Practice · Moving-point triangle area by subtraction · Sensim Math

Variant 1 easy

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{5}{9}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 6$ cm, $\overline{\mathrm{AB}} = 9$ cm and $\overline{\mathrm{BC}} = 12$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{5}{9}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 6$ cm, $\overline{\mathrm{AB}} = 9$ cm, $\overline{\mathrm{BC}} = 12$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{5}{9} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 9$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (6 + 12) \times 9 = 81

The trapezoid has area

81\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 9 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 6 \times (9 - x) = 27 - 3x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 12 \times x = 6x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 81 - \left(27 - 3x + 6x\right) = 54 - 3x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

54

.

4Solve the area condition

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

The shaded area must be at most

\frac{5}{9}

of

81

, which is

45

.

54 - 3x \le 81 \times \frac{5}{9} = 45 \;\Rightarrow\; 9 \le 3x \;\Rightarrow\; x \ge 3

\mathrm{P}

has to be at least

3

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

9

cm long, so

x

cannot exceed

9

.

3 \le x \le 9

Every position from

3

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

3

cm and at most

9

cm (

3 \le \overline{\mathrm{BP}} \le 9

)

4 · Reviewdoes it hold up?

At $x = 3$ the shaded area is $54 - 9 = 45 = \frac{5}{9} \times 81$ , exactly the limit. At $x = 9$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $54 - 27 = 27$ , which is $\frac{1}{2} \times 6 \times 9$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $54 = \frac{1}{2} \times 12 \times 9$ , too large.

Another way: Since the shaded area is at most $45$ , the two right triangles together must cover at least $81 - 45 = 36$ : $27 - 3x + 6x \ge 36$ , so $3x \ge 9$ and $x \ge 3$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $54 - 3x \le 45$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 2 easy

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{6}{13}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 5$ cm, $\overline{\mathrm{AB}} = 6$ cm and $\overline{\mathrm{BC}} = 8$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{6}{13}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 5$ cm, $\overline{\mathrm{AB}} = 6$ cm, $\overline{\mathrm{BC}} = 8$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{6}{13} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 6$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (5 + 8) \times 6 = 39

The trapezoid has area

39\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 6 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 5 \times (6 - x) = 15 - \frac{5}{2}x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 8 \times x = 4x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 39 - \left(15 - \frac{5}{2}x + 4x\right) = 24 - \frac{3}{2}x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

24

.

4Solve the area condition

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

The shaded area must be at most

\frac{6}{13}

of

39

, which is

18

.

24 - \frac{3}{2}x \le 39 \times \frac{6}{13} = 18 \;\Rightarrow\; 6 \le \frac{3}{2}x \;\Rightarrow\; x \ge 4

\mathrm{P}

has to be at least

4

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

6

cm long, so

x

cannot exceed

6

.

4 \le x \le 6

Every position from

4

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

4

cm and at most

6

cm (

4 \le \overline{\mathrm{BP}} \le 6

)

4 · Reviewdoes it hold up?

At $x = 4$ the shaded area is $24 - 6 = 18 = \frac{6}{13} \times 39$ , exactly the limit. At $x = 6$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $24 - 9 = 15$ , which is $\frac{1}{2} \times 5 \times 6$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $24 = \frac{1}{2} \times 8 \times 6$ , too large.

Another way: Since the shaded area is at most $18$ , the two right triangles together must cover at least $39 - 18 = 21$ : $15 - \frac{5}{2}x + 4x \ge 21$ , so $\frac{3}{2}x \ge 6$ and $x \ge 4$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $24 - \frac{3}{2}x \le 18$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 3 easy

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{4}{7}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 7$ cm, $\overline{\mathrm{AB}} = 14$ cm and $\overline{\mathrm{BC}} = 14$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{4}{7}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 7$ cm, $\overline{\mathrm{AB}} = 14$ cm, $\overline{\mathrm{BC}} = 14$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{4}{7} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 14$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (7 + 14) \times 14 = 147

The trapezoid has area

147\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 7 \times (14 - x) = 49 - \frac{7}{2}x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 14 \times x = 7x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 147 - \left(49 - \frac{7}{2}x + 7x\right) = 98 - \frac{7}{2}x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

98

.

4Solve the area condition

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

The shaded area must be at most

\frac{4}{7}

of

147

, which is

84

.

98 - \frac{7}{2}x \le 147 \times \frac{4}{7} = 84 \;\Rightarrow\; 14 \le \frac{7}{2}x \;\Rightarrow\; x \ge 4

\mathrm{P}

has to be at least

4

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

14

cm long, so

x

cannot exceed

14

.

4 \le x \le 14

Every position from

4

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

4

cm and at most

14

cm (

4 \le \overline{\mathrm{BP}} \le 14

)

4 · Reviewdoes it hold up?

At $x = 4$ the shaded area is $98 - 14 = 84 = \frac{4}{7} \times 147$ , exactly the limit. At $x = 14$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $98 - 49 = 49$ , which is $\frac{1}{2} \times 7 \times 14$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $98 = \frac{1}{2} \times 14 \times 14$ , too large.

Another way: Since the shaded area is at most $84$ , the two right triangles together must cover at least $147 - 84 = 63$ : $49 - \frac{7}{2}x + 7x \ge 63$ , so $\frac{7}{2}x \ge 14$ and $x \ge 4$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $98 - \frac{7}{2}x \le 84$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 4 easy

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{3}{7}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 5$ cm, $\overline{\mathrm{AB}} = 14$ cm and $\overline{\mathrm{BC}} = 10$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{3}{7}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 5$ cm, $\overline{\mathrm{AB}} = 14$ cm, $\overline{\mathrm{BC}} = 10$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{3}{7} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 14$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (5 + 10) \times 14 = 105

The trapezoid has area

105\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 5 \times (14 - x) = 35 - \frac{5}{2}x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 10 \times x = 5x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 105 - \left(35 - \frac{5}{2}x + 5x\right) = 70 - \frac{5}{2}x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

70

.

4Solve the area condition

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

The shaded area must be at most

\frac{3}{7}

of

105

, which is

45

.

70 - \frac{5}{2}x \le 105 \times \frac{3}{7} = 45 \;\Rightarrow\; 25 \le \frac{5}{2}x \;\Rightarrow\; x \ge 10

\mathrm{P}

has to be at least

10

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

14

cm long, so

x

cannot exceed

14

.

10 \le x \le 14

Every position from

10

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

10

cm and at most

14

cm (

10 \le \overline{\mathrm{BP}} \le 14

)

4 · Reviewdoes it hold up?

At $x = 10$ the shaded area is $70 - 25 = 45 = \frac{3}{7} \times 105$ , exactly the limit. At $x = 14$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $70 - 35 = 35$ , which is $\frac{1}{2} \times 5 \times 14$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $70 = \frac{1}{2} \times 10 \times 14$ , too large.

Another way: Since the shaded area is at most $45$ , the two right triangles together must cover at least $105 - 45 = 60$ : $35 - \frac{5}{2}x + 5x \ge 60$ , so $\frac{5}{2}x \ge 25$ and $x \ge 10$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $70 - \frac{5}{2}x \le 45$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 5 hard

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{7}{15}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 8$ cm, $\overline{\mathrm{AB}} = 9$ cm and $\overline{\mathrm{BC}} = 12$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{7}{15}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 8$ cm, $\overline{\mathrm{AB}} = 9$ cm, $\overline{\mathrm{BC}} = 12$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{7}{15} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 9$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (8 + 12) \times 9 = 90

The trapezoid has area

90\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 9 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 8 \times (9 - x) = 36 - 4x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 12 \times x = 6x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 90 - \left(36 - 4x + 6x\right) = 54 - 2x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

54

.

4Solve the area condition

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

The shaded area must be at most

\frac{7}{15}

of

90

, which is

42

.

54 - 2x \le 90 \times \frac{7}{15} = 42 \;\Rightarrow\; 12 \le 2x \;\Rightarrow\; x \ge 6

\mathrm{P}

has to be at least

6

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

9

cm long, so

x

cannot exceed

9

.

6 \le x \le 9

Every position from

6

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

6

cm and at most

9

cm (

6 \le \overline{\mathrm{BP}} \le 9

)

4 · Reviewdoes it hold up?

At $x = 6$ the shaded area is $54 - 12 = 42 = \frac{7}{15} \times 90$ , exactly the limit. At $x = 9$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $54 - 18 = 36$ , which is $\frac{1}{2} \times 8 \times 9$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $54 = \frac{1}{2} \times 12 \times 9$ , too large.

Another way: Since the shaded area is at most $42$ , the two right triangles together must cover at least $90 - 42 = 48$ : $36 - 4x + 6x \ge 48$ , so $2x \ge 12$ and $x \ge 6$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $54 - 2x \le 42$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 6 medium

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{7}{15}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 6$ cm and $\overline{\mathrm{BC}} = 15$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{7}{15}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 6$ cm, $\overline{\mathrm{BC}} = 15$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{7}{15} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 6$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (10 + 15) \times 6 = 75

The trapezoid has area

75\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 6 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 10 \times (6 - x) = 30 - 5x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 15 \times x = \frac{15}{2}x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 75 - \left(30 - 5x + \frac{15}{2}x\right) = 45 - \frac{5}{2}x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

45

.

4Solve the area condition

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

The shaded area must be at most

\frac{7}{15}

of

75

, which is

35

.

45 - \frac{5}{2}x \le 75 \times \frac{7}{15} = 35 \;\Rightarrow\; 10 \le \frac{5}{2}x \;\Rightarrow\; x \ge 4

\mathrm{P}

has to be at least

4

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

6

cm long, so

x

cannot exceed

6

.

4 \le x \le 6

Every position from

4

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

4

cm and at most

6

cm (

4 \le \overline{\mathrm{BP}} \le 6

)

4 · Reviewdoes it hold up?

At $x = 4$ the shaded area is $45 - 10 = 35 = \frac{7}{15} \times 75$ , exactly the limit. At $x = 6$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $45 - 15 = 30$ , which is $\frac{1}{2} \times 10 \times 6$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $45 = \frac{1}{2} \times 15 \times 6$ , too large.

Another way: Since the shaded area is at most $35$ , the two right triangles together must cover at least $75 - 35 = 40$ : $30 - 5x + \frac{15}{2}x \ge 40$ , so $\frac{5}{2}x \ge 10$ and $x \ge 4$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $45 - \frac{5}{2}x \le 35$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 7 medium

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{8}{15}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 12$ cm and $\overline{\mathrm{BC}} = 15$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{8}{15}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 12$ cm, $\overline{\mathrm{BC}} = 15$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{8}{15} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 12$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (10 + 15) \times 12 = 150

The trapezoid has area

150\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 10 \times (12 - x) = 60 - 5x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 15 \times x = \frac{15}{2}x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 150 - \left(60 - 5x + \frac{15}{2}x\right) = 90 - \frac{5}{2}x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

90

.

4Solve the area condition

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

The shaded area must be at most

\frac{8}{15}

of

150

, which is

80

.

90 - \frac{5}{2}x \le 150 \times \frac{8}{15} = 80 \;\Rightarrow\; 10 \le \frac{5}{2}x \;\Rightarrow\; x \ge 4

\mathrm{P}

has to be at least

4

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

12

cm long, so

x

cannot exceed

12

.

4 \le x \le 12

Every position from

4

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

4

cm and at most

12

cm (

4 \le \overline{\mathrm{BP}} \le 12

)

4 · Reviewdoes it hold up?

At $x = 4$ the shaded area is $90 - 10 = 80 = \frac{8}{15} \times 150$ , exactly the limit. At $x = 12$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $90 - 30 = 60$ , which is $\frac{1}{2} \times 10 \times 12$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $90 = \frac{1}{2} \times 15 \times 12$ , too large.

Another way: Since the shaded area is at most $80$ , the two right triangles together must cover at least $150 - 80 = 70$ : $60 - 5x + \frac{15}{2}x \ge 70$ , so $\frac{5}{2}x \ge 10$ and $x \ge 4$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $90 - \frac{5}{2}x \le 80$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 8 medium

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{4}{7}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 9$ cm, $\overline{\mathrm{AB}} = 14$ cm and $\overline{\mathrm{BC}} = 15$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{4}{7}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 9$ cm, $\overline{\mathrm{AB}} = 14$ cm, $\overline{\mathrm{BC}} = 15$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{4}{7} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 14$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (9 + 15) \times 14 = 168

The trapezoid has area

168\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 9 \times (14 - x) = 63 - \frac{9}{2}x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 15 \times x = \frac{15}{2}x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 168 - \left(63 - \frac{9}{2}x + \frac{15}{2}x\right) = 105 - 3x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

105

.

4Solve the area condition

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

The shaded area must be at most

\frac{4}{7}

of

168

, which is

96

.

105 - 3x \le 168 \times \frac{4}{7} = 96 \;\Rightarrow\; 9 \le 3x \;\Rightarrow\; x \ge 3

\mathrm{P}

has to be at least

3

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

14

cm long, so

x

cannot exceed

14

.

3 \le x \le 14

Every position from

3

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

3

cm and at most

14

cm (

3 \le \overline{\mathrm{BP}} \le 14

)

4 · Reviewdoes it hold up?

At $x = 3$ the shaded area is $105 - 9 = 96 = \frac{4}{7} \times 168$ , exactly the limit. At $x = 14$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $105 - 42 = 63$ , which is $\frac{1}{2} \times 9 \times 14$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $105 = \frac{1}{2} \times 15 \times 14$ , too large.

Another way: Since the shaded area is at most $96$ , the two right triangles together must cover at least $168 - 96 = 72$ : $63 - \frac{9}{2}x + \frac{15}{2}x \ge 72$ , so $3x \ge 9$ and $x \ge 3$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $105 - 3x \le 96$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 9 medium

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{8}{15}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 4$ cm, $\overline{\mathrm{AB}} = 15$ cm and $\overline{\mathrm{BC}} = 8$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{8}{15}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 4$ cm, $\overline{\mathrm{AB}} = 15$ cm, $\overline{\mathrm{BC}} = 8$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{8}{15} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 15$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (4 + 8) \times 15 = 90

The trapezoid has area

90\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 15 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 4 \times (15 - x) = 30 - 2x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 8 \times x = 4x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 90 - \left(30 - 2x + 4x\right) = 60 - 2x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

60

.

4Solve the area condition

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

The shaded area must be at most

\frac{8}{15}

of

90

, which is

48

.

60 - 2x \le 90 \times \frac{8}{15} = 48 \;\Rightarrow\; 12 \le 2x \;\Rightarrow\; x \ge 6

\mathrm{P}

has to be at least

6

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

15

cm long, so

x

cannot exceed

15

.

6 \le x \le 15

Every position from

6

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

6

cm and at most

15

cm (

6 \le \overline{\mathrm{BP}} \le 15

)

4 · Reviewdoes it hold up?

At $x = 6$ the shaded area is $60 - 12 = 48 = \frac{8}{15} \times 90$ , exactly the limit. At $x = 15$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $60 - 30 = 30$ , which is $\frac{1}{2} \times 4 \times 15$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $60 = \frac{1}{2} \times 8 \times 15$ , too large.

Another way: Since the shaded area is at most $48$ , the two right triangles together must cover at least $90 - 48 = 42$ : $30 - 2x + 4x \ge 42$ , so $2x \ge 12$ and $x \ge 6$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $60 - 2x \le 48$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 10 hard

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{5}{12}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 16$ cm and $\overline{\mathrm{BC}} = 20$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{5}{12}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 16$ cm, $\overline{\mathrm{BC}} = 20$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{5}{12} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 16$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (10 + 20) \times 16 = 240

The trapezoid has area

240\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 10 \times (16 - x) = 80 - 5x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 20 \times x = 10x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 240 - \left(80 - 5x + 10x\right) = 160 - 5x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

160

.

4Solve the area condition

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

The shaded area must be at most

\frac{5}{12}

of

240

, which is

100

.

160 - 5x \le 240 \times \frac{5}{12} = 100 \;\Rightarrow\; 60 \le 5x \;\Rightarrow\; x \ge 12

\mathrm{P}

has to be at least

12

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

16

cm long, so

x

cannot exceed

16

.

12 \le x \le 16

Every position from

12

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

12

cm and at most

16

cm (

12 \le \overline{\mathrm{BP}} \le 16

)

4 · Reviewdoes it hold up?

At $x = 12$ the shaded area is $160 - 60 = 100 = \frac{5}{12} \times 240$ , exactly the limit. At $x = 16$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $160 - 80 = 80$ , which is $\frac{1}{2} \times 10 \times 16$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $160 = \frac{1}{2} \times 20 \times 16$ , too large.

Another way: Since the shaded area is at most $100$ , the two right triangles together must cover at least $240 - 100 = 140$ : $80 - 5x + 10x \ge 140$ , so $5x \ge 60$ and $x \ge 12$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $160 - 5x \le 100$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 11 hard

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{5}{12}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 8$ cm and $\overline{\mathrm{BC}} = 20$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{5}{12}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 8$ cm, $\overline{\mathrm{BC}} = 20$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{5}{12} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 8$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (10 + 20) \times 8 = 120

The trapezoid has area

120\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

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

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 10 \times (8 - x) = 40 - 5x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 20 \times x = 10x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 120 - \left(40 - 5x + 10x\right) = 80 - 5x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

80

.

4Solve the area condition

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

The shaded area must be at most

\frac{5}{12}

of

120

, which is

50

.

80 - 5x \le 120 \times \frac{5}{12} = 50 \;\Rightarrow\; 30 \le 5x \;\Rightarrow\; x \ge 6

\mathrm{P}

has to be at least

6

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

8

cm long, so

x

cannot exceed

8

.

6 \le x \le 8

Every position from

6

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

6

cm and at most

8

cm (

6 \le \overline{\mathrm{BP}} \le 8

)

4 · Reviewdoes it hold up?

At $x = 6$ the shaded area is $80 - 30 = 50 = \frac{5}{12} \times 120$ , exactly the limit. At $x = 8$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $80 - 40 = 40$ , which is $\frac{1}{2} \times 10 \times 8$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $80 = \frac{1}{2} \times 20 \times 8$ , too large.

Another way: Since the shaded area is at most $50$ , the two right triangles together must cover at least $120 - 50 = 70$ : $40 - 5x + 10x \ge 70$ , so $5x \ge 30$ and $x \ge 6$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $80 - 5x \le 50$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Variant 12 hard

In trapezoid $\mathrm{ABCD}$ shown at the right, point $\mathrm{P}$ moves along side $\mathrm{AB}$ . We want the area of $\triangle \mathrm{DPC}$ to be at most $\frac{4}{9}$ of the area of trapezoid $\mathrm{ABCD}$ . Find the range of the length of $\overline{\mathrm{BP}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a right trapezoid $\mathrm{ABCD}$ with right angles at $\mathrm{A}$ and $\mathrm{B}$ , $\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 15$ cm and $\overline{\mathrm{BC}} = 20$ cm. Point $\mathrm{P}$ slides along $\mathrm{AB}$ . Find all lengths $\overline{\mathrm{BP}}$ for which $\triangle \mathrm{DPC}$ has at most $\frac{4}{9}$ of the trapezoid's area.

Givens

$\angle \mathrm{A}$ and $\angle \mathrm{B}$ are right angles.
$\overline{\mathrm{AD}} = 10$ cm, $\overline{\mathrm{AB}} = 15$ cm, $\overline{\mathrm{BC}} = 20$ cm.
$\mathrm{P}$ is a point on side $\mathrm{AB}$ .
Area of $\triangle \mathrm{DPC} \le \frac{4}{9} \times$ area of $\mathrm{ABCD}$ .

Unknowns

The range of $\overline{\mathrm{BP}}$ .

Constraints

$\mathrm{P}$ stays on side $\mathrm{AB}$ , so $0 \le \overline{\mathrm{BP}} \le 15$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#13 Convert to Algebra

The slanted triangle $\mathrm{DPC}$ is awkward to measure directly, but what is left of the trapezoid is two right triangles whose legs lie along the sides. Subtract them from the trapezoid to get the shaded area as an expression in $\overline{\mathrm{BP}}$ , then solve the area inequality.

3 · Execute5 carry out the plan

1Area of the trapezoid

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

The parallel sides are

\mathrm{AD}

and

\mathrm{BC}

, and the height is

\mathrm{AB}

.

\frac{1}{2} \times (10 + 20) \times 15 = 225

The trapezoid has area

225\ \text{cm}^2

.

2The two right triangles left over

#16 Count the Complement 6.G.A.1

Let

\overline{\mathrm{BP}} = x

cm, so

\overline{\mathrm{AP}} = 15 - x

cm. Triangle

\mathrm{APD}

has a right angle at

\mathrm{A}

and triangle

\mathrm{PBC}

has a right angle at

\mathrm{B}

, so their legs are their base and height.

\triangle \mathrm{APD} = \frac{1}{2} \times 10 \times (15 - x) = 75 - 5x, \qquad \triangle \mathrm{PBC} = \frac{1}{2} \times 20 \times x = 10x

These two triangles plus

\triangle \mathrm{DPC}

fill the trapezoid exactly.

3Area of the shaded triangle

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

Subtract the two right triangles from the trapezoid.

\triangle \mathrm{DPC} = 225 - \left(75 - 5x + 10x\right) = 150 - 5x

As

\mathrm{P}

moves up from

\mathrm{B}

, the shaded area shrinks steadily from

150

.

4Solve the area condition

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

The shaded area must be at most

\frac{4}{9}

of

225

, which is

100

.

150 - 5x \le 225 \times \frac{4}{9} = 100 \;\Rightarrow\; 50 \le 5x \;\Rightarrow\; x \ge 10

\mathrm{P}

has to be at least

10

cm above

\mathrm{B}

.

5Combine with where P can be

#1 Draw a Diagram 7.EE.B.4

\mathrm{P}

lies on side

\mathrm{AB}

, which is

15

cm long, so

x

cannot exceed

15

.

10 \le x \le 15

Every position from

10

cm above

\mathrm{B}

up to

\mathrm{A}

works.

Answer: at least

10

cm and at most

15

cm (

10 \le \overline{\mathrm{BP}} \le 15

)

4 · Reviewdoes it hold up?

At $x = 10$ the shaded area is $150 - 50 = 100 = \frac{4}{9} \times 225$ , exactly the limit. At $x = 15$ ( $\mathrm{P}$ at $\mathrm{A}$ ) it is $150 - 75 = 75$ , which is $\frac{1}{2} \times 10 \times 15$ , the area of triangle $\mathrm{DAC}$ . At $x = 0$ ( $\mathrm{P}$ at $\mathrm{B}$ ) it is $150 = \frac{1}{2} \times 20 \times 15$ , too large.

Another way: Since the shaded area is at most $100$ , the two right triangles together must cover at least $225 - 100 = 125$ : $75 - 5x + 10x \ge 125$ , so $5x \ge 50$ and $x \ge 10$ .

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Finding the trapezoid's area and the two right triangles' areas, then decomposing the trapezoid.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the shaded area in terms of $x$ and solving $150 - 5x \le 100$.

💡Takeaway. If a shape is hard to measure, measure what is left around it and subtract.

Moving-point triangle area by subtraction · 12 practice problems

Generated variants — 12

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