2002 AMC 10A Problem 25

P21-25~6 minPrint

Quadrilaterals & Polygon Areas Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Cut the figure into known shapes (triangles, rectangles, sectors)

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

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Key insight

Slide leg DA over to start at C. It cuts off a 5-12-13 right triangle (13 = 52 - 39), and that triangle's height is the trapezoid's height.

Solution

For the area we need the height, which is not given. The trick is to move one slanted side so that both slanted sides end up in one triangle.

1. Cut off a triangle. From $C$ , draw a segment parallel to $DA$ , meeting $AB$ at $E$ . Now $AECD$ has both pairs of opposite sides parallel, so it is a parallelogram, and
$AE = CD = 39, \qquad EC = DA = 5.$

2. The leftover triangle $EBC$ has sides
$EB = 52 - 39 = 13, \qquad EC = 5, \qquad BC = 12.$
These are $5$ , $12$ , $13$ : since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ , it is a right triangle with the right angle at $C$ .

Its area. The legs $5$ and $12$ are a base and a height: area $= \frac12 \times 5 \times 12 = 30$ .

4. The height of the trapezoid. The side $EB = 13$ lies along $AB$ , and the height from $C$ down to $AB$ is the trapezoid's height $h$ . Using $EB$ as the base, the same area is $\frac12 \times 13 \times h$ . So
$\frac12 \times 13 \times h = 30, \qquad h = \frac{60}{13}.$

5. Area of the trapezoid. Average the two bases and multiply by the height:
$\frac{52 + 39}{2} \times \frac{60}{13} = \frac{91}{2} \times \frac{60}{13} = 7 \times 30 = 210.$
(Check another way: the parallelogram has area $39 \times \frac{60}{13} = 180$ , and adding the triangle's $30$ gives $210$ .)

The answer is $\boxed{\textbf{(C)}\ 210}$ .

Why this works

A trapezoid is a parallelogram plus a triangle. Sliding one leg across puts both legs into one triangle whose third side is the difference of the bases. Here that triangle is the $5$ - $12$ - $13$ right triangle, and computing its area two ways gives the height. When legs and the difference of the bases form a familiar right triangle, expect this trick.

Alternative approach

Extend the two legs until they meet at a point $P$ above $CD$ . Triangle $PDC$ is a smaller copy of triangle $PAB$ , with sides $\frac{39}{52} = \frac34$ as long. Since $AD = 5$ is the missing quarter of $PA$ , $PA = 20$ and $PD = 15$ ; in the same way $PB = 48$ and $PC = 36$ . Triangle $PAB$ has sides $20$ , $48$ , $52$ , which is $5$ - $12$ - $13$ times $4$ , so it has a right angle at $P$ . Then
$\text{area} = \frac12 \times 20 \times 48 - \frac12 \times 15 \times 36 = 480 - 270 = 210.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Through $C$ draw the segment parallel to $DA$ , meeting $AB$ at $E$ . Then $AECD$ is a parallelogram, so $AE = CD = 39$ and $CE = DA = 5$ . The leftover piece is triangle $EBC$ with
$EB = 52 - 39 = 13,\qquad CE = 5,\qquad BC = 12.$
Since $5^2 + 12^2 = 13^2$ , this triangle is right-angled at $C$ , and its area is $\frac{1}{2}\cdot 5 \cdot 12 = 30$ .

The height of the trapezoid equals the altitude of this triangle onto its hypotenuse $EB$ (which lies along $AB$ ):
$h = \frac{2 \cdot 30}{13} = \frac{60}{13}.$

Now apply the trapezoid area formula:
$[ABCD] = \frac{52 + 39}{2}\cdot\frac{60}{13} = \frac{91}{2}\cdot\frac{60}{13} = 7 \cdot 30 = 210.$

The answer is $\boxed{\textbf{(C)}\ 210}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 9: Polygons (§9.3)