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# Area of trapezoid | AMC 8, 2011|Problem 20 Try this beautiful problem from Geometry: The area of trapezoid.

## The area of the trapezoid - AMC-8, 2011 - Problem 20

Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

• $700$
• $750$
• $800$

### Key Concepts

Geometry

Trapezoid

Area of Triangle

Answer:$750$

AMC-8(2011) Problem 20

Pre College Mathematics

## Try with Hints

Draw altitudes from the top points A and B to CD at X and Y points

Can you now finish the problem ..........

The area of the trapezoid is $\frac{1}{2} \times (AB+CD) \times$ (height between AB and CD)

can you finish the problem........

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on $\triangle ADX and \triangle BYC$ ,

$(DX)^2+(AX)^2=(AD)^2$

$\Rightarrow (a)^2+(12)^2=(15)^2$

$\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9$

and

$(BY)^2+(YC)^2=(BC)^2$

$\Rightarrow (12)^2+(b)^2=(20)^2$

$\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16$

Now ABYX is a Rectangle so $XY=AB=50$

$CD=DX+XY+YC=a+XY+b=9+50+16=75$

The area of the trapezoid is $\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750$

.

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Try this beautiful problem from Geometry: The area of trapezoid.

## The area of the trapezoid - AMC-8, 2011 - Problem 20

Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

• $700$
• $750$
• $800$

### Key Concepts

Geometry

Trapezoid

Area of Triangle

Answer:$750$

AMC-8(2011) Problem 20

Pre College Mathematics

## Try with Hints

Draw altitudes from the top points A and B to CD at X and Y points

Can you now finish the problem ..........

The area of the trapezoid is $\frac{1}{2} \times (AB+CD) \times$ (height between AB and CD)

can you finish the problem........

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on $\triangle ADX and \triangle BYC$ ,

$(DX)^2+(AX)^2=(AD)^2$

$\Rightarrow (a)^2+(12)^2=(15)^2$

$\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9$

and

$(BY)^2+(YC)^2=(BC)^2$

$\Rightarrow (12)^2+(b)^2=(20)^2$

$\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16$

Now ABYX is a Rectangle so $XY=AB=50$

$CD=DX+XY+YC=a+XY+b=9+50+16=75$

The area of the trapezoid is $\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750$

.

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