Chapter 6, Problem 34CR

To determine

To calculate:

Area of triangle by using Heron’s area formula.

Answer to Problem 34CR

Value of triangle solved by using Heron’s area formula is $AB=11.29m,BC=33.5m$

Explanation of Solution

Given information:

  $a=15,b=8,c=10$

Calculation:

Let’s consider,

  $a=15,b=8,c=10$

Diagonals $AC$ and $BD$ of the parallelogram $ABCD$ intersects at the point $O$ at angle of $34°$ ,

Other angle $\theta$ is,

  $\begin{array}{l}\theta =180°-34°\\ \theta =146°\end{array}$

Length of diagonal $AC$ is $40m$ similarly length of the diagonal $BD$ is $30m$ diagonals of parallelogram bisect each other,

  $\begin{array}{l}OA=OC=20\\ OB=OD=15\end{array}$

Now in the triangle $OAB$ , two sides, $OA,OB$ and included angle is known thus cosine law can be used to determine third side $AB$ ,

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Therefore, for the triangle $OAB$ ,

  $\begin{array}{l}A{B}^2=O{A}^2+O{B}^2-2\left(OA\right)\left(OB\right)\cos 34°\\ ={20}^2+{15}^2-2\left(20\right)\left(15\right)\left(0.829\right)\\ =127.577\\ AB=11.29m\end{array}$

Similarly, the cosine law can be applied in he triangle $OBC$ as two sides, $OB,OC$ and included angle is known thus third side $BC$ can be determined as,

  $\begin{array}{l}B{C}^2=O{B}^2+O{C}^2-2\left(OB\right)\left(OC\right)\cos 146°\\ ={15}^2+{20}^2-2\left(15\right)\left(20\right)\left(-0.829\right)\\ =1122.422\\ BC=33.5m\end{array}$

Therefore, value of triangle solved by using Heron’s area formula is $AB=11.29m,BC=33.5m$