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## Trapezoid Problem | AIME I, 1992 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on trapezoid.

## Trapezoid – AIME I, 1992

Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre P on AB is drawn tangent to BC and AD. Given that AP=$\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 164
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Trapezoid

Angle Bisectors

AIME I, 1992, Question 9

Coordinate Geometry by Loney

## Try with Hints

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

$\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}$

solving we get 120y=(70)(92)

or, AP=y=$\frac{161}{3}$

or, 161+3=164.

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## Problem on Area of Trapezoid | AMC-10A, 2002 | Problem 25

Try this beautiful problem on area of trapezoid from Geometry.

## Problem on Area of Trapezoid – AMC-10A, 2002- Problem 25

In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$. The area of $ABCD$ is

• $182$
• $195$
• $210$
• $234$
• $260$

### Key Concepts

Geometry

Trapezoid

Triangle

Answer: $210$

AMC-10A (2002) Problem 25

Pre College Mathematics

## Try with Hints

Given that $ABCD$ is a Trapezium with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$.we have to find out the area of the Trapezium.Normally the area of the trapezium is $\frac{1}{2} (AD +BC) \times$(height between CD & AB).but we don’t know the height.So another way if we extend $AD$ & $BC$ ,they will meet a point $E$.Now clearly Area of $ABCD$=Area of $\triangle ABE$ – Area of $\triangle EDC$.Can you find out the area of $\triangle EAB$ & Area of $\triangle EDC$?

Can you now finish the problem ……….

Now $AB||DC$ , Therefore $\triangle EDC \sim \triangle EAB$

$\Rightarrow \frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}$

$\Rightarrow \frac{ED}{ED+DA}=\frac{EC}{EC+BC}=\frac{DC}{AB}$

$\Rightarrow \frac{ED}{ED+5}=\frac{EC}{EC+12}=\frac{39}{52}$

Now , $\frac{ED}{ED+5}=\frac{39}{52}$

$\Rightarrow ED=15$

And $\frac{EC}{EC+12}=\frac{39}{52}$

$\Rightarrow CE=36$

Therefore $BE$=$12+36=48$ and $AE=20$

Notice that in the $\triangle EDC$, ${ED}^2 +{EC}^2=(36)^2+(15)^2=(39)^2=(DC)^2$ $\Rightarrow \triangle EDC$ is a Right-angle Triangle

Therefore Area of $\triangle EDC=\frac{1}{2} \times 36 \times 15=270$

Similarly In the $\triangle EAB$, ${EA}^2 +{EB}^2=(48)^2+(20)^2=(52)^2=(AB)^2$ $\Rightarrow \triangle EAB$ is a Right-angle Triangle

Therefore Area of $\triangle EAB=\frac{1}{2} \times 48 \times 20=480$

Now can you find out the area of $ABCD$?

can you finish the problem……..

Therefore Area of $ABCD$=Area of $\triangle ABE$ – Area of $\triangle EDC$=$480-270=210$

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## Problem on Semicircle | AMC 8, 2013 | Problem 20

Try this beautiful problem from Geometry based on Semicircle.

## Area of the Semicircle – AMC 8, 2013 – Problem 20

A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

• $\frac{\pi}{2}$
• $\pi$
• $\frac{\pi}{3}$

### Key Concepts

Geometry

Square

semi circle

Answer:$\pi$

AMC-8 (2013) Problem 20

Pre College Mathematics

## Try with Hints

At first we have to find out the radius of the semicircle for the area of the semicircle.Now in the diagram,AC is the radius of the semicircle and also AC is the hypotenuse of the right Triangle ABC.

Can you now finish the problem ……….

Now AB=1 AND AC=1 (As ABDE $1\times 2$ rectangle .So using pythagorean theorm we can eassily get the value of AC .and area of semicircle =$\frac{\pi r^2}{2}$

can you finish the problem……..

Given that ABDE is a square whose AB=1 and BD=2

Therefore BC=1

Clearly AC be the radius of the given semi circle

From the $\triangle ABC$,$(AB)^2+(BC)^2=(AC)^2\Rightarrow AC=\sqrt{(1^2+1^2)}=\sqrt2$

Therefore the area of the semicircle=$\frac{1}{2}\times \pi(\sqrt 2)^2$=$\pi$

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## Area of the Trapezoid | AMC 8, 2002 | Problem 20

Try this beautiful problem from Geometry based on Area of Trapezoid.

## Area of the Trapezoid – AMC- 8, 2002 – Problem 20

The area of triangle XYZ is 8  square inches. Points  A and B  are midpoints of congruent segments  XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?

• $6$
• $4$
• $3$

### Key Concepts

Geometry

Triangle

Trapezoid

Answer:$3$

AMC-8 (2002) Problem 20

Pre College Mathematics

## Try with Hints

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Let us assume that the length of YZ=$x$ and length of $XC$= $y$

Can you now finish the problem ……….

Therefore area of the trapezoid= $\frac{1}{2} \times (YC+AO) \times OC$

can you finish the problem……..

Let us assume that the length of YZ=$x$ and length of $XC$= $y$

Given that area of $\triangle xyz$=8

Therefore $\frac{1}{2} \times YZ \times XC$=8

$\Rightarrow \frac{1}{2} \times x \times y$ =8

$\Rightarrow xy=16$

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Then by the mid point theorm we can say that $AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}$ and $OC=\frac{1}{2} XC=\frac{y}{2}$

Therefore area of the trapezoid shaded area = $\frac{1}{2} \times (YC+AO) \times OC$= $\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}$ =$\frac{3xy}{16}=3$ (as $xy$=16)

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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 $\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750$

.