Depth · one concept from first principles · step 2 of 4

Graph a moving point's area piece by piece

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=6$ , $\overline{\text{BC}}=8$ , and $\angle \text{C}=90^\circ$ . A point $\text{P}$ starts at $\text{B}$ and moves along the sides of the triangle through $\text{C}$ to $\text{A}$ . Let $x$ be the distance $\text{P}$ has traveled from $\text{B}$ , and let $y$ be the area of $\triangle \text{ABP}$ at that moment. Find the area of the region enclosed by the graph of the relationship between $x$ and $y$ and the $x$ -axis. ( $0<x<14$ )

FunctionsGeometry

Your answer

How to solve

Strategy Identify Subproblems — Split the trip at the corner C. On each side, △ ABP has a fixed height and a base that grows or shrinks steadily, so y is a linear function of x on each piece. Then draw the graph: it is a triangle whose area is easy.

1STEP 1

P on BC

For 0 ＜ x ＜ 8, base BP=x lies on BC and the height is AC=6.

y=1/2 × x × 6=3x

2STEP 2

P on CA

For 8 ≤ x ＜ 14, P is on CA with AP=14-x; take AP as the base, and the height is BC=8.

y=1/2×(14-x) × 8=56-4x

3STEP 3

Draw the graph

At x=8 both rules give 3 × 8=24 and 56-32=24, so the graph rises in a straight line from (0, 0) to the peak (8, 24) and falls to (14, 0).

4STEP 4

Area under the graph

The region is a triangle with base 14 on the x-axis and height 24.

1/2 × 14 × 24=168

Answer

168

The peak 24 equals the area of △ ABC, 1/2 × 8 × 6=24, as it should when P=C. As x→ 14, P approaches A and the area approaches 0, matching 56-4 × 14=0.

Takeaway

When a moving point turns a corner, start a new formula -- each straight stretch gives a straight piece of graph.

P on BC
P on CA
Draw the graph
Area under the graph

Where next?

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▶ Practice — 12 problems