Try this beautiful problem from Geometry: The area of trapezoid
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?
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Geometry
trapezoid
Triangle
But try the problem first...
Answer: $ 10 $
AMC-8 (2003) Problem 21
Pre College Mathematics
First hint
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 ..........
Second Hint
Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm
can you finish the problem........
Final Step
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