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

Try this beautiful problem from Geometry: Area of Trapezium from 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

But try the problem first…

Answer: $75$

Source

AMC-10A (2018) Problem 24

Pre College Mathematics

## Try with Hints

First hint

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

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

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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