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AMC 8 Geometry Math Olympiad USA Math Olympiad

The area of trapezoid | AMC 8, 2003 | Problem 21

Try this beautiful problem from Geometry: The area of a trapezoid from AMC-8 (2003). You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: The area of trapezoid

The area of trapezoid – AMC-8, 2003- Problem 21


The area of trapezoid ABCD is 164 \(cm^2\). The altitude is  8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters?

The area of trapezoid - problem figure

,

 i

  • $8$
  • $ 10 $
  • $15$

Key Concepts


Geometry

trapezoid

Triangle

Check the Answer


Answer: $ 10 $

AMC-8 (2003) Problem 21

Pre College Mathematics

Try with Hints


Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Can you now finish the problem ……….

finding the area of trapezoid

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

can you finish the problem……..

finding the area of trapezoid

Given that the area of the trapezoid is 164 sq.unit

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

Using Pythagorean rules on the triangle ABD,we have…

\((AD)^2 + (BD)^2 =(AB)^2\)

\( \Rightarrow (AD)^2 + (8)^2 =(10)^2\)

\( \Rightarrow (AD)^2 =(10)^2 – (8)^2 \)

\( \Rightarrow (AD)^2 = 36\)

\( \Rightarrow (AD) =6\)

Using Pythagorean rules on the triangle CED,we have…

\((CE)^2 + (DE)^2 =(DC)^2\)

\( \Rightarrow (CE)^2 + (8)^2 =(17)^2\)

\( \Rightarrow (CE)^2 =(17)^2 – (8)^2 \)

\( \Rightarrow (CE)^2 = 225\)

\( \Rightarrow (CE) =15\)

Let BC= DE=x

Therefore area of the trapezoid=\(\frac{1}{2} \times (AD+BC) \times 8\)=164

\(\Rightarrow \frac{1}{2} \times (6+x+15) \times 8\) =164

\(\Rightarrow x=10\)

Therefore BC=10 cm

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