Intersection Mania

Geometry Level 3

Give ABC is a triangle have $AB=k;\quad \widehat { BAC } ={ 60 }^{ \circ };\quad \widehat { ACB } ={ 45 }^{ \circ };\quad AD\bot BC\quad (D\in BC);\quad \\ BE\bot AC\quad (E\in AC);\quad CF\bot AB\quad (F\in BA)$ . Calculated according to $k$ , the perimeter of $\Delta ABC$ is:

{ p }_{ \Delta ABC }=\frac { k(3+\sqrt { 5 } +\sqrt { 2 } ) }{ 3 }

{ p }_{ \Delta ABC }=12k\sqrt { 35 }

None of these choices

{ p }_{ \Delta ABC }=\frac { k(3+\sqrt { 6 } +\sqrt { 3 } ) }{ 2 }

1 solution

Tôn Ngọc Minh Quân
Dec 14, 2015

$*\quad In\quad \Delta ABE\quad have:\quad \frac { AE }{ AB } =cosBAE=cos{ 60 }^{ 0 }=\frac { 1 }{ 2 } \Rightarrow AE=\frac { AB }{ 2 } =\frac { k }{ 2 } .\\ *\frac { BE }{ AB } =sinBAE=sin{ 60 }^{ 0 }=\frac { \sqrt { 3 } }{ 2 } \Rightarrow BE=\frac { AB\sqrt { 3 } }{ 2 } =\frac { k\sqrt { 3 } }{ 2 } .\\ *Easy\quad to\quad know\quad CE=BE=\frac { k\sqrt { 3 } }{ 2 } \Rightarrow AC=AE+CE=\frac { k(1+\sqrt { 3 } ) }{ 2 } .\\ *\frac { BE }{ CB } =sinBCE=sin{ 45 }^{ 0 }=\frac { \sqrt { 2 } }{ 2 } \Rightarrow BC=BE:\frac { \sqrt { 2 } }{ 2 } =\frac { k\sqrt { 6 } }{ 2 } .\\ \Rightarrow { p }_{ \Delta ABC }=AB+AC+BC=\frac { k(3+\sqrt { 6 } +\sqrt { 3 } ) }{ 2 } .$

Intersection Mania

1 solution

0 pending reports