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March 27, 2020

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?

Area of Rectangle Problem
  • $7.1$
  • $7.6$
  • $7.8$

Key Concepts


Geometry

Rectangle

Parallelogram

Check the Answer


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 Rectangle 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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