Practice · Graph a moving point's area piece by piece · Sensim Math

Variant 1 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=8$ , $\overline{\text{BC}}=10$ , 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. ( $0<x<18$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=10$ , $\overline{\text{AC}}=8$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<18$ .

Givens

$\overline{\text{AC}}=8$ , $\overline{\text{BC}}=10$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<18$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=10$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<10

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=8

.

y=\frac{1}{2}\times x\times 8=4x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

10\le x<18

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=18-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=10

.

y=\frac{1}{2}\times(18-x)\times 10=90-5x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=10

both rules give

4\times 10=40

and

90-50=40

, so the graph rises in a straight line from

(0, 0)

to the peak

(10, 40)

and falls to

(18, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

18

on the

x

-axis and height

40

.

\frac{1}{2}\times 18\times 40=360

Once the graph is drawn, the question is plain triangle area.

Answer:

360

4 · Reviewdoes it hold up?

The peak $40$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 10\times 8=40$ , as it should when $\text{P}=\text{C}$ . As $x\to 18$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $90-5\times 18=0$ .

Another way: Split the region at $x=10$ instead: the left triangle has area $\frac{1}{2}\times 10\times 40=200$ and the right one $\frac{1}{2}\times 8\times 40=160$ , and $200+160=360$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 2 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=10$ , $\overline{\text{BC}}=6$ , 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. ( $0<x<16$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=6$ , $\overline{\text{AC}}=10$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<16$ .

Givens

$\overline{\text{AC}}=10$ , $\overline{\text{BC}}=6$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<16$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=6$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<6

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=10

.

y=\frac{1}{2}\times x\times 10=5x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

6\le x<16

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=16-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=6

.

y=\frac{1}{2}\times(16-x)\times 6=48-3x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=6

both rules give

5\times 6=30

and

48-18=30

, so the graph rises in a straight line from

(0, 0)

to the peak

(6, 30)

and falls to

(16, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

16

on the

x

-axis and height

30

.

\frac{1}{2}\times 16\times 30=240

Once the graph is drawn, the question is plain triangle area.

Answer:

240

4 · Reviewdoes it hold up?

The peak $30$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 6\times 10=30$ , as it should when $\text{P}=\text{C}$ . As $x\to 16$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $48-3\times 16=0$ .

Another way: Split the region at $x=6$ instead: the left triangle has area $\frac{1}{2}\times 6\times 30=90$ and the right one $\frac{1}{2}\times 10\times 30=150$ , and $90+150=240$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 3 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=2$ , $\overline{\text{BC}}=12$ , 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. ( $0<x<14$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=12$ , $\overline{\text{AC}}=2$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<14$ .

Givens

$\overline{\text{AC}}=2$ , $\overline{\text{BC}}=12$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<14$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=12$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<12

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=2

.

y=\frac{1}{2}\times x\times 2=1x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

12\le x<14

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=14-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=12

.

y=\frac{1}{2}\times(14-x)\times 12=84-6x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=12

both rules give

1\times 12=12

and

84-72=12

, so the graph rises in a straight line from

(0, 0)

to the peak

(12, 12)

and falls to

(14, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

14

on the

x

-axis and height

12

.

\frac{1}{2}\times 14\times 12=84

Once the graph is drawn, the question is plain triangle area.

Answer:

84

4 · Reviewdoes it hold up?

The peak $12$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 12\times 2=12$ , as it should when $\text{P}=\text{C}$ . As $x\to 14$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $84-6\times 14=0$ .

Another way: Split the region at $x=12$ instead: the left triangle has area $\frac{1}{2}\times 12\times 12=72$ and the right one $\frac{1}{2}\times 2\times 12=12$ , and $72+12=84$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 4 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=2$ , $\overline{\text{BC}}=8$ , 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. ( $0<x<10$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=8$ , $\overline{\text{AC}}=2$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<10$ .

Givens

$\overline{\text{AC}}=2$ , $\overline{\text{BC}}=8$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<10$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=8$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<8

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=2

.

y=\frac{1}{2}\times x\times 2=1x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

8\le x<10

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=10-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=8

.

y=\frac{1}{2}\times(10-x)\times 8=40-4x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=8

both rules give

1\times 8=8

and

40-32=8

, so the graph rises in a straight line from

(0, 0)

to the peak

(8, 8)

and falls to

(10, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

10

on the

x

-axis and height

8

.

\frac{1}{2}\times 10\times 8=40

Once the graph is drawn, the question is plain triangle area.

Answer:

40

4 · Reviewdoes it hold up?

The peak $8$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 8\times 2=8$ , as it should when $\text{P}=\text{C}$ . As $x\to 10$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $40-4\times 10=0$ .

Another way: Split the region at $x=8$ instead: the left triangle has area $\frac{1}{2}\times 8\times 8=32$ and the right one $\frac{1}{2}\times 2\times 8=8$ , and $32+8=40$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 5 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=4$ , $\overline{\text{BC}}=6$ , 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. ( $0<x<10$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=6$ , $\overline{\text{AC}}=4$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<10$ .

Givens

$\overline{\text{AC}}=4$ , $\overline{\text{BC}}=6$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<10$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=6$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<6

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=4

.

y=\frac{1}{2}\times x\times 4=2x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

6\le x<10

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=10-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=6

.

y=\frac{1}{2}\times(10-x)\times 6=30-3x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=6

both rules give

2\times 6=12

and

30-18=12

, so the graph rises in a straight line from

(0, 0)

to the peak

(6, 12)

and falls to

(10, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

10

on the

x

-axis and height

12

.

\frac{1}{2}\times 10\times 12=60

Once the graph is drawn, the question is plain triangle area.

Answer:

60

4 · Reviewdoes it hold up?

The peak $12$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 6\times 4=12$ , as it should when $\text{P}=\text{C}$ . As $x\to 10$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $30-3\times 10=0$ .

Another way: Split the region at $x=6$ instead: the left triangle has area $\frac{1}{2}\times 6\times 12=36$ and the right one $\frac{1}{2}\times 4\times 12=24$ , and $36+24=60$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 6 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=2$ , $\overline{\text{BC}}=6$ , 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. ( $0<x<8$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=6$ , $\overline{\text{AC}}=2$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<8$ .

Givens

$\overline{\text{AC}}=2$ , $\overline{\text{BC}}=6$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<8$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=6$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<6

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=2

.

y=\frac{1}{2}\times x\times 2=1x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

6\le x<8

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=8-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=6

.

y=\frac{1}{2}\times(8-x)\times 6=24-3x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=6

both rules give

1\times 6=6

and

24-18=6

, so the graph rises in a straight line from

(0, 0)

to the peak

(6, 6)

and falls to

(8, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

8

on the

x

-axis and height

6

.

\frac{1}{2}\times 8\times 6=24

Once the graph is drawn, the question is plain triangle area.

Answer:

24

4 · Reviewdoes it hold up?

The peak $6$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 6\times 2=6$ , as it should when $\text{P}=\text{C}$ . As $x\to 8$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $24-3\times 8=0$ .

Another way: Split the region at $x=6$ instead: the left triangle has area $\frac{1}{2}\times 6\times 6=18$ and the right one $\frac{1}{2}\times 2\times 6=6$ , and $18+6=24$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 7 medium

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. ( $0<x<14$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=8$ , $\overline{\text{AC}}=6$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<14$ .

Givens

$\overline{\text{AC}}=6$ , $\overline{\text{BC}}=8$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<14$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=8$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<8

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=6

.

y=\frac{1}{2}\times x\times 6=3x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

8\le x<14

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=14-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=8

.

y=\frac{1}{2}\times(14-x)\times 8=56-4x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=8

both rules give

3\times 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)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

14

on the

x

-axis and height

24

.

\frac{1}{2}\times 14\times 24=168

Once the graph is drawn, the question is plain triangle area.

Answer:

168

4 · Reviewdoes it hold up?

The peak $24$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 8\times 6=24$ , as it should when $\text{P}=\text{C}$ . As $x\to 14$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $56-4\times 14=0$ .

Another way: Split the region at $x=8$ instead: the left triangle has area $\frac{1}{2}\times 8\times 24=96$ and the right one $\frac{1}{2}\times 6\times 24=72$ , and $96+72=168$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 8 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=4$ , $\overline{\text{BC}}=10$ , 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. ( $0<x<14$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=10$ , $\overline{\text{AC}}=4$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<14$ .

Givens

$\overline{\text{AC}}=4$ , $\overline{\text{BC}}=10$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<14$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=10$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<10

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=4

.

y=\frac{1}{2}\times x\times 4=2x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

10\le x<14

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=14-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=10

.

y=\frac{1}{2}\times(14-x)\times 10=70-5x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=10

both rules give

2\times 10=20

and

70-50=20

, so the graph rises in a straight line from

(0, 0)

to the peak

(10, 20)

and falls to

(14, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

14

on the

x

-axis and height

20

.

\frac{1}{2}\times 14\times 20=140

Once the graph is drawn, the question is plain triangle area.

Answer:

140

4 · Reviewdoes it hold up?

The peak $20$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 10\times 4=20$ , as it should when $\text{P}=\text{C}$ . As $x\to 14$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $70-5\times 14=0$ .

Another way: Split the region at $x=10$ instead: the left triangle has area $\frac{1}{2}\times 10\times 20=100$ and the right one $\frac{1}{2}\times 4\times 20=40$ , and $100+40=140$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 9 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=6$ , $\overline{\text{BC}}=12$ , 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. ( $0<x<18$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=12$ , $\overline{\text{AC}}=6$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<18$ .

Givens

$\overline{\text{AC}}=6$ , $\overline{\text{BC}}=12$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<18$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=12$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<12

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=6

.

y=\frac{1}{2}\times x\times 6=3x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

12\le x<18

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=18-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=12

.

y=\frac{1}{2}\times(18-x)\times 12=108-6x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=12

both rules give

3\times 12=36

and

108-72=36

, so the graph rises in a straight line from

(0, 0)

to the peak

(12, 36)

and falls to

(18, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

18

on the

x

-axis and height

36

.

\frac{1}{2}\times 18\times 36=324

Once the graph is drawn, the question is plain triangle area.

Answer:

324

4 · Reviewdoes it hold up?

The peak $36$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 12\times 6=36$ , as it should when $\text{P}=\text{C}$ . As $x\to 18$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $108-6\times 18=0$ .

Another way: Split the region at $x=12$ instead: the left triangle has area $\frac{1}{2}\times 12\times 36=216$ and the right one $\frac{1}{2}\times 6\times 36=108$ , and $216+108=324$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 10 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=12$ , $\overline{\text{BC}}=8$ , 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. ( $0<x<20$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=8$ , $\overline{\text{AC}}=12$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<20$ .

Givens

$\overline{\text{AC}}=12$ , $\overline{\text{BC}}=8$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<20$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=8$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<8

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=12

.

y=\frac{1}{2}\times x\times 12=6x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

8\le x<20

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=20-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=8

.

y=\frac{1}{2}\times(20-x)\times 8=80-4x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=8

both rules give

6\times 8=48

and

80-32=48

, so the graph rises in a straight line from

(0, 0)

to the peak

(8, 48)

and falls to

(20, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

20

on the

x

-axis and height

48

.

\frac{1}{2}\times 20\times 48=480

Once the graph is drawn, the question is plain triangle area.

Answer:

480

4 · Reviewdoes it hold up?

The peak $48$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 8\times 12=48$ , as it should when $\text{P}=\text{C}$ . As $x\to 20$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $80-4\times 20=0$ .

Another way: Split the region at $x=8$ instead: the left triangle has area $\frac{1}{2}\times 8\times 48=192$ and the right one $\frac{1}{2}\times 12\times 48=288$ , and $192+288=480$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 11 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=4$ , $\overline{\text{BC}}=8$ , 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. ( $0<x<12$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=8$ , $\overline{\text{AC}}=4$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<12$ .

Givens

$\overline{\text{AC}}=4$ , $\overline{\text{BC}}=8$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<12$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=8$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<8

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=4

.

y=\frac{1}{2}\times x\times 4=2x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

8\le x<12

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=12-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=8

.

y=\frac{1}{2}\times(12-x)\times 8=48-4x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=8

both rules give

2\times 8=16

and

48-32=16

, so the graph rises in a straight line from

(0, 0)

to the peak

(8, 16)

and falls to

(12, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

12

on the

x

-axis and height

16

.

\frac{1}{2}\times 12\times 16=96

Once the graph is drawn, the question is plain triangle area.

Answer:

96

4 · Reviewdoes it hold up?

The peak $16$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 8\times 4=16$ , as it should when $\text{P}=\text{C}$ . As $x\to 12$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $48-4\times 12=0$ .

Another way: Split the region at $x=8$ instead: the left triangle has area $\frac{1}{2}\times 8\times 16=64$ and the right one $\frac{1}{2}\times 4\times 16=32$ , and $64+32=96$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Variant 12 medium

As shown in the figure, $\text{ABC}$ is a right triangle with $\overline{\text{AC}}=4$ , $\overline{\text{BC}}=12$ , 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. ( $0<x<16$ )

Your answer

Show solution

1 · Understandwhat's really being asked

Point $\text{P}$ travels from $\text{B}$ to $\text{C}$ to $\text{A}$ along a right triangle with $\overline{\text{BC}}=12$ , $\overline{\text{AC}}=4$ . The area $y$ of $\triangle \text{ABP}$ depends on the distance $x$ traveled. Find the area enclosed by the graph of $y$ against $x$ and the $x$ -axis for $0<x<16$ .

Givens

$\overline{\text{AC}}=4$ , $\overline{\text{BC}}=12$ , $\angle \text{C}=90^\circ$ .
$\text{P}$ moves $\text{B}\to\text{C}\to\text{A}$ ; $x$ is the distance traveled.
$y$ is the area of $\triangle \text{ABP}$ , and $0<x<16$ .

Unknowns

The area between the graph of $y$ and the $x$ -axis.

Constraints

The rule for $y$ changes when $\text{P}$ turns the corner at $\text{C}$ ( $x=12$ ).

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#1 Draw a Diagram

Split the trip at the corner $\text{C}$ . On each side, $\triangle \text{ABP}$ 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.

3 · Execute4 carry out the plan

1P on BC

#13 Convert to Algebra 8.F.B.4

For

0<x<12

, base

\overline{\text{BP}}=x

lies on

\text{BC}

and the height is

\overline{\text{AC}}=4

.

y=\frac{1}{2}\times x\times 4=2x

Fixed height, growing base: the area grows at a constant rate.

2P on CA

#7 Identify Subproblems 8.F.B.4

For

12\le x<16

,

\text{P}

is on

\text{CA}

with

\overline{\text{AP}}=16-x

; take

\text{AP}

as the base, and the height is

\overline{\text{BC}}=12

.

y=\frac{1}{2}\times(16-x)\times 12=96-6x

Now the base shrinks while the height stays fixed.

3Draw the graph

#1 Draw a Diagram 8.F.B.5

At

x=12

both rules give

2\times 12=24

and

96-72=24

, so the graph rises in a straight line from

(0, 0)

to the peak

(12, 24)

and falls to

(16, 0)

.

The area is largest when

\text{P}

is at the corner

\text{C}

, where the triangle is all of

\triangle \text{ABC}

.

4Area under the graph

#7 Identify Subproblems 6.G.A.1

The region is a triangle with base

16

on the

x

-axis and height

24

.

\frac{1}{2}\times 16\times 24=192

Once the graph is drawn, the question is plain triangle area.

Answer:

192

4 · Reviewdoes it hold up?

The peak $24$ equals the area of $\triangle \text{ABC}$ , $\frac{1}{2}\times 12\times 4=24$ , as it should when $\text{P}=\text{C}$ . As $x\to 16$ , $\text{P}$ approaches $\text{A}$ and the area approaches $0$ , matching $96-6\times 16=0$ .

Another way: Split the region at $x=12$ instead: the left triangle has area $\frac{1}{2}\times 12\times 24=144$ and the right one $\frac{1}{2}\times 4\times 24=48$ , and $144+48=192$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the area as a linear function of distance on each side.
8.F.B.5 Describe qualitatively the functional relationship between two quantities — Sketching the rise-then-fall graph of the area function.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding triangle areas in the figure and under the graph.

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

Graph a moving point's area piece by piece · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?