Question Number 204062 by BaliramKumar last updated on 05/Feb/24
$$\mathrm{I}.\:\:\:\:\:\:\:\mathrm{A}\left(−\mathrm{5},\:−\mathrm{1}\right);\:\mathrm{B}\left(\mathrm{3},\:−\mathrm{5}\right);\:\mathrm{C}\left(\mathrm{5},\:\mathrm{2}\right)\:\:\:\:\:\:{ar}\left(\bigtriangleup\mathrm{ABC}\right)\:=\:? \\ $$$$\mathrm{II}.\:\:\:\:\:\mathrm{A}\left(\mathrm{5},\:\mathrm{3}\right);\:\mathrm{B}\left(\mathrm{2},\:\mathrm{5}\right);\:\mathrm{C}\left(−\mathrm{5},\:\mathrm{3}\right);\:\mathrm{D}\left(−\mathrm{4},\:−\mathrm{3}\right)\:\:\:\:\:\:\:{ar}\left(\Box\mathrm{ABCD}\right)\:=\:? \\ $$$$\mathrm{shortest}\:\mathrm{solution}\: \\ $$
Answered by mr W last updated on 05/Feb/24
$${I}. \\ $$$${A}=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{\mathrm{3}−\left(−\mathrm{5}\right)}&{−\mathrm{5}−\left(−\mathrm{1}\right)}\\{\mathrm{5}−\left(−\mathrm{5}\right)}&{\mathrm{2}−\left(−\mathrm{1}\right)}\end{vmatrix}\mid \\ $$$$\:=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{\mathrm{8}}&{−\mathrm{4}}\\{\mathrm{10}}&{\mathrm{3}}\end{vmatrix}\mid \\ $$$$=\frac{\mathrm{24}+\mathrm{40}}{\mathrm{2}}=\mathrm{32} \\ $$$${II}. \\ $$$${A}=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{−\mathrm{3}}&{\mathrm{2}}\\{−\mathrm{10}}&{\mathrm{0}}\end{vmatrix}\mid+\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{−\mathrm{10}}&{\mathrm{0}}\\{−\mathrm{9}}&{−\mathrm{6}}\end{vmatrix}\mid \\ $$$$\:\:\:=\frac{\mathrm{20}}{\mathrm{2}}+\frac{\mathrm{60}}{\mathrm{2}}=\mathrm{40} \\ $$
Commented by BaliramKumar last updated on 05/Feb/24
$$\mathrm{any}\:\mathrm{other}\:\mathrm{solution}\: \\ $$
Commented by mr W last updated on 05/Feb/24
$${surely}\:{there}\:{are}\:{other}\:{methods}. \\ $$$${but}\:{this}\:{is}\:{the}\:{shortest}\:{solution} \\ $$$${i}\:{think}. \\ $$
Answered by AST last updated on 05/Feb/24
![AB=(√(80));BC=(√(53));AC=(√(109)) [ABC]=(√((((√(80))+(√(53))+(√(109)))((√(80))+(√(53))−(√(109)((√(80))+(√(109))−(√(53)))((√(109))+(√(53))−(√(80))))))/(16))) =(√(((24+2(√(80×53)))(−24+2(√(53×80))))/(16)))=(√((4×53×80−24^2 )/(16))) =32](https://www.tinkutara.com/question/Q204069.png)
$${AB}=\sqrt{\mathrm{80}};{BC}=\sqrt{\mathrm{53}};{AC}=\sqrt{\mathrm{109}} \\ $$$$\left[{ABC}\right]=\sqrt{\frac{\left(\sqrt{\mathrm{80}}+\sqrt{\mathrm{53}}+\sqrt{\mathrm{109}}\right)\left(\sqrt{\mathrm{80}}+\sqrt{\mathrm{53}}−\sqrt{\left.\mathrm{109}\right)\left(\sqrt{\mathrm{80}}+\sqrt{\mathrm{109}}−\sqrt{\mathrm{53}}\right)\left(\sqrt{\mathrm{109}}+\sqrt{\mathrm{53}}−\sqrt{\mathrm{80}}\right)}\right.}{\mathrm{16}}} \\ $$$$=\sqrt{\frac{\left(\mathrm{24}+\mathrm{2}\sqrt{\mathrm{80}×\mathrm{53}}\right)\left(−\mathrm{24}+\mathrm{2}\sqrt{\mathrm{53}×\mathrm{80}}\right)}{\mathrm{16}}}=\sqrt{\frac{\mathrm{4}×\mathrm{53}×\mathrm{80}−\mathrm{24}^{\mathrm{2}} }{\mathrm{16}}} \\ $$$$=\mathrm{32} \\ $$
Answered by AST last updated on 05/Feb/24
![Translate any vertex say C(5,2)→C(0,0) ⇒A(−5,−1)→A(−10,−3)∧B(3,−5)→B(−2,−7) ⇒[ABC]=∣(1/2) determinant (((−10),(−2)),((−3),(−7)))∣=((64)/2)=32](https://www.tinkutara.com/question/Q204070.png)
$${Translate}\:{any}\:{vertex}\:{say}\:{C}\left(\mathrm{5},\mathrm{2}\right)\rightarrow{C}\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\Rightarrow{A}\left(−\mathrm{5},−\mathrm{1}\right)\rightarrow{A}\left(−\mathrm{10},−\mathrm{3}\right)\wedge{B}\left(\mathrm{3},−\mathrm{5}\right)\rightarrow{B}\left(−\mathrm{2},−\mathrm{7}\right) \\ $$$$\Rightarrow\left[{ABC}\right]=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{−\mathrm{10}}&{−\mathrm{2}}\\{−\mathrm{3}}&{−\mathrm{7}}\end{vmatrix}\mid=\frac{\mathrm{64}}{\mathrm{2}}=\mathrm{32} \\ $$
Answered by AST last updated on 05/Feb/24
![Let C(−5,3)→(0,0);[ABCD]=[ABD]+[BCD] ⇒A(5,3)→A(10,0);B(2,5)→B(7,2);D→(1,−6) ⇒[BCD]=∣(1/2) determinant ((7,1),(2,(−6)))∣=22 Let A→(0,0)⇒B→(−3,2);D→(−9,−6) ⇒[ABD]=∣(1/2) determinant (((−3),(−9)),(2,(−6)))∣=18⇒[ABCD]=40](https://www.tinkutara.com/question/Q204076.png)
$${Let}\:{C}\left(−\mathrm{5},\mathrm{3}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right);\left[{ABCD}\right]=\left[{ABD}\right]+\left[{BCD}\right] \\ $$$$\Rightarrow{A}\left(\mathrm{5},\mathrm{3}\right)\rightarrow{A}\left(\mathrm{10},\mathrm{0}\right);{B}\left(\mathrm{2},\mathrm{5}\right)\rightarrow{B}\left(\mathrm{7},\mathrm{2}\right);{D}\rightarrow\left(\mathrm{1},−\mathrm{6}\right) \\ $$$$\Rightarrow\left[{BCD}\right]=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{\mathrm{7}}&{\mathrm{1}}\\{\mathrm{2}}&{−\mathrm{6}}\end{vmatrix}\mid=\mathrm{22} \\ $$$${Let}\:{A}\rightarrow\left(\mathrm{0},\mathrm{0}\right)\Rightarrow{B}\rightarrow\left(−\mathrm{3},\mathrm{2}\right);{D}\rightarrow\left(−\mathrm{9},−\mathrm{6}\right) \\ $$$$\Rightarrow\left[{ABD}\right]=\mid\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{−\mathrm{3}}&{−\mathrm{9}}\\{\mathrm{2}}&{−\mathrm{6}}\end{vmatrix}\mid=\mathrm{18}\Rightarrow\left[{ABCD}\right]=\mathrm{40} \\ $$