2022 AMC 10B Problem 2

P1-10~2 minPrint

Quadrilaterals & Polygon Areas Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed

In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ , $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)

Show solution without answering

Key insight

All four sides of a rhombus are equal, so AB = 5. That makes triangle ABP a 3-4-5 right triangle, and its side BP = 4 is the height.

Solution

The area of a rhombus (like any parallelogram) is base times height. The segment $BP$ is perpendicular to $AD$ , so it is the height when $AD$ is the base. We need the lengths $AD$ and $BP$ .

1. The base. $P$ lies on $AD$ , so $AD = AP + PD = 3 + 2 = 5$ .
2. Another side. All four sides of a rhombus are equal, so $AB = 5$ too.
3. The height. Triangle $ABP$ has a right angle at $P$ , a long side $AB = 5$ and a leg $AP = 3$ . By the Pythagorean theorem,
$BP = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.$
(This is the familiar $3$ - $4$ - $5$ right triangle.)
4. The area. Base times height is $5 \cdot 4 = 20$ .

The answer is $\boxed{\textbf{(D)}\ 20}$ .

Why this works

A line dropped from a corner straight down to the opposite side is exactly the height of a parallelogram. The fact that every side of a rhombus has the same length turns the pieces $3$ and $2$ into a right triangle whose long side you already know. When a right angle appears inside a rhombus, look for a right triangle that uses one full side of the rhombus as its longest side.

Concise solution

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

Since $P$ lies on $\overline{AD}$ , the side $AD = AP + PD = 3 + 2 = 5$ . A rhombus has four equal sides, so $AB = 5$ as well.

Triangle $ABP$ is right-angled at $P$ with hypotenuse $AB = 5$ and leg $AP = 3$ , so
$BP = \sqrt{5^2 - 3^2} = 4.$

A rhombus is a parallelogram, and $BP$ is the height onto base $AD$ . Therefore the area is $AD \cdot BP = 5 \cdot 4 = 20$ .

The answer is $\boxed{\textbf{(D)}\ 20}$ .

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 9: Polygons (§9.3)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)