A geometry problem by Wildan Bagus Wicaksono

Geometry Level 3

Build flat A B C D ABCD below is trapezoidal with A B AB parallel to D C DC . Point E E and F F lie on the C D CD so that A D AD is parallel to B E BE and A F AF parallel to B C BC . Point H H is the intersection of A C AC and B E BE . If the length A B = 4 AB = 4 cm and length C D = 10 CD = 10 cm, determine the ratio of triangle A G H AGH and trapezoidal area.

9 : 107 9 : 107 8 : 105 8 : 105 3 : 103 3 : 103 7 : 113 7 : 113 1 : 106 1 : 106

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

We get it

h 1 + h 2 = h h 3 + h 4 = h { h }_{ 1 }+{ h }_{ 2 }=h\\ { h }_{ 3 }+{ h }_{ 4 }=h

The E F H EFH triangle is similar to D F A DFA triangle.

6 h = 2 h 1 3 h 1 = h h 1 = 1 3 h h 2 = 2 3 h \frac { 6 }{ h } =\frac { 2 }{ { h }_{ 1 } } \\ 3{ h }_{ 1 }=h\\ { h }_{ 1 }=\frac { 1 }{ 3 } h\\ { h }_{ 2 }=\frac { 2 }{ 3 } h

The E C G ECG triangle is similar to D C A DCA triangle.

10 h = 6 h 3 3 h = 5 h 3 h 3 = 3 5 h h 4 = 2 5 \frac { 10 }{ h } =\frac { 6 }{ { h }_{ 3 } } \\ 3h=5{ h }_{ 3 }\\ { h }_{ 3 }=\frac { 3 }{ 5 } h\\ { h }_{ 4 }=\frac { 2 }{ 5 }

Area of triangle A G H AGH is

Ratio of triangle A G H AGH and trapezoidal A B C D ABCD area is

[ A G H ] [ A B C D ] = [ A B H ] − [ A B G ] A B + C D 2 h [ A G H ] [ A B C D ] = 1 2 ⋅ 4 ⋅ 2 3 t − 1 2 ⋅ 4 ⋅ 2 5 h 4 + 10 2 h [ A G H ] [ A B C D ] = 4 3 t − 4 5 h 14 2 h [ A G H ] [ A B C D ] = 20 − 12 15 h 7 h [ A G H ] [ A B C D ] = 8 15 7 [ A G H ] [ A B C D ] = 8 105 \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\frac { \left[ ABH \right] -\left[ ABG \right] }{ \frac { AB+CD }{ 2 } h } \\ \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\frac { \frac { 1 }{ 2 } \cdot 4\cdot \frac { 2 }{ 3 } t-\frac { 1 }{ 2 } \cdot 4\cdot \frac { 2 }{ 5 } h }{ \frac { 4+10 }{ 2 } h } \\ \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\frac { \frac { 4 }{ 3 } t-\frac { 4 }{ 5 } h }{ \frac { 14 }{ 2 } h } \\ \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\frac { \frac { 20-12 }{ 15 } h }{ 7h } \\ \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\frac { \frac { 8 }{ 15 } }{ 7 } \\ \frac { \left[ AGH \right] }{ \left[ ABCD \right] } =\boxed { \frac { 8 }{ 105 } }

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