Practice · Decide which side the halving line crosses · Sensim Math

Variant 1 easy

In the figure, point C lies on the graph of the direct proportion $y=3x$ and has $x$ -coordinate $2$ , and A is the point $(6, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=3x$ with $x$ -coordinate $2$ .
A $(6, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=3x

at

x=2

, and B is level with C directly above A.

y = 3 \times 2 = 6 \Rightarrow C(2, 6),\ B(6, 6)

Plugging

x

into

y=3x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=6

and CB

=6-2=4

; height

6

.

\frac{1}{2} \times (6 + 4) \times 6 = 30$, half is $15

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 6 \times 6 = 18 > 15

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=6

) at D

(6, 6a)

. Triangle OAD has base

6

and height

6a

.

\frac{1}{2} \times 6 \times 6a = 15 \Rightarrow 18a = 15 \Rightarrow a = \frac{5}{6}

One unknown, one equation.

Answer:

\frac{5}{6}

4 · Reviewdoes it hold up?

D $= (6, 5)$ lies on AB between $0$ and $6$ , and triangle OAD has area $\frac{1}{2} \times 6 \times 5 = 15$ , half of $30$ . Also $\frac{5}{6} < \frac{6}{6}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $30 - 18a$ ; setting $30 - 18a = 18a$ gives the same $a = \frac{5}{6}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=3x$ and the point $(6, 6a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 2 easy

In the figure, point C lies on the graph of the direct proportion $y=2x$ and has $x$ -coordinate $2$ , and A is the point $(8, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=2x$ with $x$ -coordinate $2$ .
A $(8, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=2x

at

x=2

, and B is level with C directly above A.

y = 2 \times 2 = 4 \Rightarrow C(2, 4),\ B(8, 4)

Plugging

x

into

y=2x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=8

and CB

=8-2=6

; height

4

.

\frac{1}{2} \times (8 + 6) \times 4 = 28$, half is $14

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 8 \times 4 = 16 > 14

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=8

) at D

(8, 8a)

. Triangle OAD has base

8

and height

8a

.

\frac{1}{2} \times 8 \times 8a = 14 \Rightarrow 32a = 14 \Rightarrow a = \frac{7}{16}

One unknown, one equation.

Answer:

\frac{7}{16}

4 · Reviewdoes it hold up?

D $= (8, 3.5)$ lies on AB between $0$ and $4$ , and triangle OAD has area $\frac{1}{2} \times 8 \times 3.5 = 14$ , half of $28$ . Also $\frac{7}{16} < \frac{4}{8}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $28 - 32a$ ; setting $28 - 32a = 32a$ gives the same $a = \frac{7}{16}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=2x$ and the point $(8, 8a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 3 easy

In the figure, point C lies on the graph of the direct proportion $y=3x$ and has $x$ -coordinate $4$ , and A is the point $(8, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=3x$ with $x$ -coordinate $4$ .
A $(8, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=3x

at

x=4

, and B is level with C directly above A.

y = 3 \times 4 = 12 \Rightarrow C(4, 12),\ B(8, 12)

Plugging

x

into

y=3x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=8

and CB

=8-4=4

; height

12

.

\frac{1}{2} \times (8 + 4) \times 12 = 72$, half is $36

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 8 \times 12 = 48 > 36

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=8

) at D

(8, 8a)

. Triangle OAD has base

8

and height

8a

.

\frac{1}{2} \times 8 \times 8a = 36 \Rightarrow 32a = 36 \Rightarrow a = \frac{9}{8}

One unknown, one equation.

Answer:

\frac{9}{8}

4 · Reviewdoes it hold up?

D $= (8, 9)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 8 \times 9 = 36$ , half of $72$ . Also $\frac{9}{8} < \frac{12}{8}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $72 - 32a$ ; setting $72 - 32a = 32a$ gives the same $a = \frac{9}{8}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=3x$ and the point $(8, 8a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 4 medium

In the figure, point C lies on the graph of the direct proportion $y=4x$ and has $x$ -coordinate $2$ , and A is the point $(8, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=4x$ with $x$ -coordinate $2$ .
A $(8, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=4x

at

x=2

, and B is level with C directly above A.

y = 4 \times 2 = 8 \Rightarrow C(2, 8),\ B(8, 8)

Plugging

x

into

y=4x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=8

and CB

=8-2=6

; height

8

.

\frac{1}{2} \times (8 + 6) \times 8 = 56$, half is $28

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 8 \times 8 = 32 > 28

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=8

) at D

(8, 8a)

. Triangle OAD has base

8

and height

8a

.

\frac{1}{2} \times 8 \times 8a = 28 \Rightarrow 32a = 28 \Rightarrow a = \frac{7}{8}

One unknown, one equation.

Answer:

\frac{7}{8}

4 · Reviewdoes it hold up?

D $= (8, 7)$ lies on AB between $0$ and $8$ , and triangle OAD has area $\frac{1}{2} \times 8 \times 7 = 28$ , half of $56$ . Also $\frac{7}{8} < \frac{8}{8}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $56 - 32a$ ; setting $56 - 32a = 32a$ gives the same $a = \frac{7}{8}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=4x$ and the point $(8, 8a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 5 easy

In the figure, point C lies on the graph of the direct proportion $y=2x$ and has $x$ -coordinate $4$ , and A is the point $(8, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=2x$ with $x$ -coordinate $4$ .
A $(8, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=2x

at

x=4

, and B is level with C directly above A.

y = 2 \times 4 = 8 \Rightarrow C(4, 8),\ B(8, 8)

Plugging

x

into

y=2x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=8

and CB

=8-4=4

; height

8

.

\frac{1}{2} \times (8 + 4) \times 8 = 48$, half is $24

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 8 \times 8 = 32 > 24

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=8

) at D

(8, 8a)

. Triangle OAD has base

8

and height

8a

.

\frac{1}{2} \times 8 \times 8a = 24 \Rightarrow 32a = 24 \Rightarrow a = \frac{3}{4}

One unknown, one equation.

Answer:

\frac{3}{4}

4 · Reviewdoes it hold up?

D $= (8, 6)$ lies on AB between $0$ and $8$ , and triangle OAD has area $\frac{1}{2} \times 8 \times 6 = 24$ , half of $48$ . Also $\frac{3}{4} < \frac{8}{8}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $48 - 32a$ ; setting $48 - 32a = 32a$ gives the same $a = \frac{3}{4}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=2x$ and the point $(8, 8a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 6 medium

In the figure, point C lies on the graph of the direct proportion $y=2x$ and has $x$ -coordinate $4$ , and A is the point $(10, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=2x$ with $x$ -coordinate $4$ .
A $(10, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=2x

at

x=4

, and B is level with C directly above A.

y = 2 \times 4 = 8 \Rightarrow C(4, 8),\ B(10, 8)

Plugging

x

into

y=2x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=10

and CB

=10-4=6

; height

8

.

\frac{1}{2} \times (10 + 6) \times 8 = 64$, half is $32

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 10 \times 8 = 40 > 32

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=10

) at D

(10, 10a)

. Triangle OAD has base

10

and height

10a

.

\frac{1}{2} \times 10 \times 10a = 32 \Rightarrow 50a = 32 \Rightarrow a = \frac{16}{25}

One unknown, one equation.

Answer:

\frac{16}{25}

4 · Reviewdoes it hold up?

D $= (10, 6.4)$ lies on AB between $0$ and $8$ , and triangle OAD has area $\frac{1}{2} \times 10 \times 6.4 = 32$ , half of $64$ . Also $\frac{16}{25} < \frac{8}{10}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $64 - 50a$ ; setting $64 - 50a = 50a$ gives the same $a = \frac{16}{25}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=2x$ and the point $(10, 10a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 7 medium

In the figure, point C lies on the graph of the direct proportion $y=3x$ and has $x$ -coordinate $4$ , and A is the point $(10, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=3x$ with $x$ -coordinate $4$ .
A $(10, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=3x

at

x=4

, and B is level with C directly above A.

y = 3 \times 4 = 12 \Rightarrow C(4, 12),\ B(10, 12)

Plugging

x

into

y=3x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=10

and CB

=10-4=6

; height

12

.

\frac{1}{2} \times (10 + 6) \times 12 = 96$, half is $48

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 10 \times 12 = 60 > 48

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=10

) at D

(10, 10a)

. Triangle OAD has base

10

and height

10a

.

\frac{1}{2} \times 10 \times 10a = 48 \Rightarrow 50a = 48 \Rightarrow a = \frac{24}{25}

One unknown, one equation.

Answer:

\frac{24}{25}

4 · Reviewdoes it hold up?

D $= (10, 9.6)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 10 \times 9.6 = 48$ , half of $96$ . Also $\frac{24}{25} < \frac{12}{10}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $96 - 50a$ ; setting $96 - 50a = 50a$ gives the same $a = \frac{24}{25}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=3x$ and the point $(10, 10a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 8 hard

In the figure, point C lies on the graph of the direct proportion $y=4x$ and has $x$ -coordinate $2$ , and A is the point $(10, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=4x$ with $x$ -coordinate $2$ .
A $(10, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=4x

at

x=2

, and B is level with C directly above A.

y = 4 \times 2 = 8 \Rightarrow C(2, 8),\ B(10, 8)

Plugging

x

into

y=4x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=10

and CB

=10-2=8

; height

8

.

\frac{1}{2} \times (10 + 8) \times 8 = 72$, half is $36

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 10 \times 8 = 40 > 36

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=10

) at D

(10, 10a)

. Triangle OAD has base

10

and height

10a

.

\frac{1}{2} \times 10 \times 10a = 36 \Rightarrow 50a = 36 \Rightarrow a = \frac{18}{25}

One unknown, one equation.

Answer:

\frac{18}{25}

4 · Reviewdoes it hold up?

D $= (10, 7.2)$ lies on AB between $0$ and $8$ , and triangle OAD has area $\frac{1}{2} \times 10 \times 7.2 = 36$ , half of $72$ . Also $\frac{18}{25} < \frac{8}{10}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $72 - 50a$ ; setting $72 - 50a = 50a$ gives the same $a = \frac{18}{25}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=4x$ and the point $(10, 10a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 9 medium

In the figure, point C lies on the graph of the direct proportion $y=2x$ and has $x$ -coordinate $6$ , and A is the point $(10, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=2x$ with $x$ -coordinate $6$ .
A $(10, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=2x

at

x=6

, and B is level with C directly above A.

y = 2 \times 6 = 12 \Rightarrow C(6, 12),\ B(10, 12)

Plugging

x

into

y=2x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=10

and CB

=10-6=4

; height

12

.

\frac{1}{2} \times (10 + 4) \times 12 = 84$, half is $42

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 10 \times 12 = 60 > 42

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=10

) at D

(10, 10a)

. Triangle OAD has base

10

and height

10a

.

\frac{1}{2} \times 10 \times 10a = 42 \Rightarrow 50a = 42 \Rightarrow a = \frac{21}{25}

One unknown, one equation.

Answer:

\frac{21}{25}

4 · Reviewdoes it hold up?

D $= (10, 8.4)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 10 \times 8.4 = 42$ , half of $84$ . Also $\frac{21}{25} < \frac{12}{10}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $84 - 50a$ ; setting $84 - 50a = 50a$ gives the same $a = \frac{21}{25}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=2x$ and the point $(10, 10a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 10 hard

In the figure, point C lies on the graph of the direct proportion $y=4x$ and has $x$ -coordinate $3$ , and A is the point $(12, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=4x$ with $x$ -coordinate $3$ .
A $(12, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=4x

at

x=3

, and B is level with C directly above A.

y = 4 \times 3 = 12 \Rightarrow C(3, 12),\ B(12, 12)

Plugging

x

into

y=4x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=12

and CB

=12-3=9

; height

12

.

\frac{1}{2} \times (12 + 9) \times 12 = 126$, half is $63

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 12 \times 12 = 72 > 63

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=12

) at D

(12, 12a)

. Triangle OAD has base

12

and height

12a

.

\frac{1}{2} \times 12 \times 12a = 63 \Rightarrow 72a = 63 \Rightarrow a = \frac{7}{8}

One unknown, one equation.

Answer:

\frac{7}{8}

4 · Reviewdoes it hold up?

D $= (12, 10.5)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 12 \times 10.5 = 63$ , half of $126$ . Also $\frac{7}{8} < \frac{12}{12}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $126 - 72a$ ; setting $126 - 72a = 72a$ gives the same $a = \frac{7}{8}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=4x$ and the point $(12, 12a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 11 hard

In the figure, point C lies on the graph of the direct proportion $y=2x$ and has $x$ -coordinate $6$ , and A is the point $(12, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=2x$ with $x$ -coordinate $6$ .
A $(12, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=2x

at

x=6

, and B is level with C directly above A.

y = 2 \times 6 = 12 \Rightarrow C(6, 12),\ B(12, 12)

Plugging

x

into

y=2x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=12

and CB

=12-6=6

; height

12

.

\frac{1}{2} \times (12 + 6) \times 12 = 108$, half is $54

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 12 \times 12 = 72 > 54

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=12

) at D

(12, 12a)

. Triangle OAD has base

12

and height

12a

.

\frac{1}{2} \times 12 \times 12a = 54 \Rightarrow 72a = 54 \Rightarrow a = \frac{3}{4}

One unknown, one equation.

Answer:

\frac{3}{4}

4 · Reviewdoes it hold up?

D $= (12, 9)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 12 \times 9 = 54$ , half of $108$ . Also $\frac{3}{4} < \frac{12}{12}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $108 - 72a$ ; setting $108 - 72a = 72a$ gives the same $a = \frac{3}{4}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=2x$ and the point $(12, 12a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Variant 12 hard

In the figure, point C lies on the graph of the direct proportion $y=3x$ and has $x$ -coordinate $4$ , and A is the point $(12, 0)$ . Side CB is parallel to the $x$ -axis and side BA is perpendicular to it, so OABC is a trapezoid. If the graph of the direct proportion $y=ax$ divides the area of trapezoid OABC into two equal parts, find the constant $a$ . (O is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

Find the trapezoid's vertices and area. A line $y=ax$ through O must cut off exactly half the area. First decide which side of the trapezoid the line must hit, then use a triangle-area equation to find $a$ .

Givens

C is on $y=3x$ with $x$ -coordinate $4$ .
A $(12, 0)$ ; CB is parallel to the $x$ -axis and BA is vertical.
$y=ax$ halves the area of trapezoid OABC.

Unknowns

The constant $a$ .

Constraints

The halving line passes through the vertex O.

2 · Planchoose the strategy

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

Break it into subproblems: coordinates, total area, which side the line crosses (compare the diagonal OB's triangle with half the area), then one area equation.

3 · Execute4 carry out the plan

1Locate C and B

#1 Draw a Diagram 7.RP.A.2

C is on

y=3x

at

x=4

, and B is level with C directly above A.

y = 3 \times 4 = 12 \Rightarrow C(4, 12),\ B(12, 12)

Plugging

x

into

y=3x

gives the height of C.

2Area of the trapezoid

#7 Identify Subproblems 7.G.B.6

Parallel sides: OA

=12

and CB

=12-4=8

; height

12

.

\frac{1}{2} \times (12 + 8) \times 12 = 120$, half is $60

Trapezoid area is the average of the parallel sides times the height.

3Which side does the line hit?

#7 Identify Subproblems 7.G.B.6

The diagonal OB cuts off triangle OAB. Compare it with half the area.

\frac{1}{2} \times 12 \times 12 = 72 > 60

Even the line through B takes too much, so the halving line must be flatter and cross side AB below B.

4Area equation on side AB

#13 Convert to Algebra 7.EE.B.4

The line meets AB (where

x=12

) at D

(12, 12a)

. Triangle OAD has base

12

and height

12a

.

\frac{1}{2} \times 12 \times 12a = 60 \Rightarrow 72a = 60 \Rightarrow a = \frac{5}{6}

One unknown, one equation.

Answer:

\frac{5}{6}

4 · Reviewdoes it hold up?

D $= (12, 10)$ lies on AB between $0$ and $12$ , and triangle OAD has area $\frac{1}{2} \times 12 \times 10 = 60$ , half of $120$ . Also $\frac{5}{6} < \frac{12}{12}$ , the slope of OB, as expected.

Another way: Measure the other piece: the region above the line is the quadrilateral ODBC with area $120 - 72a$ ; setting $120 - 72a = 72a$ gives the same $a = \frac{5}{6}$ .

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding a point on $y=3x$ and the point $(12, 12a)$ on $y=ax$.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Areas of the trapezoid and of the triangles.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the area equation for $a$.

💡Takeaway. Before solving, test the easy line through a corner -- it tells you which side the answer lives on.

Decide which side the halving line crosses · 12 practice problems

Generated variants — 12

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