Chapter 11, Problem 17RP

To determine

To calculate:Thearea oftriangleABC .

Answer to Problem 17RP

The area of triangleABCis $360$ .

Explanation of Solution

Given information:

In triangleABC ,

Side AC =18,

Side AB = $41$ ,

Side BC = $41$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

The perpendicular BD to base AC forms two $9-40-41\Delta \text{'}s$

Side BD can be calculated by applying Pythagoras Theorem.

In right angled triangle BDA , we get

  $\begin{array}{l}AD=DC=\frac{AC}{2}=\frac{18}{2}=9\\ {\left(BA\right)}^2={\left(BD\right)}^2+{\left(DA\right)}^2\end{array}$

  $\begin{array}{l}{\left(BA\right)}^2={\left(40\right)}^2+{\left(9\right)}^2\\ =1600+81\\ =1681\\ \therefore BA=\sqrt{1681}=41\end{array}$

Area of triangle:

  $\begin{array}{l}A(△ABC)=\frac{1}{2}\ast AC\ast BD\\ A(△ABC)=\frac{1}{2}\ast 18\ast 40\\ A(△ABC)=360\end{array}$