Chapter 11.1, Problem 3BCYP

To determine

Find the perimeter and area of the triangle ABC.

Answer to Problem 3BCYP

  $A\approx 207.6c{m}^2$

  $P\approx 75.6cm$

Explanation of Solution

Given:

  

Formula Used:

Area of a triangle with base b and height h is :

  $A=\frac{1}{2}bh$

Calculation:

We find the height h of triangle ABC , by Pythagoras theorem in right angle triangle ABD:

  $\begin{array}{l}A{B}^2=A{D}^2-B{D}^2\\ \Rightarrow h^2={13}^2-6^2\\ \Rightarrow h=\sqrt{169+36}\\ \Rightarrow h\approx 14.3cm\end{array}$

Find AC in right angle triangle ABC using Pythagoras theorem:

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2\\ \Rightarrow A{C}^2={\left(\sqrt{205}\right)}^2+{29}^2\\ \Rightarrow AC=\sqrt{205+841}\\ \Rightarrow AC\approx 32.3cm\end{array}$

The perimeter of the triangle ABC :

  $P=AB+BC+AC\approx 14.3+29+32.3=75.6cm$

The base of the triangle ABC = BC = b = 29 cm.

So, the area of triangle ABC:

  $\begin{array}{l}A=\frac{1}{2}bh\\ \Rightarrow A=\frac{1}{2}(29)(14.3)\\ \Rightarrow A\approx 207.6c{m}^2\end{array}$