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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

Geometry

Triangle

Trapezium

Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

## Try with Hints

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 ……….

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……..

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