How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Area of a Triangle | AMC-8, 2000 | Problem 25

Try this beautiful problem from Geometry: Area of a Triangle

## Area of the Triangle- AMC-8, 2000- Problem 25

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $BC$ and $CD$ are joined to form a triangle, the area of that triangle is

• $25$
• $27$
• $29$

### Key Concepts

Geometry

Triangle

square

Answer: $27$

AMC-8 (2000) Problem 25

Pre College Mathematics

## Try with Hints

Area of the triangle =$\frac{1}{2} \times base \times height$

Can you now finish the problem ..........

Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)$

can you finish the problem........

Given that area of the rectangle ABCD=72

Let length AB=$x$ and length of CD=$y$

Therefore DE=EC=$\frac{y}{2}$ and BF=FC=$\frac{x}{2}$

Area of ABCD=$xy$=72

Area of the $\triangle ADE=\frac{1}{2}\times DE \times AD= \frac{1}{2}\times \frac{x}{2} \times y =\frac{xy}{4}=\frac{72}{4}=18$

Area of the $\triangle EFC=\frac{1}{2}\times EC \times FC= \frac{1}{2}\times \frac{x}{2} \times \frac{y}{2} =\frac{xy}{8}=\frac{72}{8}=9$

Area of the $\triangle ABF=\frac{1}{2}\times AB \times BF= \frac{1}{2}\times y\times \frac{x}{2}=\frac{xy}{4}=\frac{72}{4}=18$

Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)=72-(18+8+18)=27$