MR-696 / 15. problem

Calculate the area of an isosceles triangle with a base of 12 cm, if the height corresponding to the base is equal to the segment joining the midpoints of the base and the leg.

The angle EFC is also α because the leg of the angle AF is normal to the leg CF and leg AE  is normal to leg EF (Angles with normal legs).

${b}^{2}={6}^{2}+{{h}_{a}}^{2}$

${{h}_{a}}^{2}=\frac{{b}^{2}}{4}$

${h}_{a}=\frac{b}{2}$

${b}^{2}={6}^{2}+\frac{{b}^{2}}{4}{|}{·}{4}$

$4{b}^{2}=4·{6}^{2}+\overline{)4}·\frac{{b}^{2}}{\overline{)4}}$

$4{b}^{2}-{b}^{2}=4·{6}^{2}$

$3{b}^{2}=4·{6}^{2}$

${b}^{2}=\frac{4·{6}^{2}}{3}$

$b=\sqrt{\frac{4·{6}^{2}}{3}}=\frac{\sqrt{4}·\sqrt{{6}^{2}}}{\sqrt{3}}=\frac{2·6}{\sqrt{3}}·\frac{\sqrt{3}}{\sqrt{3}}=\frac{2·6·\sqrt{3}}{3}=\frac{2·2·\overline{)3}·\sqrt{3}}{\overline{)3}}$

$b=4\sqrt{3}$

${h}_{a}=\frac{b}{2}=\frac{4\sqrt{3}}{2}=\frac{2·\overline{){2}}·\sqrt{3}}{\overline{)2}}$

${h}_{a}=2\sqrt{3}$

$A=\frac{a·{h}_{a}}{2}=\frac{12·\overline{)2}\sqrt{3}}{\overline{)2}}$

$A=12\sqrt{3}$