Try this beautiful problem from Geometry based on the Area of the Trapezium

Area of the Trapezium – AMC-10A, 2018- Problem 24


Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

  • $79$
  • $ 75$
  • $82$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


But try the problem first…

Answer: $75$

Source
Suggested Reading

AMC-10A (2018) Problem 24

Pre College Mathematics

Try with Hints


First hint

Area of Trapezium - Problem

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)

can you find out the value of \(BG,DF \) and height between them….?

Can you now finish the problem ……….

Second Hint

Area of Trapezium - figure

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)

From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)

Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)

now can you find out the area of Trapezium ?

can you finish the problem……..

Final Step

Area of shaded Trapezium

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)

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