MR-697 / 16. problem

Two sides of a triangle are 10 cm and 14 cm, and the angle opposite the first is 45°. Calculate the area of the triangle.

${h}_{c}=14·\frac{\sqrt{2}}{2}=7·\overline{)2}·\frac{\sqrt{2}}{\overline{)2}}$

${h}_{c}=7\sqrt{2}$

${10}^{2}={14}^{2}+{c}^{2}-\overline{)2}·14·c·\frac{\sqrt{2}}{\overline{)2}}$

$100=196+{c}^{2}-14\sqrt{2}c$

$196+{c}^{2}-14\sqrt{2}c-100=0$

${c}^{2}-14\sqrt{2}c+96=0$

${c}_{1,2}=\frac{-\left(-14\sqrt{2}\right)±\sqrt{{\left(14\sqrt{2}\right)}^{2}-4·1·96}}{2}$

${c}_{1,2}=\frac{14\sqrt{2}±\sqrt{196·2-384}}{2}$

${c}_{1,2}=\frac{14\sqrt{2}±\sqrt{8}}{2}$

${c}_{1,2}=\frac{14\sqrt{2}±\sqrt{2·4}}{2}=\frac{14\sqrt{2}±\sqrt{2}\sqrt{4}}{2}$

${c}_{1,2}=\frac{14\sqrt{2}±2\sqrt{2}}{2}$

${c}_{1}=\frac{14\sqrt{2}+2\sqrt{2}}{2}=\frac{16\sqrt{2}}{2}=\frac{8·\overline{)2}\sqrt{2}}{\overline{)2}}$

${c}_{1}=8\sqrt{2}$

${c}_{2}=\frac{14\sqrt{2}-2\sqrt{2}}{2}=\frac{12\sqrt{2}}{2}=\frac{6·\overline{)2}\sqrt{2}}{\overline{)2}}$

${c}_{2}=6\sqrt{2}$

${A}_{1}=\frac{{c}_{1}·{h}_{c}}{2}=\frac{8\overline{)\sqrt{2}}·7\overline{)\sqrt{2}}}{\overline{)2}}=8·7$

${A}_{1}=56$

${A}_{2}=\frac{{c}_{2}·{h}_{c}}{2}=\frac{6\overline{)\sqrt{2}}·7\overline{)\sqrt{2}}}{\overline{)2}}=6·7$

${A}_{2}=42$