of }\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}{\text{ = 405 in}}^2\\ \text{Area of }\mathrm{\Delta }{\text{ACD = 690 in}}^2\end{array}$

Explanation of Solution

Given information :

There are three triangles in below figure

  $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{D},\mathrm{\Delta }\mathrm{A}\mathrm{C}\mathrm{D}$ and $\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}$

  

Concept used:

Area of the triangle $=\frac{1}{2}\mathrm{b}\mathrm{h}$

Pythagoras Theorem:

In a right-angle triangle: the square of the hypotenuse is equal to sum of the square of the base and altitude(opposite)

Calculation:

First, find side $\mathrm{B}\mathrm{D}$ by Pythagoras Theorem

  $\begin{array}{l}\mathrm{B}{\mathrm{D}}^2=\mathrm{B}{\mathrm{C}}^2+\mathrm{C}{\mathrm{D}}^2\\ \mathrm{B}{\mathrm{D}}^2={27}^2+{30}^2\\ \mathrm{B}{\mathrm{D}}^2=729+900\\ \mathrm{B}\mathrm{D}=\sqrt{1629}\\ \mathrm{B}\mathrm{D}\approx 40.36\end{array}$

Area of $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{D}$

  $\begin{array}{l}\mathrm{A}=\frac{1}{2}\mathrm{b}\mathrm{h}\\ =\frac{1}{2}\times 19\times 30\\ =285{\text{ in}}^2\end{array}$

Area of $\mathrm{\Delta }\mathrm{B}\mathrm{C}\mathrm{D}$

  $\begin{array}{l}\mathrm{A}=\frac{1}{2}\mathrm{b}\mathrm{h}\\ =\frac{1}{2}\times 27\times 30\\ =405{\text{ in}}^2\end{array}$

Area of $\mathrm{\Delta }\mathrm{A}\mathrm{C}\mathrm{D}$

  $\begin{array}{l}\mathrm{A}=\frac{1}{2}\mathrm{b}\mathrm{h}\\ =\frac{1}{2}\times \left(27+19\right)\times 30\\ =690{\text{ in}}^2\end{array}$