Practice · Trapezoid minus two right triangles, in letters · Sensim Math

Variant 1 easy

For the points $P(a, b)$ and $Q(-a, 3b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $16$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-a, 3b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $16$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 3b

, and its height is

H'H = a-(-a) = 2a

.

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

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = a

and

QH' = 3b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times a \times 3b = \frac{3}{2}ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

4ab - \left(\frac{3}{2}ab + \frac{ab}{2}\right) = 2ab

Combine like terms:

4 - \frac{3}{2} - \frac{1}{2} = 2

.

5Use the given area

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

Set the expression equal to

16

and solve for

ab

.

2ab = 16 \Rightarrow ab = 8

Divide both sides by

2

.

Answer:

8

4 · Reviewdoes it hold up?

Try $a=2$ , $b=4$ (so $ab=8$ ): $P(2, 4)$ , $Q(-2, 12)$ . The trapezoid is $\frac{1}{2} \times (4+12) \times 4 = 32$ , minus $\frac{1}{2} \times 2 \times 12 = 12$ and $\frac{1}{2} \times 2 \times 4 = 4$ , leaves $16$ .

Another way: Enclose the triangle in the rectangle from $x=-a$ to $x=a$ and $y=0$ to $y=3b$ , area $6ab$ , and subtract the three right triangles $\frac{3}{2}ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 2a \times 2b = 2ab$ . Together they take away $4ab$ , and $6 - 4 = 2$ , so the triangle is again $2ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $2ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $2ab = 16$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 2 easy

For the points $P(a, b)$ and $Q(-a, 2b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $18$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-a, 2b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $18$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 2b

, and its height is

H'H = a-(-a) = 2a

.

\frac{1}{2} \times (b + 2b) \times 2a = 3ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = a

and

QH' = 2b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times a \times 2b = ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

3ab - \left(ab + \frac{ab}{2}\right) = \frac{3}{2}ab

Combine like terms:

3 - 1 - \frac{1}{2} = \frac{3}{2}

.

5Use the given area

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

Set the expression equal to

18

and solve for

ab

.

\frac{3}{2}ab = 18 \Rightarrow ab = 12

Multiply both sides by

\frac{2}{3}

.

Answer:

12

4 · Reviewdoes it hold up?

Try $a=3$ , $b=4$ (so $ab=12$ ): $P(3, 4)$ , $Q(-3, 8)$ . The trapezoid is $\frac{1}{2} \times (4+8) \times 6 = 36$ , minus $\frac{1}{2} \times 3 \times 8 = 12$ and $\frac{1}{2} \times 3 \times 4 = 6$ , leaves $18$ .

Another way: Enclose the triangle in the rectangle from $x=-a$ to $x=a$ and $y=0$ to $y=2b$ , area $4ab$ , and subtract the three right triangles $ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 2a \times b = ab$ . Together they take away $\frac{5}{2}ab$ , and $4 - \frac{5}{2} = \frac{3}{2}$ , so the triangle is again $\frac{3}{2}ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $\frac{3}{2}ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $\frac{3}{2}ab = 18$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 3 easy

For the points $P(a, b)$ and $Q(-2a, 2b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $10$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-2a, 2b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $10$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-2a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 2b

, and its height is

H'H = a-(-2a) = 3a

.

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

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 2a

and

QH' = 2b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 2a \times 2b = 2ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

\frac{9}{2}ab - \left(2ab + \frac{ab}{2}\right) = 2ab

Combine like terms:

\frac{9}{2} - 2 - \frac{1}{2} = 2

.

5Use the given area

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

Set the expression equal to

10

and solve for

ab

.

2ab = 10 \Rightarrow ab = 5

Divide both sides by

2

.

Answer:

5

4 · Reviewdoes it hold up?

Try $a=1$ , $b=5$ (so $ab=5$ ): $P(1, 5)$ , $Q(-2, 10)$ . The trapezoid is $\frac{1}{2} \times (5+10) \times 3 = 22\frac{1}{2}$ , minus $\frac{1}{2} \times 2 \times 10 = 10$ and $\frac{1}{2} \times 1 \times 5 = 2\frac{1}{2}$ , leaves $10$ .

Another way: Enclose the triangle in the rectangle from $x=-2a$ to $x=a$ and $y=0$ to $y=2b$ , area $6ab$ , and subtract the three right triangles $2ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 3a \times b = \frac{3}{2}ab$ . Together they take away $4ab$ , and $6 - 4 = 2$ , so the triangle is again $2ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $2ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $2ab = 10$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 4 easy

For the points $P(a, b)$ and $Q(-a, 5b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $21$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-a, 5b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $21$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 5b

, and its height is

H'H = a-(-a) = 2a

.

\frac{1}{2} \times (b + 5b) \times 2a = 6ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = a

and

QH' = 5b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times a \times 5b = \frac{5}{2}ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

6ab - \left(\frac{5}{2}ab + \frac{ab}{2}\right) = 3ab

Combine like terms:

6 - \frac{5}{2} - \frac{1}{2} = 3

.

5Use the given area

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

Set the expression equal to

21

and solve for

ab

.

3ab = 21 \Rightarrow ab = 7

Divide both sides by

3

.

Answer:

7

4 · Reviewdoes it hold up?

Try $a=1$ , $b=7$ (so $ab=7$ ): $P(1, 7)$ , $Q(-1, 35)$ . The trapezoid is $\frac{1}{2} \times (7+35) \times 2 = 42$ , minus $\frac{1}{2} \times 1 \times 35 = 17\frac{1}{2}$ and $\frac{1}{2} \times 1 \times 7 = 3\frac{1}{2}$ , leaves $21$ .

Another way: Enclose the triangle in the rectangle from $x=-a$ to $x=a$ and $y=0$ to $y=5b$ , area $10ab$ , and subtract the three right triangles $\frac{5}{2}ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 2a \times 4b = 4ab$ . Together they take away $7ab$ , and $10 - 7 = 3$ , so the triangle is again $3ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $3ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $3ab = 21$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 5 medium

For the points $P(a, b)$ and $Q(-4a, 4b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $12$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-4a, 4b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $12$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-4a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 4b

, and its height is

H'H = a-(-4a) = 5a

.

\frac{1}{2} \times (b + 4b) \times 5a = \frac{25}{2}ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 4a

and

QH' = 4b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 4a \times 4b = 8ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

\frac{25}{2}ab - \left(8ab + \frac{ab}{2}\right) = 4ab

Combine like terms:

\frac{25}{2} - 8 - \frac{1}{2} = 4

.

5Use the given area

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

Set the expression equal to

12

and solve for

ab

.

4ab = 12 \Rightarrow ab = 3

Divide both sides by

4

.

Answer:

3

4 · Reviewdoes it hold up?

Try $a=1$ , $b=3$ (so $ab=3$ ): $P(1, 3)$ , $Q(-4, 12)$ . The trapezoid is $\frac{1}{2} \times (3+12) \times 5 = 37\frac{1}{2}$ , minus $\frac{1}{2} \times 4 \times 12 = 24$ and $\frac{1}{2} \times 1 \times 3 = 1\frac{1}{2}$ , leaves $12$ .

Another way: Enclose the triangle in the rectangle from $x=-4a$ to $x=a$ and $y=0$ to $y=4b$ , area $20ab$ , and subtract the three right triangles $8ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 5a \times 3b = \frac{15}{2}ab$ . Together they take away $16ab$ , and $20 - 16 = 4$ , so the triangle is again $4ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $4ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $4ab = 12$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 6 medium

For the points $P(a, b)$ and $Q(-2a, 3b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $15$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-2a, 3b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $15$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-2a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 3b

, and its height is

H'H = a-(-2a) = 3a

.

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

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 2a

and

QH' = 3b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 2a \times 3b = 3ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

6ab - \left(3ab + \frac{ab}{2}\right) = \frac{5}{2}ab

Combine like terms:

6 - 3 - \frac{1}{2} = \frac{5}{2}

.

5Use the given area

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

Set the expression equal to

15

and solve for

ab

.

\frac{5}{2}ab = 15 \Rightarrow ab = 6

Multiply both sides by

\frac{2}{5}

.

Answer:

6

4 · Reviewdoes it hold up?

Try $a=2$ , $b=3$ (so $ab=6$ ): $P(2, 3)$ , $Q(-4, 9)$ . The trapezoid is $\frac{1}{2} \times (3+9) \times 6 = 36$ , minus $\frac{1}{2} \times 4 \times 9 = 18$ and $\frac{1}{2} \times 2 \times 3 = 3$ , leaves $15$ .

Another way: Enclose the triangle in the rectangle from $x=-2a$ to $x=a$ and $y=0$ to $y=3b$ , area $9ab$ , and subtract the three right triangles $3ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 3a \times 2b = 3ab$ . Together they take away $\frac{13}{2}ab$ , and $9 - \frac{13}{2} = \frac{5}{2}$ , so the triangle is again $\frac{5}{2}ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $\frac{5}{2}ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $\frac{5}{2}ab = 15$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 7 medium

For the points $P(a, b)$ and $Q(-4a, 3b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $21$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-4a, 3b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $21$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-4a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 3b

, and its height is

H'H = a-(-4a) = 5a

.

\frac{1}{2} \times (b + 3b) \times 5a = 10ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 4a

and

QH' = 3b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 4a \times 3b = 6ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

10ab - \left(6ab + \frac{ab}{2}\right) = \frac{7}{2}ab

Combine like terms:

10 - 6 - \frac{1}{2} = \frac{7}{2}

.

5Use the given area

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

Set the expression equal to

21

and solve for

ab

.

\frac{7}{2}ab = 21 \Rightarrow ab = 6

Multiply both sides by

\frac{2}{7}

.

Answer:

6

4 · Reviewdoes it hold up?

Try $a=2$ , $b=3$ (so $ab=6$ ): $P(2, 3)$ , $Q(-8, 9)$ . The trapezoid is $\frac{1}{2} \times (3+9) \times 10 = 60$ , minus $\frac{1}{2} \times 8 \times 9 = 36$ and $\frac{1}{2} \times 2 \times 3 = 3$ , leaves $21$ .

Another way: Enclose the triangle in the rectangle from $x=-4a$ to $x=a$ and $y=0$ to $y=3b$ , area $15ab$ , and subtract the three right triangles $6ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 5a \times 2b = 5ab$ . Together they take away $\frac{23}{2}ab$ , and $15 - \frac{23}{2} = \frac{7}{2}$ , so the triangle is again $\frac{7}{2}ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $\frac{7}{2}ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $\frac{7}{2}ab = 21$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 8 medium

For the points $P(a, b)$ and $Q(-3a, 2b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $25$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-3a, 2b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $25$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-3a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 2b

, and its height is

H'H = a-(-3a) = 4a

.

\frac{1}{2} \times (b + 2b) \times 4a = 6ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 3a

and

QH' = 2b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 3a \times 2b = 3ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

6ab - \left(3ab + \frac{ab}{2}\right) = \frac{5}{2}ab

Combine like terms:

6 - 3 - \frac{1}{2} = \frac{5}{2}

.

5Use the given area

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

Set the expression equal to

25

and solve for

ab

.

\frac{5}{2}ab = 25 \Rightarrow ab = 10

Multiply both sides by

\frac{2}{5}

.

Answer:

10

4 · Reviewdoes it hold up?

Try $a=2$ , $b=5$ (so $ab=10$ ): $P(2, 5)$ , $Q(-6, 10)$ . The trapezoid is $\frac{1}{2} \times (5+10) \times 8 = 60$ , minus $\frac{1}{2} \times 6 \times 10 = 30$ and $\frac{1}{2} \times 2 \times 5 = 5$ , leaves $25$ .

Another way: Enclose the triangle in the rectangle from $x=-3a$ to $x=a$ and $y=0$ to $y=2b$ , area $8ab$ , and subtract the three right triangles $3ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 4a \times b = 2ab$ . Together they take away $\frac{11}{2}ab$ , and $8 - \frac{11}{2} = \frac{5}{2}$ , so the triangle is again $\frac{5}{2}ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $\frac{5}{2}ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $\frac{5}{2}ab = 25$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 9 hard

For the points $P(a, b)$ and $Q(-2a, 4b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $27$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-2a, 4b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $27$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-2a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 4b

, and its height is

H'H = a-(-2a) = 3a

.

\frac{1}{2} \times (b + 4b) \times 3a = \frac{15}{2}ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 2a

and

QH' = 4b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 2a \times 4b = 4ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

\frac{15}{2}ab - \left(4ab + \frac{ab}{2}\right) = 3ab

Combine like terms:

\frac{15}{2} - 4 - \frac{1}{2} = 3

.

5Use the given area

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

Set the expression equal to

27

and solve for

ab

.

3ab = 27 \Rightarrow ab = 9

Divide both sides by

3

.

Answer:

9

4 · Reviewdoes it hold up?

Try $a=3$ , $b=3$ (so $ab=9$ ): $P(3, 3)$ , $Q(-6, 12)$ . The trapezoid is $\frac{1}{2} \times (3+12) \times 9 = 67\frac{1}{2}$ , minus $\frac{1}{2} \times 6 \times 12 = 36$ and $\frac{1}{2} \times 3 \times 3 = 4\frac{1}{2}$ , leaves $27$ .

Another way: Enclose the triangle in the rectangle from $x=-2a$ to $x=a$ and $y=0$ to $y=4b$ , area $12ab$ , and subtract the three right triangles $4ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 3a \times 3b = \frac{9}{2}ab$ . Together they take away $9ab$ , and $12 - 9 = 3$ , so the triangle is again $3ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $3ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $3ab = 27$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 10 hard

For the points $P(a, b)$ and $Q(-3a, 5b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $28$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-3a, 5b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $28$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-3a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 5b

, and its height is

H'H = a-(-3a) = 4a

.

\frac{1}{2} \times (b + 5b) \times 4a = 12ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 3a

and

QH' = 5b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 3a \times 5b = \frac{15}{2}ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

12ab - \left(\frac{15}{2}ab + \frac{ab}{2}\right) = 4ab

Combine like terms:

12 - \frac{15}{2} - \frac{1}{2} = 4

.

5Use the given area

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

Set the expression equal to

28

and solve for

ab

.

4ab = 28 \Rightarrow ab = 7

Divide both sides by

4

.

Answer:

7

4 · Reviewdoes it hold up?

Try $a=1$ , $b=7$ (so $ab=7$ ): $P(1, 7)$ , $Q(-3, 35)$ . The trapezoid is $\frac{1}{2} \times (7+35) \times 4 = 84$ , minus $\frac{1}{2} \times 3 \times 35 = 52\frac{1}{2}$ and $\frac{1}{2} \times 1 \times 7 = 3\frac{1}{2}$ , leaves $28$ .

Another way: Enclose the triangle in the rectangle from $x=-3a$ to $x=a$ and $y=0$ to $y=5b$ , area $20ab$ , and subtract the three right triangles $\frac{15}{2}ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 4a \times 4b = 8ab$ . Together they take away $16ab$ , and $20 - 16 = 4$ , so the triangle is again $4ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $4ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $4ab = 28$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 11 hard

For the points $P(a, b)$ and $Q(-5a, 2b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $28$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-5a, 2b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $28$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-5a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 2b

, and its height is

H'H = a-(-5a) = 6a

.

\frac{1}{2} \times (b + 2b) \times 6a = 9ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 5a

and

QH' = 2b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 5a \times 2b = 5ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

9ab - \left(5ab + \frac{ab}{2}\right) = \frac{7}{2}ab

Combine like terms:

9 - 5 - \frac{1}{2} = \frac{7}{2}

.

5Use the given area

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

Set the expression equal to

28

and solve for

ab

.

\frac{7}{2}ab = 28 \Rightarrow ab = 8

Multiply both sides by

\frac{2}{7}

.

Answer:

8

4 · Reviewdoes it hold up?

Try $a=2$ , $b=4$ (so $ab=8$ ): $P(2, 4)$ , $Q(-10, 8)$ . The trapezoid is $\frac{1}{2} \times (4+8) \times 12 = 72$ , minus $\frac{1}{2} \times 10 \times 8 = 40$ and $\frac{1}{2} \times 2 \times 4 = 4$ , leaves $28$ .

Another way: Enclose the triangle in the rectangle from $x=-5a$ to $x=a$ and $y=0$ to $y=2b$ , area $12ab$ , and subtract the three right triangles $5ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 6a \times b = 3ab$ . Together they take away $\frac{17}{2}ab$ , and $12 - \frac{17}{2} = \frac{7}{2}$ , so the triangle is again $\frac{7}{2}ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $\frac{7}{2}ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $\frac{7}{2}ab = 28$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Variant 12 hard

For the points $P(a, b)$ and $Q(-3a, 3b)$ , where $a>0$ and $b>0$ , the area of triangle $OPQ$ is $30$ . Find the value of the product $ab$ of the constants $a$ and $b$ . ( $O$ is the origin.)

Your answer

Show solution

1 · Understandwhat's really being asked

$P$ is in Quadrant I and $Q$ is in Quadrant II, with coordinates written in letters. Express the area of triangle $OPQ$ in terms of $a$ and $b$ and use the given area to find $ab$ .

Givens

$P(a, b)$ and $Q(-3a, 3b)$ with $a>0$ , $b>0$ .
The area of triangle $OPQ$ is $30$ .
$O$ is the origin.

Unknowns

The value of $ab$ .

Constraints

$a$ and $b$ themselves cannot be found, only their product.

2 · Planchoose the strategy

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

No side of triangle $OPQ$ lies along a grid line, so split the job into pieces whose areas are easy: drop perpendiculars from $P$ and $Q$ to the $x$ -axis to get a trapezoid, which is the triangle plus two right triangles that share the corner $O$ . Then subtract and solve the resulting equation.

3 · Execute5 carry out the plan

1Drop perpendiculars

#1 Draw a Diagram 7.G.B.6

Let

H

and

H'

be the feet of the perpendiculars from

P

and

Q

to the

x

-axis. Since

a>0

,

H

is right of

O

and

H'

is left of

O

, so

O

lies on segment

H'H

.

H(a, 0), \quad H'(-3a, 0)

A perpendicular to the

x

-axis keeps the

x

-coordinate and sets

y

to

0

.

2Area of trapezoid $PQH'H$

#7 Identify Subproblems 7.G.B.6

Its parallel sides are the vertical segments

PH = b

and

QH' = 3b

, and its height is

H'H = a-(-3a) = 4a

.

\frac{1}{2} \times (b + 3b) \times 4a = 8ab

Trapezoid area is half the sum of the parallel sides times the height.

3Areas of the two right triangles

#7 Identify Subproblems 7.G.B.6

Triangle

OQH'

has legs

OH' = 3a

and

QH' = 3b

; triangle

OPH

has legs

OH = a

and

PH = b

.

\triangle OQH' = \frac{1}{2} \times 3a \times 3b = \frac{9}{2}ab, \quad \triangle OPH = \frac{1}{2} \times a \times b = \frac{ab}{2}

Right triangles with legs along the grid are half of a rectangle.

4Subtract to get triangle $OPQ$

#13 Convert to Algebra 6.EE.A.3

Because

O

is on the bottom side, the trapezoid is exactly triangle

OPQ

plus the two right triangles.

8ab - \left(\frac{9}{2}ab + \frac{ab}{2}\right) = 3ab

Combine like terms:

8 - \frac{9}{2} - \frac{1}{2} = 3

.

5Use the given area

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

Set the expression equal to

30

and solve for

ab

.

3ab = 30 \Rightarrow ab = 10

Divide both sides by

3

.

Answer:

10

4 · Reviewdoes it hold up?

Try $a=2$ , $b=5$ (so $ab=10$ ): $P(2, 5)$ , $Q(-6, 15)$ . The trapezoid is $\frac{1}{2} \times (5+15) \times 8 = 80$ , minus $\frac{1}{2} \times 6 \times 15 = 45$ and $\frac{1}{2} \times 2 \times 5 = 5$ , leaves $30$ .

Another way: Enclose the triangle in the rectangle from $x=-3a$ to $x=a$ and $y=0$ to $y=3b$ , area $12ab$ , and subtract the three right triangles $\frac{9}{2}ab$ , $\frac{ab}{2}$ , and $\frac{1}{2} \times 4a \times 2b = 4ab$ . Together they take away $9ab$ , and $12 - 9 = 3$ , so the triangle is again $3ab$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding areas of trapezoids and right triangles by composing and decomposing.
6.EE.A.3 Apply properties to generate equivalent expressions — Combining like terms in $ab$ to get $3ab$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation $3ab = 30$.

💡Takeaway. Even with letters for coordinates, cut-and-subtract works -- and the answer can be a product you never split apart.

Trapezoid minus two right triangles, in letters · 12 practice problems

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