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| 1. |
ABC is a triangle where angleB=135. Prove that AC sq=AB sq+BCsq+4arABC |
| Answer» Construction:- Draw\xa0{tex}A D \\perp B C{/tex}Proof:-In\xa0{tex}\\triangle ADB{/tex},\xa0By using\xa0Pythagoras theorem, we get\xa0AD2\xa0=\xa0AB2 - BD2\xa0....(i)In\xa0{tex}\\triangle ADC{/tex},\xa0By using\xa0Pythagoras theorem, we get\xa0AC2 = AD2 + DC2{tex}\\Rightarrow{/tex}\xa0AC2 = AB2 - BD2 +(BD + BC)2 [from (i)]{tex}\\Rightarrow{/tex}\xa0AC2 = AB2 - BD2 + BD2 + BC2 + 2BC\xa0{tex}\\times{/tex}\xa0BD{tex}\\Rightarrow{/tex}\xa0AC2 = AB2 + BC2 + 2BC\xa0{tex}\\times{/tex}\xa0BD ...(ii){tex}\\therefore{/tex}\xa0Area of\xa0{tex}\\triangle ABC{/tex}\xa0=\xa0{tex}\\frac { 1 } { 2 } B C \\times A D{/tex}{tex}\\Rightarrow \\quad 2 \\operatorname { ar } ( \\Delta A B C ) = B C \\times A D{/tex}\xa0...(iii){tex}\\because \\angle A B C = 135{/tex}Then,\xa0{tex}\\angle A B D = 180 ^ { \\circ } - 135 ^ { \\circ } = 45 ^ { \\circ }{/tex}In\xa0{tex}\\triangle ADB{/tex}{tex}\\tan 45 ^ { \\circ } = \\frac { A D } { D B }{/tex}{tex}\\Rightarrow \\quad 1 = \\frac { A D } { B D }{/tex}{tex}\\Rightarrow{/tex}{tex}\xa0BD = AD{/tex}\xa0...(iv)from (ii) and (iv){tex}2 a r ( \\triangle A B C ) = B C \\times B D{/tex}\xa0...(v)From (ii) and (v)AC2 = AB2 + BC2 +\xa0{tex}2 \\times 2 a r ( \\triangle A B C ){/tex}{tex}\\Rightarrow{/tex}\xa0{tex}AC^2 = AB^2 + BC^2 + 4ar{/tex}\xa0({tex}\\triangle ABC{/tex}) | |