# Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2004 ,Problem-24, based on area of Rectangle.

## Rectangle | AMC-8, 2004 | Problem 24

In the figure ABCD is a rectangle and EFGH  is a parallelogram. Using the measurements given in the figure, what is the length  d  of the segment that is perpendicular to  HE and FG?

• $7.1$
• $7.6$
• $7.8$

### Key Concepts

Geometry

Rectangle

Parallelogram

Answer:$7.6$

AMC-8, 2004 problem 24

Pre College Mathematics

## Try with Hints

Find Area of the Rectangle and area of the Triangles i.e $(\triangle AHE ,\triangle EBF , \triangle FCG , \triangle DHG)$

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

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of$(\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG)$

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

Area of the Rectangle =$CD \times AD$=$10 \times 8$=80 sq.unit

Area of the $\triangle AHE$ =$\frac{1}{2} \times AH \times AE$= $\frac{1}{2} \times 4 \times 3$ =6 sq.unit

Area of the $\triangle EBF$ =$\frac{1}{2} \times EB \times BE$= $\frac{1}{2} \times 6 \times 5$ =15 sq.unit

Area of the $\triangle FCG$ =$\frac{1}{2} \times GC \times FC$= $\frac{1}{2} \times 4\times 3$ =6 sq.unit

Area of the $\triangle DHG$ =$\frac{1}{2} \times DG \times DH$= $\frac{1}{2} \times 6 \times 5$ =15 sq.unit

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of$(\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG)$=$80-(6+15+6+15)=80-42=38$

As ABCD is a Rectangle ,$\triangle GCF$ is a Right-angle triangle,

Therefore GF=$\sqrt{4^2 + 3^2}$=5 sq.unit

Now Area of the parallelogram EFGH=$GF \times d$=38

$\Rightarrow 5 \times d$=38

$\Rightarrow d=7.6$

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