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

Try this beautiful problem from Geometry:Area of Trapezium.AMC-10A, 2018. You may use sequential hints to solve the problem

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




Check the Answer

Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

Try with Hints

Area of the Trapezium

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 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 ……….

Area of the Trapezium- Problem

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 and area of Triangle?

can you finish the problem……..

Area of the shaded portion

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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