Chapter 6.5, Problem 17CYU

To determine

The altitude of the triangle whose base and area are $12+\sqrt{3}cm$ and $189+4\sqrt{3}c{m}^2$ respectively.

Answer to Problem 17CYU

The altitude of the triangle is $\frac{378+8\sqrt{3}}{12+\sqrt{3}}cm$ .

Explanation of Solution

Given information: The base of the triangle is $12+\sqrt{3}cm$ and area is $189+4\sqrt{3}c{m}^2$ .

Calculation: From the formula of area of triangle,

  $\begin{array}{c}\text{Area of triangle}\left(A\right)=\frac{1}{2}\left(\text{base}\right)\left(\text{altitude}\right)\\ 189+4\sqrt{3}=\frac{1}{2}\left(12+\sqrt{3}\right)\left(\text{altitude}\right)\\ \text{or, 2}\left(189+4\sqrt{3}\right)=\left(12+\sqrt{3}\right)\left(\text{altitude}\right)\\ 378+8\sqrt{3}=\left(12+\sqrt{3}\right)\left(\text{altitude}\right)\\ \text{or, altitude}=\frac{378+8\sqrt{3}}{12+\sqrt{3}}cm\end{array}$