Question Number 221298 by ajfour last updated on 29/May/25
$${Find}\:{the}\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$$${sides}\:{are}\:\sqrt{\mathrm{20}},\:\sqrt{\mathrm{26}}.\:{and}\:\sqrt{\mathrm{34}}\:. \\ $$
Answered by Ghisom last updated on 29/May/25
$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}:\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula} \\ $$$$\mathrm{area}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{11} \\ $$
Commented by ajfour last updated on 29/May/25
Commented by ajfour last updated on 29/May/25
$${yes}! \\ $$
Commented by ajfour last updated on 31/May/25
$${With}\:{help}\:{of}\:{diagram}\:{we}\:{see} \\ $$$$\bigtriangleup=\mathrm{24}−\mathrm{4}−\mathrm{3}−\mathrm{6}=\mathrm{11} \\ $$$${e}.{g}.\:\:\mathrm{4}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\right)\left(\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$
Answered by mr W last updated on 29/May/25
$$\mathrm{cos}\:{A}=\frac{\mathrm{26}+\mathrm{34}−\mathrm{20}}{\mathrm{2}\sqrt{\mathrm{26}×\mathrm{34}}}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{221}}}\:\Rightarrow\mathrm{sin}\:{A}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{221}}} \\ $$$${Area}=\frac{\sqrt{\mathrm{26}×\mathrm{34}}×\mathrm{11}}{\mathrm{2}×\sqrt{\mathrm{221}}}=\mathrm{11} \\ $$
Answered by fantastic last updated on 30/May/25
$${We}\:{can}\:{use}\:{Heron}'{s}\:{formula}\:{to}\:{calculate}\:{the}\:{area} \\ $$
Answered by fantastic last updated on 30/May/25
$${we}\:{will}\:{use}\:{Heron}'{s}\:{formula} \\ $$$$\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\: \\ $$$${where}\:{s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:{and}\:{a},{b},{c}\:{are}\:{the}\:{side}\:{length}\:{of}\:{the}\:{triangle} \\ $$$${here}\:{s}=\left(\frac{\sqrt{\mathrm{20}}+\sqrt{\mathrm{26}}+\sqrt{\mathrm{34}}}{\mathrm{2}}\right). \\ $$$${Calulating}\:{this}\:{we}\:{get}\:\mathrm{11}{sq}.{unit} \\ $$
Answered by MathematicalUser2357 last updated on 31/May/25
$$\mathrm{11} \\ $$