Practice · Bisect a pentagon by a line from one vertex · Sensim Math

Variant 1 easy

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 3)$ , $\text{B}(-2, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , and $\text{E}(3, 1)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 3)$ , $\text{B}(-2, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , $\text{E}(3, 1)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-2, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=1

and width

\text{PD}=2

.

\frac{1}{2}\times(3+1)\times 2=4

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

3

) minus triangle

\text{BQC}

(legs

1

and

1

).

\frac{1}{2}\times(1+3)\times 3-\frac{1}{2}\times 1\times 1=6-\frac{1}{2}=\frac{11}{2}

The whole pentagon is

4+\frac{11}{2}=\frac{19}{2}

, so each half must be

\frac{19}{4}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

\frac{11}{2}-\frac{19}{4}=\frac{3}{4}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

3

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 3=\frac{3}{4},\ 1-t=\frac{1}{2},\ t=\frac{1}{2}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 3)

and

\text{M}\left(\frac{1}{2}, 0\right)

.

\frac{3-0}{1-\frac{1}{2}}=6,\quad 3=6\times 1+b,\ b=-3

Two points give the slope, then one point gives the intercept.

Answer:

y=6x-3

4 · Reviewdoes it hold up?

The line $y=6x-3$ meets $y=0$ at $x=\frac{1}{2}$ , which is on side $\text{CD}$ (between $-1$ and $3$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 3)$ , $\text{M}\left(\frac{1}{2}, 0\right)$ , $\text{D}(3, 0)$ , $\text{E}(3, 1)$ : $\frac{1}{2}\times\left(11-3\times \frac{1}{2}\right)=\frac{19}{4}$ , exactly half of $\frac{19}{2}$ .

Another way: Box method: the rectangle $-2\le x\le 3$ , $0\le y\le 3$ has area $15$ ; removing the three corner triangles of areas $3$ , $\frac{1}{2}$ , $2$ gives the pentagon's area $\frac{19}{2}$ . Then the right piece through $\text{M}(t, 0)$ has area $4+\frac{3}{2}(1-t)$ ; setting it to $\frac{19}{4}$ again gives $t=\frac{1}{2}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 2 medium

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 4)$ , $\text{B}(-3, 2)$ , $\text{C}(-2, 0)$ , $\text{D}(4, 0)$ , and $\text{E}(4, 2)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 4)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 4)$ , $\text{B}(-3, 2)$ , $\text{C}(-2, 0)$ , $\text{D}(4, 0)$ , $\text{E}(4, 2)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=4

,

\text{ED}=2

and width

\text{PD}=3

.

\frac{1}{2}\times(4+2)\times 3=9

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=2

,

\text{AP}=4

, width

4

) minus triangle

\text{BQC}

(legs

2

and

1

).

\frac{1}{2}\times(2+4)\times 4-\frac{1}{2}\times 2\times 1=12-1=11

The whole pentagon is

9+11=20

, so each half must be

10

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

11-10=1

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

4

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 4=1,\ 1-t=\frac{1}{2},\ t=\frac{1}{2}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 4)

and

\text{M}\left(\frac{1}{2}, 0\right)

.

\frac{4-0}{1-\frac{1}{2}}=8,\quad 4=8\times 1+b,\ b=-4

Two points give the slope, then one point gives the intercept.

Answer:

y=8x-4

4 · Reviewdoes it hold up?

The line $y=8x-4$ meets $y=0$ at $x=\frac{1}{2}$ , which is on side $\text{CD}$ (between $-2$ and $4$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 4)$ , $\text{M}\left(\frac{1}{2}, 0\right)$ , $\text{D}(4, 0)$ , $\text{E}(4, 2)$ : $\frac{1}{2}\times\left(22-4\times \frac{1}{2}\right)=10$ , exactly half of $20$ .

Another way: Box method: the rectangle $-3\le x\le 4$ , $0\le y\le 4$ has area $28$ ; removing the three corner triangles of areas $4$ , $1$ , $3$ gives the pentagon's area $20$ . Then the right piece through $\text{M}(t, 0)$ has area $9+2(1-t)$ ; setting it to $10$ again gives $t=\frac{1}{2}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 3 easy

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 4)$ , $\text{B}(-2, 2)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , and $\text{E}(3, 3)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 4)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 4)$ , $\text{B}(-2, 2)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , $\text{E}(3, 3)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-2, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=4

,

\text{ED}=3

and width

\text{PD}=2

.

\frac{1}{2}\times(4+3)\times 2=7

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=2

,

\text{AP}=4

, width

3

) minus triangle

\text{BQC}

(legs

2

and

1

).

\frac{1}{2}\times(2+4)\times 3-\frac{1}{2}\times 2\times 1=9-1=8

The whole pentagon is

7+8=15

, so each half must be

\frac{15}{2}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

8-\frac{15}{2}=\frac{1}{2}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

4

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 4=\frac{1}{2},\ 1-t=\frac{1}{4},\ t=\frac{3}{4}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 4)

and

\text{M}\left(\frac{3}{4}, 0\right)

.

\frac{4-0}{1-\frac{3}{4}}=16,\quad 4=16\times 1+b,\ b=-12

Two points give the slope, then one point gives the intercept.

Answer:

y=16x-12

4 · Reviewdoes it hold up?

The line $y=16x-12$ meets $y=0$ at $x=\frac{3}{4}$ , which is on side $\text{CD}$ (between $-1$ and $3$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 4)$ , $\text{M}\left(\frac{3}{4}, 0\right)$ , $\text{D}(3, 0)$ , $\text{E}(3, 3)$ : $\frac{1}{2}\times\left(18-4\times \frac{3}{4}\right)=\frac{15}{2}$ , exactly half of $15$ .

Another way: Box method: the rectangle $-2\le x\le 3$ , $0\le y\le 4$ has area $20$ ; removing the three corner triangles of areas $3$ , $1$ , $1$ gives the pentagon's area $15$ . Then the right piece through $\text{M}(t, 0)$ has area $7+2(1-t)$ ; setting it to $\frac{15}{2}$ again gives $t=\frac{3}{4}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 4 easy

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 2)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(4, 0)$ , and $\text{E}(4, 1)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 2)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 2)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(4, 0)$ , $\text{E}(4, 1)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=2

,

\text{ED}=1

and width

\text{PD}=3

.

\frac{1}{2}\times(2+1)\times 3=\frac{9}{2}

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=2

, width

4

) minus triangle

\text{BQC}

(legs

1

and

2

).

\frac{1}{2}\times(1+2)\times 4-\frac{1}{2}\times 1\times 2=6-1=5

The whole pentagon is

\frac{9}{2}+5=\frac{19}{2}

, so each half must be

\frac{19}{4}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

5-\frac{19}{4}=\frac{1}{4}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

2

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 2=\frac{1}{4},\ 1-t=\frac{1}{4},\ t=\frac{3}{4}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 2)

and

\text{M}\left(\frac{3}{4}, 0\right)

.

\frac{2-0}{1-\frac{3}{4}}=8,\quad 2=8\times 1+b,\ b=-6

Two points give the slope, then one point gives the intercept.

Answer:

y=8x-6

4 · Reviewdoes it hold up?

The line $y=8x-6$ meets $y=0$ at $x=\frac{3}{4}$ , which is on side $\text{CD}$ (between $-1$ and $4$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 2)$ , $\text{M}\left(\frac{3}{4}, 0\right)$ , $\text{D}(4, 0)$ , $\text{E}(4, 1)$ : $\frac{1}{2}\times\left(11-2\times \frac{3}{4}\right)=\frac{19}{4}$ , exactly half of $\frac{19}{2}$ .

Another way: Box method: the rectangle $-3\le x\le 4$ , $0\le y\le 2$ has area $14$ ; removing the three corner triangles of areas $2$ , $1$ , $\frac{3}{2}$ gives the pentagon's area $\frac{19}{2}$ . Then the right piece through $\text{M}(t, 0)$ has area $\frac{9}{2}+(1-t)$ ; setting it to $\frac{19}{4}$ again gives $t=\frac{3}{4}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 5 easy

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 3)$ , $\text{B}(-4, 1)$ , $\text{C}(-3, 0)$ , $\text{D}(4, 0)$ , and $\text{E}(4, 2)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 3)$ , $\text{B}(-4, 1)$ , $\text{C}(-3, 0)$ , $\text{D}(4, 0)$ , $\text{E}(4, 2)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-4, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=2

and width

\text{PD}=3

.

\frac{1}{2}\times(3+2)\times 3=\frac{15}{2}

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

5

) minus triangle

\text{BQC}

(legs

1

and

1

).

\frac{1}{2}\times(1+3)\times 5-\frac{1}{2}\times 1\times 1=10-\frac{1}{2}=\frac{19}{2}

The whole pentagon is

\frac{15}{2}+\frac{19}{2}=17

, so each half must be

\frac{17}{2}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

\frac{19}{2}-\frac{17}{2}=1

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

3

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 3=1,\ 1-t=\frac{2}{3},\ t=\frac{1}{3}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 3)

and

\text{M}\left(\frac{1}{3}, 0\right)

.

\frac{3-0}{1-\frac{1}{3}}=\frac{9}{2},\quad 3=\frac{9}{2}\times 1+b,\ b=-\frac{3}{2}

Two points give the slope, then one point gives the intercept.

Answer:

y=\frac{9}{2}x-\frac{3}{2}

4 · Reviewdoes it hold up?

The line $y=\frac{9}{2}x-\frac{3}{2}$ meets $y=0$ at $x=\frac{1}{3}$ , which is on side $\text{CD}$ (between $-3$ and $4$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 3)$ , $\text{M}\left(\frac{1}{3}, 0\right)$ , $\text{D}(4, 0)$ , $\text{E}(4, 2)$ : $\frac{1}{2}\times\left(18-3\times \frac{1}{3}\right)=\frac{17}{2}$ , exactly half of $17$ .

Another way: Box method: the rectangle $-4\le x\le 4$ , $0\le y\le 3$ has area $24$ ; removing the three corner triangles of areas $5$ , $\frac{1}{2}$ , $\frac{3}{2}$ gives the pentagon's area $17$ . Then the right piece through $\text{M}(t, 0)$ has area $\frac{15}{2}+\frac{3}{2}(1-t)$ ; setting it to $\frac{17}{2}$ again gives $t=\frac{1}{3}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 6 medium

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(4, 0)$ , and $\text{E}(4, 1)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(4, 0)$ , $\text{E}(4, 1)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=1

and width

\text{PD}=3

.

\frac{1}{2}\times(3+1)\times 3=6

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

4

) minus triangle

\text{BQC}

(legs

1

and

2

).

\frac{1}{2}\times(1+3)\times 4-\frac{1}{2}\times 1\times 2=8-1=7

The whole pentagon is

6+7=13

, so each half must be

\frac{13}{2}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

7-\frac{13}{2}=\frac{1}{2}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

3

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 3=\frac{1}{2},\ 1-t=\frac{1}{3},\ t=\frac{2}{3}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 3)

and

\text{M}\left(\frac{2}{3}, 0\right)

.

\frac{3-0}{1-\frac{2}{3}}=9,\quad 3=9\times 1+b,\ b=-6

Two points give the slope, then one point gives the intercept.

Answer:

y=9x-6

4 · Reviewdoes it hold up?

The line $y=9x-6$ meets $y=0$ at $x=\frac{2}{3}$ , which is on side $\text{CD}$ (between $-1$ and $4$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 3)$ , $\text{M}\left(\frac{2}{3}, 0\right)$ , $\text{D}(4, 0)$ , $\text{E}(4, 1)$ : $\frac{1}{2}\times\left(15-3\times \frac{2}{3}\right)=\frac{13}{2}$ , exactly half of $13$ .

Another way: Box method: the rectangle $-3\le x\le 4$ , $0\le y\le 3$ has area $21$ ; removing the three corner triangles of areas $4$ , $1$ , $3$ gives the pentagon's area $13$ . Then the right piece through $\text{M}(t, 0)$ has area $6+\frac{3}{2}(1-t)$ ; setting it to $\frac{13}{2}$ again gives $t=\frac{2}{3}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 7 medium

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 5)$ , $\text{B}(-2, 2)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , and $\text{E}(3, 2)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 5)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 5)$ , $\text{B}(-2, 2)$ , $\text{C}(-1, 0)$ , $\text{D}(3, 0)$ , $\text{E}(3, 2)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-2, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=5

,

\text{ED}=2

and width

\text{PD}=2

.

\frac{1}{2}\times(5+2)\times 2=7

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=2

,

\text{AP}=5

, width

3

) minus triangle

\text{BQC}

(legs

2

and

1

).

\frac{1}{2}\times(2+5)\times 3-\frac{1}{2}\times 2\times 1=\frac{21}{2}-1=\frac{19}{2}

The whole pentagon is

7+\frac{19}{2}=\frac{33}{2}

, so each half must be

\frac{33}{4}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

\frac{19}{2}-\frac{33}{4}=\frac{5}{4}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

5

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 5=\frac{5}{4},\ 1-t=\frac{1}{2},\ t=\frac{1}{2}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 5)

and

\text{M}\left(\frac{1}{2}, 0\right)

.

\frac{5-0}{1-\frac{1}{2}}=10,\quad 5=10\times 1+b,\ b=-5

Two points give the slope, then one point gives the intercept.

Answer:

y=10x-5

4 · Reviewdoes it hold up?

The line $y=10x-5$ meets $y=0$ at $x=\frac{1}{2}$ , which is on side $\text{CD}$ (between $-1$ and $3$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 5)$ , $\text{M}\left(\frac{1}{2}, 0\right)$ , $\text{D}(3, 0)$ , $\text{E}(3, 2)$ : $\frac{1}{2}\times\left(19-5\times \frac{1}{2}\right)=\frac{33}{4}$ , exactly half of $\frac{33}{2}$ .

Another way: Box method: the rectangle $-2\le x\le 3$ , $0\le y\le 5$ has area $25$ ; removing the three corner triangles of areas $\frac{9}{2}$ , $1$ , $3$ gives the pentagon's area $\frac{33}{2}$ . Then the right piece through $\text{M}(t, 0)$ has area $7+\frac{5}{2}(1-t)$ ; setting it to $\frac{33}{4}$ again gives $t=\frac{1}{2}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 8 medium

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(1, 5)$ , $\text{B}(-4, 1)$ , $\text{C}(-2, 0)$ , $\text{D}(4, 0)$ , and $\text{E}(4, 3)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(1, 5)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(1, 5)$ , $\text{B}(-4, 1)$ , $\text{C}(-2, 0)$ , $\text{D}(4, 0)$ , $\text{E}(4, 3)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=1$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 1

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(1, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-4, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=5

,

\text{ED}=3

and width

\text{PD}=3

.

\frac{1}{2}\times(5+3)\times 3=12

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=5

, width

5

) minus triangle

\text{BQC}

(legs

1

and

2

).

\frac{1}{2}\times(1+5)\times 5-\frac{1}{2}\times 1\times 2=15-1=14

The whole pentagon is

12+14=26

, so each half must be

13

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

14-13=1

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<1

; this moves triangle

\text{AMP}

, with height

5

and base

1-t

, to the right piece.

\frac{1}{2}\times(1-t)\times 5=1,\ 1-t=\frac{2}{5},\ t=\frac{3}{5}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(1, 5)

and

\text{M}\left(\frac{3}{5}, 0\right)

.

\frac{5-0}{1-\frac{3}{5}}=\frac{25}{2},\quad 5=\frac{25}{2}\times 1+b,\ b=-\frac{15}{2}

Two points give the slope, then one point gives the intercept.

Answer:

y=\frac{25}{2}x-\frac{15}{2}

4 · Reviewdoes it hold up?

The line $y=\frac{25}{2}x-\frac{15}{2}$ meets $y=0$ at $x=\frac{3}{5}$ , which is on side $\text{CD}$ (between $-2$ and $4$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(1, 5)$ , $\text{M}\left(\frac{3}{5}, 0\right)$ , $\text{D}(4, 0)$ , $\text{E}(4, 3)$ : $\frac{1}{2}\times\left(29-5\times \frac{3}{5}\right)=13$ , exactly half of $26$ .

Another way: Box method: the rectangle $-4\le x\le 4$ , $0\le y\le 5$ has area $40$ ; removing the three corner triangles of areas $10$ , $1$ , $3$ gives the pentagon's area $26$ . Then the right piece through $\text{M}(t, 0)$ has area $12+\frac{5}{2}(1-t)$ ; setting it to $13$ again gives $t=\frac{3}{5}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 9 hard

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(2, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , and $\text{E}(5, 2)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(2, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(2, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 2)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=2$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 2

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(2, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=2

and width

\text{PD}=3

.

\frac{1}{2}\times(3+2)\times 3=\frac{15}{2}

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

5

) minus triangle

\text{BQC}

(legs

1

and

2

).

\frac{1}{2}\times(1+3)\times 5-\frac{1}{2}\times 1\times 2=10-1=9

The whole pentagon is

\frac{15}{2}+9=\frac{33}{2}

, so each half must be

\frac{33}{4}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

9-\frac{33}{4}=\frac{3}{4}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<2

; this moves triangle

\text{AMP}

, with height

3

and base

2-t

, to the right piece.

\frac{1}{2}\times(2-t)\times 3=\frac{3}{4},\ 2-t=\frac{1}{2},\ t=\frac{3}{2}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(2, 3)

and

\text{M}\left(\frac{3}{2}, 0\right)

.

\frac{3-0}{2-\frac{3}{2}}=6,\quad 3=6\times 2+b,\ b=-9

Two points give the slope, then one point gives the intercept.

Answer:

y=6x-9

4 · Reviewdoes it hold up?

The line $y=6x-9$ meets $y=0$ at $x=\frac{3}{2}$ , which is on side $\text{CD}$ (between $-1$ and $5$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(2, 3)$ , $\text{M}\left(\frac{3}{2}, 0\right)$ , $\text{D}(5, 0)$ , $\text{E}(5, 2)$ : $\frac{1}{2}\times\left(21-3\times \frac{3}{2}\right)=\frac{33}{4}$ , exactly half of $\frac{33}{2}$ .

Another way: Box method: the rectangle $-3\le x\le 5$ , $0\le y\le 3$ has area $24$ ; removing the three corner triangles of areas $5$ , $1$ , $\frac{3}{2}$ gives the pentagon's area $\frac{33}{2}$ . Then the right piece through $\text{M}(t, 0)$ has area $\frac{15}{2}+\frac{3}{2}(2-t)$ ; setting it to $\frac{33}{4}$ again gives $t=\frac{3}{2}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 10 hard

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(2, 3)$ , $\text{B}(-4, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , and $\text{E}(5, 2)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(2, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(2, 3)$ , $\text{B}(-4, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 2)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=2$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 2

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(2, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-4, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=2

and width

\text{PD}=3

.

\frac{1}{2}\times(3+2)\times 3=\frac{15}{2}

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

6

) minus triangle

\text{BQC}

(legs

1

and

3

).

\frac{1}{2}\times(1+3)\times 6-\frac{1}{2}\times 1\times 3=12-\frac{3}{2}=\frac{21}{2}

The whole pentagon is

\frac{15}{2}+\frac{21}{2}=18

, so each half must be

9

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

\frac{21}{2}-9=\frac{3}{2}

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<2

; this moves triangle

\text{AMP}

, with height

3

and base

2-t

, to the right piece.

\frac{1}{2}\times(2-t)\times 3=\frac{3}{2},\ 2-t=1,\ t=1

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(2, 3)

and

\text{M}(1, 0)

.

\frac{3-0}{2-1}=3,\quad 3=3\times 2+b,\ b=-3

Two points give the slope, then one point gives the intercept.

Answer:

y=3x-3

4 · Reviewdoes it hold up?

The line $y=3x-3$ meets $y=0$ at $x=1$ , which is on side $\text{CD}$ (between $-1$ and $5$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(2, 3)$ , $\text{M}(1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 2)$ : $\frac{1}{2}\times\left(21-3\times 1\right)=9$ , exactly half of $18$ .

Another way: Box method: the rectangle $-4\le x\le 5$ , $0\le y\le 3$ has area $27$ ; removing the three corner triangles of areas $6$ , $\frac{3}{2}$ , $\frac{3}{2}$ gives the pentagon's area $18$ . Then the right piece through $\text{M}(t, 0)$ has area $\frac{15}{2}+\frac{3}{2}(2-t)$ ; setting it to $9$ again gives $t=1$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 11 hard

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(2, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(1, 0)$ , $\text{D}(5, 0)$ , and $\text{E}(5, 1)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through vertex $\text{A}(2, 3)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(2, 3)$ , $\text{B}(-3, 1)$ , $\text{C}(1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 1)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=2$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 2

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(2, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=3

,

\text{ED}=1

and width

\text{PD}=3

.

\frac{1}{2}\times(3+1)\times 3=6

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=3

, width

5

) minus triangle

\text{BQC}

(legs

1

and

4

).

\frac{1}{2}\times(1+3)\times 5-\frac{1}{2}\times 1\times 4=10-2=8

The whole pentagon is

6+8=14

, so each half must be

7

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

8-7=1

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<2

; this moves triangle

\text{AMP}

, with height

3

and base

2-t

, to the right piece.

\frac{1}{2}\times(2-t)\times 3=1,\ 2-t=\frac{2}{3},\ t=\frac{4}{3}

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(2, 3)

and

\text{M}\left(\frac{4}{3}, 0\right)

.

\frac{3-0}{2-\frac{4}{3}}=\frac{9}{2},\quad 3=\frac{9}{2}\times 2+b,\ b=-6

Two points give the slope, then one point gives the intercept.

Answer:

y=\frac{9}{2}x-6

4 · Reviewdoes it hold up?

The line $y=\frac{9}{2}x-6$ meets $y=0$ at $x=\frac{4}{3}$ , which is on side $\text{CD}$ (between $1$ and $5$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(2, 3)$ , $\text{M}\left(\frac{4}{3}, 0\right)$ , $\text{D}(5, 0)$ , $\text{E}(5, 1)$ : $\frac{1}{2}\times\left(18-3\times \frac{4}{3}\right)=7$ , exactly half of $14$ .

Another way: Box method: the rectangle $-3\le x\le 5$ , $0\le y\le 3$ has area $24$ ; removing the three corner triangles of areas $5$ , $2$ , $3$ gives the pentagon's area $14$ . Then the right piece through $\text{M}(t, 0)$ has area $6+\frac{3}{2}(2-t)$ ; setting it to $7$ again gives $t=\frac{4}{3}$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Variant 12 hard

As shown in the figure, pentagon $\text{ABCDE}$ has vertices $\text{A}(2, 2)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , and $\text{E}(5, 1)$ . A line through point $\text{A}$ divides the area of pentagon $\text{ABCDE}$ into two equal parts. Find the equation of this line.

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1 · Understandwhat's really being asked

A line through vertex $\text{A}(2, 2)$ splits pentagon $\text{ABCDE}$ into two regions of equal area. Find its equation.

Givens

Vertices $\text{A}(2, 2)$ , $\text{B}(-3, 1)$ , $\text{C}(-1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 1)$ .
The line passes through $\text{A}$ and halves the pentagon's area.

Unknowns

The equation of the bisecting line.

Constraints

The line's second point is unknown; it must lie on a side of the pentagon.

2 · Planchoose the strategy

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

Split the pentagon with the vertical line $x=2$ through $\text{A}$ into two pieces whose areas are easy to find. They come out close but not equal, so the bisector only needs to tilt a little from vertical, crossing side $\text{CD}$ at an unknown point $\text{M}$ . A thin triangle makes up the difference, and its area fixes $\text{M}$ .

3 · Execute4 carry out the plan

1Cut along x = 2

#7 Identify Subproblems 6.G.A.1

Let

\text{P}(2, 0)

be the foot of the perpendicular from

\text{A}

, and

\text{Q}(-3, 0)

the foot from

\text{B}

. The right piece

\text{APDE}

is a trapezoid with parallel sides

\text{AP}=2

,

\text{ED}=1

and width

\text{PD}=3

.

\frac{1}{2}\times(2+1)\times 3=\frac{9}{2}

Vertical cuts turn the pieces into trapezoids and triangles.

2Area of the left piece

#7 Identify Subproblems 6.G.A.1

The left piece

\text{ABCP}

is trapezoid

\text{ABQP}

(sides

\text{BQ}=1

,

\text{AP}=2

, width

5

) minus triangle

\text{BQC}

(legs

1

and

2

).

\frac{1}{2}\times(1+2)\times 5-\frac{1}{2}\times 1\times 2=\frac{15}{2}-1=\frac{13}{2}

The whole pentagon is

\frac{9}{2}+\frac{13}{2}=11

, so each half must be

\frac{11}{2}

.

3Shift area with a thin triangle

#1 Draw a Diagram 6.G.A.1

The left piece is too big by

\frac{13}{2}-\frac{11}{2}=1

. Tilt the cut so it meets

\text{CD}

at

\text{M}(t, 0)

with

t<2

; this moves triangle

\text{AMP}

, with height

2

and base

2-t

, to the right piece.

\frac{1}{2}\times(2-t)\times 2=1,\ 2-t=1,\ t=1

Only the small difference has to be moved, not the whole area recomputed.

4Equation of line AM

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

The line passes through

\text{A}(2, 2)

and

\text{M}(1, 0)

.

\frac{2-0}{2-1}=2,\quad 2=2\times 2+b,\ b=-2

Two points give the slope, then one point gives the intercept.

Answer:

y=2x-2

4 · Reviewdoes it hold up?

The line $y=2x-2$ meets $y=0$ at $x=1$ , which is on side $\text{CD}$ (between $-1$ and $5$ ). Right piece $\text{AMDE}$ by the shoelace formula with $\text{A}(2, 2)$ , $\text{M}(1, 0)$ , $\text{D}(5, 0)$ , $\text{E}(5, 1)$ : $\frac{1}{2}\times\left(13-2\times 1\right)=\frac{11}{2}$ , exactly half of $11$ .

Another way: Box method: the rectangle $-3\le x\le 5$ , $0\le y\le 2$ has area $16$ ; removing the three corner triangles of areas $\frac{5}{2}$ , $1$ , $\frac{3}{2}$ gives the pentagon's area $11$ . Then the right piece through $\text{M}(t, 0)$ has area $\frac{9}{2}+(2-t)$ ; setting it to $\frac{11}{2}$ again gives $t=1$ .

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the equation of the line through two known points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding polygon areas by splitting into trapezoids and triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Reading side lengths of a polygon from vertex coordinates.

💡Takeaway. Cut where the areas are easy, then slide the cut by just the leftover amount.

Bisect a pentagon by a line from one vertex · 12 practice problems

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