Practice · Box the triangle, subtract the corner triangles · Sensim Math

Variant 1 easy

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(1, 1)$ , $B(-2, 0)$ , and $C(2, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(1, 1)$ , $B(-2, 0)$ , $C(2, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-2

to

2

and the

y

-coordinates from

-1

to

1

. The rectangle with those sides has corners

P(-2, 1)

,

Q(-2, -1)

,

C(2, -1)

,

R(2, 1)

; each vertex of the triangle is on its boundary.

P(-2, 1), \quad Q(-2, -1), \quad R(2, 1)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

2-(-2)=4

and its height is

1-(-1)=2

.

PQCR: \ (2-(-2)) \times (1-(-1)) = 4 \times 2 = 8

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 1-(-2) = 3

and

PB = 1-0 = 1

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 2-(-2) = 4

. Triangle

ACR

has legs

AR = 2-1 = 1

and

RC = 1-(-1) = 2

.

\triangle APB = \frac{1}{2} \times 3 \times 1 = \frac{3}{2}, \quad \triangle BQC = \frac{1}{2} \times 1 \times 4 = 2, \quad \triangle ACR = \frac{1}{2} \times 1 \times 2 = 1

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

8 - \left(\frac{3}{2} + 2 + 1\right) = 8 - \frac{9}{2} = \frac{7}{2}

Whole minus the parts we do not want.

Answer:

\frac{7}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $8$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $1 \times (0-(-1)) + (-2) \times (-1-1) + 2 \times (1-0) = 1+4+2 = 7$ , halved, also gives $\frac{7}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-2, 0)$ is on it, side $AC$ crosses it at $\left(\frac{3}{2}, 0\right)$ , so the triangle splits into two triangles with base $\frac{3}{2}-(-2) = \frac{7}{2}$ and heights $1$ and $1$ , for a total of $\frac{1}{2} \times \frac{7}{2} \times 2 = \frac{7}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 2 easy

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(1, 2)$ , $B(-3, 0)$ , and $C(3, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(1, 2)$ , $B(-3, 0)$ , $C(3, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-3

to

3

and the

y

-coordinates from

-1

to

2

. The rectangle with those sides has corners

P(-3, 2)

,

Q(-3, -1)

,

C(3, -1)

,

R(3, 2)

; each vertex of the triangle is on its boundary.

P(-3, 2), \quad Q(-3, -1), \quad R(3, 2)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

3-(-3)=6

and its height is

2-(-1)=3

.

PQCR: \ (3-(-3)) \times (2-(-1)) = 6 \times 3 = 18

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 1-(-3) = 4

and

PB = 2-0 = 2

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 3-(-3) = 6

. Triangle

ACR

has legs

AR = 3-1 = 2

and

RC = 2-(-1) = 3

.

\triangle APB = \frac{1}{2} \times 4 \times 2 = 4, \quad \triangle BQC = \frac{1}{2} \times 1 \times 6 = 3, \quad \triangle ACR = \frac{1}{2} \times 2 \times 3 = 3

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

18 - \left(4 + 3 + 3\right) = 18 - 10 = 8

Whole minus the parts we do not want.

Answer:

8

4 · Reviewdoes it hold up?

The answer is less than half of the $18$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $1 \times (0-(-1)) + (-3) \times (-1-2) + 3 \times (2-0) = 1+9+6 = 16$ , halved, also gives $8$ .

Another way: Cut along the $x$ -axis instead: $B(-3, 0)$ is on it, side $AC$ crosses it at $\left(\frac{7}{3}, 0\right)$ , so the triangle splits into two triangles with base $\frac{7}{3}-(-3) = \frac{16}{3}$ and heights $2$ and $1$ , for a total of $\frac{1}{2} \times \frac{16}{3} \times 3 = 8$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 3 easy

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(-1, 3)$ , $B(-3, 0)$ , and $C(2, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(-1, 3)$ , $B(-3, 0)$ , $C(2, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-3

to

2

and the

y

-coordinates from

-1

to

3

. The rectangle with those sides has corners

P(-3, 3)

,

Q(-3, -1)

,

C(2, -1)

,

R(2, 3)

; each vertex of the triangle is on its boundary.

P(-3, 3), \quad Q(-3, -1), \quad R(2, 3)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

2-(-3)=5

and its height is

3-(-1)=4

.

PQCR: \ (2-(-3)) \times (3-(-1)) = 5 \times 4 = 20

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = -1-(-3) = 2

and

PB = 3-0 = 3

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 2-(-3) = 5

. Triangle

ACR

has legs

AR = 2-(-1) = 3

and

RC = 3-(-1) = 4

.

\triangle APB = \frac{1}{2} \times 2 \times 3 = 3, \quad \triangle BQC = \frac{1}{2} \times 1 \times 5 = \frac{5}{2}, \quad \triangle ACR = \frac{1}{2} \times 3 \times 4 = 6

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

20 - \left(3 + \frac{5}{2} + 6\right) = 20 - \frac{23}{2} = \frac{17}{2}

Whole minus the parts we do not want.

Answer:

\frac{17}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $20$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $(-1) \times (0-(-1)) + (-3) \times (-1-3) + 2 \times (3-0) = -1+12+6 = 17$ , halved, also gives $\frac{17}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-3, 0)$ is on it, side $AC$ crosses it at $\left(\frac{5}{4}, 0\right)$ , so the triangle splits into two triangles with base $\frac{5}{4}-(-3) = \frac{17}{4}$ and heights $3$ and $1$ , for a total of $\frac{1}{2} \times \frac{17}{4} \times 4 = \frac{17}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 4 medium

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(1, 2)$ , $B(-2, 0)$ , and $C(3, -3)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(1, 2)$ , $B(-2, 0)$ , $C(3, -3)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-2

to

3

and the

y

-coordinates from

-3

to

2

. The rectangle with those sides has corners

P(-2, 2)

,

Q(-2, -3)

,

C(3, -3)

,

R(3, 2)

; each vertex of the triangle is on its boundary.

P(-2, 2), \quad Q(-2, -3), \quad R(3, 2)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

3-(-2)=5

and its height is

2-(-3)=5

.

PQCR: \ (3-(-2)) \times (2-(-3)) = 5 \times 5 = 25

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 1-(-2) = 3

and

PB = 2-0 = 2

. Triangle

BQC

has legs

BQ = 0-(-3) = 3

and

QC = 3-(-2) = 5

. Triangle

ACR

has legs

AR = 3-1 = 2

and

RC = 2-(-3) = 5

.

\triangle APB = \frac{1}{2} \times 3 \times 2 = 3, \quad \triangle BQC = \frac{1}{2} \times 3 \times 5 = \frac{15}{2}, \quad \triangle ACR = \frac{1}{2} \times 2 \times 5 = 5

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

25 - \left(3 + \frac{15}{2} + 5\right) = 25 - \frac{31}{2} = \frac{19}{2}

Whole minus the parts we do not want.

Answer:

\frac{19}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $25$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $1 \times (0-(-3)) + (-2) \times (-3-2) + 3 \times (2-0) = 3+10+6 = 19$ , halved, also gives $\frac{19}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-2, 0)$ is on it, side $AC$ crosses it at $\left(\frac{9}{5}, 0\right)$ , so the triangle splits into two triangles with base $\frac{9}{5}-(-2) = \frac{19}{5}$ and heights $2$ and $3$ , for a total of $\frac{1}{2} \times \frac{19}{5} \times 5 = \frac{19}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 5 easy

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(2, 2)$ , $B(-1, 0)$ , and $C(3, -2)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(2, 2)$ , $B(-1, 0)$ , $C(3, -2)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-1

to

3

and the

y

-coordinates from

-2

to

2

. The rectangle with those sides has corners

P(-1, 2)

,

Q(-1, -2)

,

C(3, -2)

,

R(3, 2)

; each vertex of the triangle is on its boundary.

P(-1, 2), \quad Q(-1, -2), \quad R(3, 2)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

3-(-1)=4

and its height is

2-(-2)=4

.

PQCR: \ (3-(-1)) \times (2-(-2)) = 4 \times 4 = 16

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 2-(-1) = 3

and

PB = 2-0 = 2

. Triangle

BQC

has legs

BQ = 0-(-2) = 2

and

QC = 3-(-1) = 4

. Triangle

ACR

has legs

AR = 3-2 = 1

and

RC = 2-(-2) = 4

.

\triangle APB = \frac{1}{2} \times 3 \times 2 = 3, \quad \triangle BQC = \frac{1}{2} \times 2 \times 4 = 4, \quad \triangle ACR = \frac{1}{2} \times 1 \times 4 = 2

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

16 - \left(3 + 4 + 2\right) = 16 - 9 = 7

Whole minus the parts we do not want.

Answer:

7

4 · Reviewdoes it hold up?

The answer is less than half of the $16$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $2 \times (0-(-2)) + (-1) \times (-2-2) + 3 \times (2-0) = 4+4+6 = 14$ , halved, also gives $7$ .

Another way: Cut along the $x$ -axis instead: $B(-1, 0)$ is on it, side $AC$ crosses it at $\left(\frac{5}{2}, 0\right)$ , so the triangle splits into two triangles with base $\frac{5}{2}-(-1) = \frac{7}{2}$ and heights $2$ and $2$ , for a total of $\frac{1}{2} \times \frac{7}{2} \times 4 = 7$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 6 medium

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(2, 1)$ , $B(-2, 0)$ , and $C(3, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(2, 1)$ , $B(-2, 0)$ , $C(3, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-2

to

3

and the

y

-coordinates from

-1

to

1

. The rectangle with those sides has corners

P(-2, 1)

,

Q(-2, -1)

,

C(3, -1)

,

R(3, 1)

; each vertex of the triangle is on its boundary.

P(-2, 1), \quad Q(-2, -1), \quad R(3, 1)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

3-(-2)=5

and its height is

1-(-1)=2

.

PQCR: \ (3-(-2)) \times (1-(-1)) = 5 \times 2 = 10

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 2-(-2) = 4

and

PB = 1-0 = 1

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 3-(-2) = 5

. Triangle

ACR

has legs

AR = 3-2 = 1

and

RC = 1-(-1) = 2

.

\triangle APB = \frac{1}{2} \times 4 \times 1 = 2, \quad \triangle BQC = \frac{1}{2} \times 1 \times 5 = \frac{5}{2}, \quad \triangle ACR = \frac{1}{2} \times 1 \times 2 = 1

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

10 - \left(2 + \frac{5}{2} + 1\right) = 10 - \frac{11}{2} = \frac{9}{2}

Whole minus the parts we do not want.

Answer:

\frac{9}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $10$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $2 \times (0-(-1)) + (-2) \times (-1-1) + 3 \times (1-0) = 2+4+3 = 9$ , halved, also gives $\frac{9}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-2, 0)$ is on it, side $AC$ crosses it at $\left(\frac{5}{2}, 0\right)$ , so the triangle splits into two triangles with base $\frac{5}{2}-(-2) = \frac{9}{2}$ and heights $1$ and $1$ , for a total of $\frac{1}{2} \times \frac{9}{2} \times 2 = \frac{9}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 7 hard

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(2, 3)$ , $B(-3, 0)$ , and $C(4, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(2, 3)$ , $B(-3, 0)$ , $C(4, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-3

to

4

and the

y

-coordinates from

-1

to

3

. The rectangle with those sides has corners

P(-3, 3)

,

Q(-3, -1)

,

C(4, -1)

,

R(4, 3)

; each vertex of the triangle is on its boundary.

P(-3, 3), \quad Q(-3, -1), \quad R(4, 3)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

4-(-3)=7

and its height is

3-(-1)=4

.

PQCR: \ (4-(-3)) \times (3-(-1)) = 7 \times 4 = 28

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 2-(-3) = 5

and

PB = 3-0 = 3

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 4-(-3) = 7

. Triangle

ACR

has legs

AR = 4-2 = 2

and

RC = 3-(-1) = 4

.

\triangle APB = \frac{1}{2} \times 5 \times 3 = \frac{15}{2}, \quad \triangle BQC = \frac{1}{2} \times 1 \times 7 = \frac{7}{2}, \quad \triangle ACR = \frac{1}{2} \times 2 \times 4 = 4

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

28 - \left(\frac{15}{2} + \frac{7}{2} + 4\right) = 28 - 15 = 13

Whole minus the parts we do not want.

Answer:

13

4 · Reviewdoes it hold up?

The answer is less than half of the $28$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $2 \times (0-(-1)) + (-3) \times (-1-3) + 4 \times (3-0) = 2+12+12 = 26$ , halved, also gives $13$ .

Another way: Cut along the $x$ -axis instead: $B(-3, 0)$ is on it, side $AC$ crosses it at $\left(\frac{7}{2}, 0\right)$ , so the triangle splits into two triangles with base $\frac{7}{2}-(-3) = \frac{13}{2}$ and heights $3$ and $1$ , for a total of $\frac{1}{2} \times \frac{13}{2} \times 4 = 13$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 8 medium

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(-1, 2)$ , $B(-4, 0)$ , and $C(2, -2)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(-1, 2)$ , $B(-4, 0)$ , $C(2, -2)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-4

to

2

and the

y

-coordinates from

-2

to

2

. The rectangle with those sides has corners

P(-4, 2)

,

Q(-4, -2)

,

C(2, -2)

,

R(2, 2)

; each vertex of the triangle is on its boundary.

P(-4, 2), \quad Q(-4, -2), \quad R(2, 2)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

2-(-4)=6

and its height is

2-(-2)=4

.

PQCR: \ (2-(-4)) \times (2-(-2)) = 6 \times 4 = 24

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = -1-(-4) = 3

and

PB = 2-0 = 2

. Triangle

BQC

has legs

BQ = 0-(-2) = 2

and

QC = 2-(-4) = 6

. Triangle

ACR

has legs

AR = 2-(-1) = 3

and

RC = 2-(-2) = 4

.

\triangle APB = \frac{1}{2} \times 3 \times 2 = 3, \quad \triangle BQC = \frac{1}{2} \times 2 \times 6 = 6, \quad \triangle ACR = \frac{1}{2} \times 3 \times 4 = 6

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

24 - \left(3 + 6 + 6\right) = 24 - 15 = 9

Whole minus the parts we do not want.

Answer:

9

4 · Reviewdoes it hold up?

The answer is less than half of the $24$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $(-1) \times (0-(-2)) + (-4) \times (-2-2) + 2 \times (2-0) = -2+16+4 = 18$ , halved, also gives $9$ .

Another way: Cut along the $x$ -axis instead: $B(-4, 0)$ is on it, side $AC$ crosses it at $\left(\frac{1}{2}, 0\right)$ , so the triangle splits into two triangles with base $\frac{1}{2}-(-4) = \frac{9}{2}$ and heights $2$ and $2$ , for a total of $\frac{1}{2} \times \frac{9}{2} \times 4 = 9$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 9 medium

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(1, 3)$ , $B(-2, 0)$ , and $C(4, -2)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(1, 3)$ , $B(-2, 0)$ , $C(4, -2)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-2

to

4

and the

y

-coordinates from

-2

to

3

. The rectangle with those sides has corners

P(-2, 3)

,

Q(-2, -2)

,

C(4, -2)

,

R(4, 3)

; each vertex of the triangle is on its boundary.

P(-2, 3), \quad Q(-2, -2), \quad R(4, 3)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

4-(-2)=6

and its height is

3-(-2)=5

.

PQCR: \ (4-(-2)) \times (3-(-2)) = 6 \times 5 = 30

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 1-(-2) = 3

and

PB = 3-0 = 3

. Triangle

BQC

has legs

BQ = 0-(-2) = 2

and

QC = 4-(-2) = 6

. Triangle

ACR

has legs

AR = 4-1 = 3

and

RC = 3-(-2) = 5

.

\triangle APB = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}, \quad \triangle BQC = \frac{1}{2} \times 2 \times 6 = 6, \quad \triangle ACR = \frac{1}{2} \times 3 \times 5 = \frac{15}{2}

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

30 - \left(\frac{9}{2} + 6 + \frac{15}{2}\right) = 30 - 18 = 12

Whole minus the parts we do not want.

Answer:

12

4 · Reviewdoes it hold up?

The answer is less than half of the $30$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $1 \times (0-(-2)) + (-2) \times (-2-3) + 4 \times (3-0) = 2+10+12 = 24$ , halved, also gives $12$ .

Another way: Cut along the $x$ -axis instead: $B(-2, 0)$ is on it, side $AC$ crosses it at $\left(\frac{14}{5}, 0\right)$ , so the triangle splits into two triangles with base $\frac{14}{5}-(-2) = \frac{24}{5}$ and heights $3$ and $2$ , for a total of $\frac{1}{2} \times \frac{24}{5} \times 5 = 12$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 10 hard

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(1, 1)$ , $B(-3, 0)$ , and $C(4, -2)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(1, 1)$ , $B(-3, 0)$ , $C(4, -2)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-3

to

4

and the

y

-coordinates from

-2

to

1

. The rectangle with those sides has corners

P(-3, 1)

,

Q(-3, -2)

,

C(4, -2)

,

R(4, 1)

; each vertex of the triangle is on its boundary.

P(-3, 1), \quad Q(-3, -2), \quad R(4, 1)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

4-(-3)=7

and its height is

1-(-2)=3

.

PQCR: \ (4-(-3)) \times (1-(-2)) = 7 \times 3 = 21

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 1-(-3) = 4

and

PB = 1-0 = 1

. Triangle

BQC

has legs

BQ = 0-(-2) = 2

and

QC = 4-(-3) = 7

. Triangle

ACR

has legs

AR = 4-1 = 3

and

RC = 1-(-2) = 3

.

\triangle APB = \frac{1}{2} \times 4 \times 1 = 2, \quad \triangle BQC = \frac{1}{2} \times 2 \times 7 = 7, \quad \triangle ACR = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

21 - \left(2 + 7 + \frac{9}{2}\right) = 21 - \frac{27}{2} = \frac{15}{2}

Whole minus the parts we do not want.

Answer:

\frac{15}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $21$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $1 \times (0-(-2)) + (-3) \times (-2-1) + 4 \times (1-0) = 2+9+4 = 15$ , halved, also gives $\frac{15}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-3, 0)$ is on it, side $AC$ crosses it at $\left(2, 0\right)$ , so the triangle splits into two triangles with base $2-(-3) = 5$ and heights $1$ and $2$ , for a total of $\frac{1}{2} \times 5 \times 3 = \frac{15}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 11 hard

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(3, 2)$ , $B(-1, 0)$ , and $C(4, -3)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(3, 2)$ , $B(-1, 0)$ , $C(4, -3)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-1

to

4

and the

y

-coordinates from

-3

to

2

. The rectangle with those sides has corners

P(-1, 2)

,

Q(-1, -3)

,

C(4, -3)

,

R(4, 2)

; each vertex of the triangle is on its boundary.

P(-1, 2), \quad Q(-1, -3), \quad R(4, 2)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

4-(-1)=5

and its height is

2-(-3)=5

.

PQCR: \ (4-(-1)) \times (2-(-3)) = 5 \times 5 = 25

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = 3-(-1) = 4

and

PB = 2-0 = 2

. Triangle

BQC

has legs

BQ = 0-(-3) = 3

and

QC = 4-(-1) = 5

. Triangle

ACR

has legs

AR = 4-3 = 1

and

RC = 2-(-3) = 5

.

\triangle APB = \frac{1}{2} \times 4 \times 2 = 4, \quad \triangle BQC = \frac{1}{2} \times 3 \times 5 = \frac{15}{2}, \quad \triangle ACR = \frac{1}{2} \times 1 \times 5 = \frac{5}{2}

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

25 - \left(4 + \frac{15}{2} + \frac{5}{2}\right) = 25 - 14 = 11

Whole minus the parts we do not want.

Answer:

11

4 · Reviewdoes it hold up?

The answer is less than half of the $25$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $3 \times (0-(-3)) + (-1) \times (-3-2) + 4 \times (2-0) = 9+5+8 = 22$ , halved, also gives $11$ .

Another way: Cut along the $x$ -axis instead: $B(-1, 0)$ is on it, side $AC$ crosses it at $\left(\frac{17}{5}, 0\right)$ , so the triangle splits into two triangles with base $\frac{17}{5}-(-1) = \frac{22}{5}$ and heights $2$ and $3$ , for a total of $\frac{1}{2} \times \frac{22}{5} \times 5 = 11$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Variant 12 hard

As shown in the figure, find the area of triangle $ABC$ formed by the three points $A(-1, 4)$ , $B(-4, 0)$ , and $C(3, -1)$ in the coordinate plane.

Your answer

Show solution

1 · Understandwhat's really being asked

A tilted triangle has its three corners at given grid points. None of its sides runs along a grid line, so find its area another way.

Givens

$A(-1, 4)$ , $B(-4, 0)$ , $C(3, -1)$ are the vertices.

Unknowns

The area of triangle $ABC$ .

Constraints

No side of the triangle is horizontal or vertical, so no base and height can be read off directly.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #1 Draw a Diagram#7 Identify Subproblems

Change focus from the triangle to what surrounds it. A rectangle with sides parallel to the axes fits snugly around the triangle; the leftover pieces are right triangles whose legs are just differences of coordinates. Subtract them from the rectangle.

3 · Execute4 carry out the plan

1Draw the enclosing rectangle

#1 Draw a Diagram 6.G.A.3

The

x

-coordinates run from

-4

to

3

and the

y

-coordinates from

-1

to

4

. The rectangle with those sides has corners

P(-4, 4)

,

Q(-4, -1)

,

C(3, -1)

,

R(3, 4)

; each vertex of the triangle is on its boundary.

P(-4, 4), \quad Q(-4, -1), \quad R(3, 4)

Use the smallest and largest

x

and

y

.

2Area of the rectangle

#7 Identify Subproblems 7.NS.A.1

Its width is

3-(-4)=7

and its height is

4-(-1)=5

.

PQCR: \ (3-(-4)) \times (4-(-1)) = 7 \times 5 = 35

A length along an axis is a difference of coordinates.

3Areas of the three corner triangles

#7 Identify Subproblems 7.G.B.6

Triangle

APB

has legs

AP = -1-(-4) = 3

and

PB = 4-0 = 4

. Triangle

BQC

has legs

BQ = 0-(-1) = 1

and

QC = 3-(-4) = 7

. Triangle

ACR

has legs

AR = 3-(-1) = 4

and

RC = 4-(-1) = 5

.

\triangle APB = \frac{1}{2} \times 3 \times 4 = 6, \quad \triangle BQC = \frac{1}{2} \times 1 \times 7 = \frac{7}{2}, \quad \triangle ACR = \frac{1}{2} \times 4 \times 5 = 10

Each corner piece is a right triangle with legs along the grid.

4Subtract

#16 Count the Complement 7.G.B.6

The triangle is what is left of the rectangle after removing the three corner triangles.

35 - \left(6 + \frac{7}{2} + 10\right) = 35 - \frac{39}{2} = \frac{31}{2}

Whole minus the parts we do not want.

Answer:

\frac{31}{2}

4 · Reviewdoes it hold up?

The answer is less than half of the $35$ -square-unit rectangle, which is expected for a triangle that touches the rectangle's sides at only three points. The shoelace sum $(-1) \times (0-(-1)) + (-4) \times (-1-4) + 3 \times (4-0) = -1+20+12 = 31$ , halved, also gives $\frac{31}{2}$ .

Another way: Cut along the $x$ -axis instead: $B(-4, 0)$ is on it, side $AC$ crosses it at $\left(\frac{11}{5}, 0\right)$ , so the triangle splits into two triangles with base $\frac{11}{5}-(-4) = \frac{31}{5}$ and heights $4$ and $1$ , for a total of $\frac{1}{2} \times \frac{31}{5} \times 5 = \frac{31}{2}$ .

Standardsmin grade 7

7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding triangle and rectangle areas by composing and decomposing.
6.G.A.3 Draw polygons in the coordinate plane using coordinates — Placing the enclosing rectangle's corners from the given coordinates.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as differences of signed coordinates.

💡Takeaway. A slanted triangle is a rectangle with three right triangles cut off -- and those are easy.

Box the triangle, subtract the corner triangles · 12 practice problems

Generated variants — 12

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Where next?