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# 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?

,

i

• $8$
• $10$
• $15$

### Key Concepts

Geometry

trapezoid

Triangle

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 ……….

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

can you finish the problem……..

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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