Chapter 11.2, Problem 15WE

To determine

To Find: The area.

Answer to Problem 15WE

  $A=\frac{25\sqrt{3}}{2}sq.unit$

Explanation of Solution

Given: A ${30}^{°}-{60}^{°}-{90}^{°}$ triangle with hypotenuse $10cm$ .

  

Here

  $\begin{array}{l}AC=2BC\\ AB=\sqrt{3}BC\\ BC=\frac{1}{2}AC\\ AB=\sqrt{3}\left(\frac{AC}{2}\right)\\ Now\\ BC=\frac{1}{2}(10)=5\\ AB=\sqrt{3}\left(\frac{10}{2}\right)=5\sqrt{3}\end{array}$

Now in triangle $ABC$

  $\begin{array}{l}b=BC=5\\ h=AB=5\sqrt{3}\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(5)\left(5\sqrt{3}\right)\\ A=\frac{25\sqrt{3}}{2}sq.unit\end{array}$

Hence, $A=\frac{25\sqrt{3}}{2}sq.unit$