Area of Rectangle Problem | 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?

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

\(\Rightarrow d=7.6\)

AMC - AIME Boot Camp... for brilliant students.

Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

Learn More

Subscribe to Cheenta at Youtube