Area in Right Triangle

In the diagram, \angle{ABC} is a right angle. Point D is on \overline{BC}, and \overline{AD} bisects \angle{CAB}. Points E and F are on \overline{AB} and \overline{AC}, respectively, so that AE=3 and AF=10. Given that EB=9 and FC=27, find the integer closest to the area of quadrilateral DCFG.

Source: NCTM Mathematics Teacher

SOLUTION
Angle Bisector Theorem
The angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

\frac{DB}{DC}=\frac{AB}{AC}

Divided Triangle Proposition
When a triangle is divided into two smaller triangles by a segment from one of its vertices to the opposite side (whether or not this segment is an angle bisector), the ratio of the areas of the two smaller triangles is equal to the ratio of the triangles’ bases.

\frac{area\;ADB}{area\;ACD}=\frac{DB}{DC}

By the Pythagorean theorem
AB^2+BC^2=AC^2
12^2+BC^2=37^2
BC^2=\sqrt{1225}
BC=35

Area\;DCFG=area\;ADC-area\;AGF
Calculate the area of ADC
\overline{AD} is the angle bisector of \angle{A} of \triangle{ABC}. By the Angle Bisector theorem
\frac{DC}{DB}=\frac{AC}{AB}=\frac{37}{12}
By the Divided Triangle proposition
\frac{area\;ADC}{area\;ABD}=\frac{DC}{DB}=\frac{37}{12}
\frac{area\;ADC}{37}=\frac{area\;ABD}{12}=\frac{area\;ADC+\;area\;ABD}{37+12}=\frac{area\;ABC}{49}
area\;ADC=\frac{37}{49}\left (area\;ABC\right )
area\;ABC=\frac{1}{2}\left (BC\times BA\right )
=\frac{1}{2}\left (35\times 12\right )
=210
area\;ADC=\frac{37}{49}\left (210\right )
=158.57
Calculate the area of AGF
\overline{AG} is the angle bisector of \angle{A} of \triangle{FAE}. By the Angle Bisector theorem
\frac{GF}{GE}=\frac{AF}{AE}=\frac{10}{3}
By the Divided Triangle proposition
area\;AGF=\frac{10}{13}\left (area\;FAE\right )

\overline{FE} is a segment from vertex F to point E on opposite side \overline{AB} in \triangle{FAB}
area\;FAE=\frac{3}{12}\left (area\;FAB\right )

\overline{BF} is a segment from vertex B to point F on opposite side \overline{AC} in \triangle{ABC}
area\;FAB=\frac{10}{37}\left (area\;ABC\right )

Finally
area\;AGF=\frac{10}{13}\left (area\;FAE\right )
=\frac{10}{13}\left (\frac{3}{12}area\;FAB\right )
=\frac{10}{13}\cdot\frac{3}{12}\left (\frac{10}{37}area\;ABC\right )
=\frac{10}{13}\cdot\frac{3}{12}\cdot\frac{10}{37}\left (210\right )
=10.91
Area\;DCFG=area\;ADC-area\;AGF
=158.57-10.91
=147.66
Integer closest to 147.66 is 148.

Answer: 148 square units

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.

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