Practice · Find which side the bisector hits first · Sensim Math

Variant 1 easy

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-6, 5)$ , $B(-6, 0)$ , $C(2, 0)$ , and $D(2, 1)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-6, 5)$ , $B(-6, 0)$ , $C(2, 0)$ , $D(2, 1)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

5

and

1

and width

2-(-6)=8

.

\frac{1}{2}\times(5+1)\times 8=24

Each half must have area

12

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=6

and height

5

.

\frac{1}{2}\times 6\times 5=15>12

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-6, k)

. The left piece is triangle

PBO

with base

BO=6

and height

k

.

\frac{1}{2}\times 6\times k=12 \Rightarrow k=4

4

is less than

5

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-6, 4\right)

.

4=-6a \Rightarrow a=-\frac{2}{3}

Negative slope: the line rises to the left, toward

AB

.

Answer:

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

4 · Reviewdoes it hold up?

Left piece: triangle with base $6$ and height $4$ , area $12$ . Right piece: $24-12=12$ . Equal, as needed. The slope $-\frac{2}{3}$ is a bit flatter than the slope $-\frac{5}{6}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $15$ is already more than $12$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 2 easy

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-6, 6)$ , $B(-6, 0)$ , $C(2, 0)$ , and $D(2, 2)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-6, 6)$ , $B(-6, 0)$ , $C(2, 0)$ , $D(2, 2)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

6

and

2

and width

2-(-6)=8

.

\frac{1}{2}\times(6+2)\times 8=32

Each half must have area

16

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=6

and height

6

.

\frac{1}{2}\times 6\times 6=18>16

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-6, k)

. The left piece is triangle

PBO

with base

BO=6

and height

k

.

\frac{1}{2}\times 6\times k=16 \Rightarrow k=\frac{16}{3}

\frac{16}{3}

is less than

6

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-6, \frac{16}{3}\right)

.

\frac{16}{3}=-6a \Rightarrow a=-\frac{8}{9}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{8}{9}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $6$ and height $\frac{16}{3}$ , area $16$ . Right piece: $32-16=16$ . Equal, as needed. The slope $-\frac{8}{9}$ is a bit flatter than the slope $-1$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $18$ is already more than $16$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 3 medium

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-6, 5)$ , $B(-6, 0)$ , $C(1, 0)$ , and $D(1, 3)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-6, 5)$ , $B(-6, 0)$ , $C(1, 0)$ , $D(1, 3)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

5

and

3

and width

1-(-6)=7

.

\frac{1}{2}\times(5+3)\times 7=28

Each half must have area

14

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=6

and height

5

.

\frac{1}{2}\times 6\times 5=15>14

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-6, k)

. The left piece is triangle

PBO

with base

BO=6

and height

k

.

\frac{1}{2}\times 6\times k=14 \Rightarrow k=\frac{14}{3}

\frac{14}{3}

is less than

5

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-6, \frac{14}{3}\right)

.

\frac{14}{3}=-6a \Rightarrow a=-\frac{7}{9}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{7}{9}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $6$ and height $\frac{14}{3}$ , area $14$ . Right piece: $28-14=14$ . Equal, as needed. The slope $-\frac{7}{9}$ is a bit flatter than the slope $-\frac{5}{6}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $15$ is already more than $14$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 4 easy

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-4, 6)$ , $B(-4, 0)$ , $C(1, 0)$ , and $D(1, 2)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-4, 6)$ , $B(-4, 0)$ , $C(1, 0)$ , $D(1, 2)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

6

and

2

and width

1-(-4)=5

.

\frac{1}{2}\times(6+2)\times 5=20

Each half must have area

10

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=4

and height

6

.

\frac{1}{2}\times 4\times 6=12>10

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-4, k)

. The left piece is triangle

PBO

with base

BO=4

and height

k

.

\frac{1}{2}\times 4\times k=10 \Rightarrow k=5

5

is less than

6

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-4, 5\right)

.

5=-4a \Rightarrow a=-\frac{5}{4}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{5}{4}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $4$ and height $5$ , area $10$ . Right piece: $20-10=10$ . Equal, as needed. The slope $-\frac{5}{4}$ is a bit flatter than the slope $-\frac{3}{2}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $12$ is already more than $10$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 5 easy

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-5, 6)$ , $B(-5, 0)$ , $C(3, 0)$ , and $D(3, 1)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-5, 6)$ , $B(-5, 0)$ , $C(3, 0)$ , $D(3, 1)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

6

and

1

and width

3-(-5)=8

.

\frac{1}{2}\times(6+1)\times 8=28

Each half must have area

14

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=5

and height

6

.

\frac{1}{2}\times 5\times 6=15>14

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-5, k)

. The left piece is triangle

PBO

with base

BO=5

and height

k

.

\frac{1}{2}\times 5\times k=14 \Rightarrow k=\frac{28}{5}

\frac{28}{5}

is less than

6

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-5, \frac{28}{5}\right)

.

\frac{28}{5}=-5a \Rightarrow a=-\frac{28}{25}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{28}{25}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $5$ and height $\frac{28}{5}$ , area $14$ . Right piece: $28-14=14$ . Equal, as needed. The slope $-\frac{28}{25}$ is a bit flatter than the slope $-\frac{6}{5}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $15$ is already more than $14$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 6 medium

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-7, 6)$ , $B(-7, 0)$ , $C(1, 0)$ , and $D(1, 1)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-7, 6)$ , $B(-7, 0)$ , $C(1, 0)$ , $D(1, 1)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

6

and

1

and width

1-(-7)=8

.

\frac{1}{2}\times(6+1)\times 8=28

Each half must have area

14

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=7

and height

6

.

\frac{1}{2}\times 7\times 6=21>14

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-7, k)

. The left piece is triangle

PBO

with base

BO=7

and height

k

.

\frac{1}{2}\times 7\times k=14 \Rightarrow k=4

4

is less than

6

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-7, 4\right)

.

4=-7a \Rightarrow a=-\frac{4}{7}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{4}{7}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $7$ and height $4$ , area $14$ . Right piece: $28-14=14$ . Equal, as needed. The slope $-\frac{4}{7}$ is a bit flatter than the slope $-\frac{6}{7}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $21$ is already more than $14$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 7 medium

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-8, 7)$ , $B(-8, 0)$ , $C(4, 0)$ , and $D(4, 1)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-8, 7)$ , $B(-8, 0)$ , $C(4, 0)$ , $D(4, 1)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

7

and

1

and width

4-(-8)=12

.

\frac{1}{2}\times(7+1)\times 12=48

Each half must have area

24

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=8

and height

7

.

\frac{1}{2}\times 8\times 7=28>24

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-8, k)

. The left piece is triangle

PBO

with base

BO=8

and height

k

.

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

6

is less than

7

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-8, 6\right)

.

6=-8a \Rightarrow a=-\frac{3}{4}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{3}{4}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $8$ and height $6$ , area $24$ . Right piece: $48-24=24$ . Equal, as needed. The slope $-\frac{3}{4}$ is a bit flatter than the slope $-\frac{7}{8}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $28$ is already more than $24$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 8 hard

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-3, 8)$ , $B(-3, 0)$ , $C(1, 0)$ , and $D(1, 2)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-3, 8)$ , $B(-3, 0)$ , $C(1, 0)$ , $D(1, 2)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

8

and

2

and width

1-(-3)=4

.

\frac{1}{2}\times(8+2)\times 4=20

Each half must have area

10

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=3

and height

8

.

\frac{1}{2}\times 3\times 8=12>10

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-3, k)

. The left piece is triangle

PBO

with base

BO=3

and height

k

.

\frac{1}{2}\times 3\times k=10 \Rightarrow k=\frac{20}{3}

\frac{20}{3}

is less than

8

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-3, \frac{20}{3}\right)

.

\frac{20}{3}=-3a \Rightarrow a=-\frac{20}{9}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{20}{9}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $3$ and height $\frac{20}{3}$ , area $10$ . Right piece: $20-10=10$ . Equal, as needed. The slope $-\frac{20}{9}$ is a bit flatter than the slope $-\frac{8}{3}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $12$ is already more than $10$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 9 hard

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-6, 8)$ , $B(-6, 0)$ , $C(2, 0)$ , and $D(2, 2)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-6, 8)$ , $B(-6, 0)$ , $C(2, 0)$ , $D(2, 2)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

8

and

2

and width

2-(-6)=8

.

\frac{1}{2}\times(8+2)\times 8=40

Each half must have area

20

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=6

and height

8

.

\frac{1}{2}\times 6\times 8=24>20

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-6, k)

. The left piece is triangle

PBO

with base

BO=6

and height

k

.

\frac{1}{2}\times 6\times k=20 \Rightarrow k=\frac{20}{3}

\frac{20}{3}

is less than

8

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-6, \frac{20}{3}\right)

.

\frac{20}{3}=-6a \Rightarrow a=-\frac{10}{9}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{10}{9}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $6$ and height $\frac{20}{3}$ , area $20$ . Right piece: $40-20=20$ . Equal, as needed. The slope $-\frac{10}{9}$ is a bit flatter than the slope $-\frac{4}{3}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $24$ is already more than $20$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 10 medium

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-4, 8)$ , $B(-4, 0)$ , $C(1, 0)$ , and $D(1, 4)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-4, 8)$ , $B(-4, 0)$ , $C(1, 0)$ , $D(1, 4)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

8

and

4

and width

1-(-4)=5

.

\frac{1}{2}\times(8+4)\times 5=30

Each half must have area

15

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=4

and height

8

.

\frac{1}{2}\times 4\times 8=16>15

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-4, k)

. The left piece is triangle

PBO

with base

BO=4

and height

k

.

\frac{1}{2}\times 4\times k=15 \Rightarrow k=\frac{15}{2}

\frac{15}{2}

is less than

8

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-4, \frac{15}{2}\right)

.

\frac{15}{2}=-4a \Rightarrow a=-\frac{15}{8}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{15}{8}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $4$ and height $\frac{15}{2}$ , area $15$ . Right piece: $30-15=15$ . Equal, as needed. The slope $-\frac{15}{8}$ is a bit flatter than the slope $-2$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $16$ is already more than $15$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 11 hard

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-8, 5)$ , $B(-8, 0)$ , $C(2, 0)$ , and $D(2, 1)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-8, 5)$ , $B(-8, 0)$ , $C(2, 0)$ , $D(2, 1)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

5

and

1

and width

2-(-8)=10

.

\frac{1}{2}\times(5+1)\times 10=30

Each half must have area

15

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=8

and height

5

.

\frac{1}{2}\times 8\times 5=20>15

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-8, k)

. The left piece is triangle

PBO

with base

BO=8

and height

k

.

\frac{1}{2}\times 8\times k=15 \Rightarrow k=\frac{15}{4}

\frac{15}{4}

is less than

5

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-8, \frac{15}{4}\right)

.

\frac{15}{4}=-8a \Rightarrow a=-\frac{15}{32}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{15}{32}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $8$ and height $\frac{15}{4}$ , area $15$ . Right piece: $30-15=15$ . Equal, as needed. The slope $-\frac{15}{32}$ is a bit flatter than the slope $-\frac{5}{8}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $20$ is already more than $15$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Variant 12 hard

As shown in the figure, quadrilateral $ABCD$ has vertices $A(-5, 8)$ , $B(-5, 0)$ , $C(1, 0)$ , and $D(1, 4)$ . Among the lines through the origin, find the linear function whose graph cuts the area of quadrilateral $ABCD$ exactly in half.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A trapezoid-shaped quadrilateral sits on the $x$ -axis with the origin on its bottom side. Find the line $y=ax$ through the origin that splits its area into two equal parts.

Givens

Vertices $A(-5, 8)$ , $B(-5, 0)$ , $C(1, 0)$ , $D(1, 4)$ .
The line passes through the origin $O(0, 0)$ , which lies on side $BC$ .

Unknowns

The equation of the bisecting line.

Constraints

A line through the origin has the form $y=ax$ ; each piece must have area half of the whole.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#3 Eliminate Possibilities#13 Convert to Algebra

Break it into subproblems. First the total area (it is a trapezoid, since $AB$ and $DC$ are both vertical). Then decide which side the line exits through, by testing the line through vertex $A$ : if the piece it cuts off is already too big, the true line must hit side $AB$ below $A$ . Finally a single triangle area gives the exit point, and the slope follows.

3 · Execute4 carry out the plan

1Total area

#7 Identify Subproblems 6.G.A.1

AB

and

DC

are vertical, so

ABCD

is a trapezoid with parallel sides

8

and

4

and width

1-(-5)=6

.

\frac{1}{2}\times(8+4)\times 6=36

Each half must have area

18

.

2Test the line through A

#3 Eliminate Possibilities 6.G.A.3

The line through

O

and

A

cuts off triangle

ABO

with base

BO=5

and height

8

.

\frac{1}{2}\times 5\times 8=20>18

Too much on the left, so the bisector must be flatter and exit through side

AB

, below

A

.

3Find the exit point on AB

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

Let the line meet

AB

at

P(-5, k)

. The left piece is triangle

PBO

with base

BO=5

and height

k

.

\frac{1}{2}\times 5\times k=18 \Rightarrow k=\frac{36}{5}

\frac{36}{5}

is less than

8

, so

P

really is on side

AB

.

4Slope through the origin

#13 Convert to Algebra 8.F.A.3

The line

y=ax

passes through

P\left(-5, \frac{36}{5}\right)

.

\frac{36}{5}=-5a \Rightarrow a=-\frac{36}{25}

Negative slope: the line rises to the left, toward

AB

.

Answer:

y=-\frac{36}{25}x

4 · Reviewdoes it hold up?

Left piece: triangle with base $5$ and height $\frac{36}{5}$ , area $18$ . Right piece: $36-18=18$ . Equal, as needed. The slope $-\frac{36}{25}$ is a bit flatter than the slope $-\frac{8}{5}$ of line $OA$ , matching the test in step $2$ .

Another way: Organize it the other way: suppose the line exits through $AD$ instead. Then the left piece would contain all of triangle $ABO$ , whose area $20$ is already more than $18$ -- impossible -- so the line must exit through $AB$ , confirming the case used.

Standardsmin grade 8

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the area of the trapezoid and of the cut-off triangles.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Using coordinates of vertices to read lengths of sides in the coordinate plane.
8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Writing the line through the origin as $y=ax$ and finding $a$ from a point.

💡Takeaway. Before cutting a shape in half, find out which side the cut lands on -- testing a corner tells you right away.

Find which side the bisector hits first · 12 practice problems

Generated variants — 12

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Constraints

Givens

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Constraints

Givens

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Givens

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Constraints

Givens

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Givens

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Givens

Unknowns

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Where next?