← Square diagonals are equal and perpendicular bisectors · Quadrilateral Diagonal Properties

Square diagonals are equal and perpendicular bisectors · 10 practice problems

4.G.A.24.MD.A.3

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 4 cm

The figure shows a square with side length $8\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $8\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 8 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 8 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 8 cm.

BD = 8 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 8 / 2 = 4 cm.

8 \div 2 = 4 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 8 cm.

Answer: 4 cm

4 · Reviewdoes it hold up?

BM = 4 cm is half the 8 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 4 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 4 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 2 easy answer: 5 cm

The figure shows a square with side length $10\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $10\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 10 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 10 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 10 cm.

BD = 10 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 10 / 2 = 5 cm.

10 \div 2 = 5 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 10 cm.

Answer: 5 cm

4 · Reviewdoes it hold up?

BM = 5 cm is half the 10 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 5 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 5 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 3 easy answer: 6 cm

The figure shows a square with side length $12\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $12\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 12 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 12 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 12 cm.

BD = 12 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 12 / 2 = 6 cm.

12 \div 2 = 6 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 12 cm.

Answer: 6 cm

4 · Reviewdoes it hold up?

BM = 6 cm is half the 12 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 6 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 6 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 4 medium answer: 7 cm

The figure shows a square with side length $14\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $14\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 14 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 14 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 14 cm.

BD = 14 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 14 / 2 = 7 cm.

14 \div 2 = 7 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 14 cm.

Answer: 7 cm

4 · Reviewdoes it hold up?

BM = 7 cm is half the 14 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 7 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 7 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 5 medium answer: 8 cm

The figure shows a square with side length $16\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $16\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 16 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 16 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 16 cm.

BD = 16 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 16 / 2 = 8 cm.

16 \div 2 = 8 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 16 cm.

Answer: 8 cm

4 · Reviewdoes it hold up?

BM = 8 cm is half the 16 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 8 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 8 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 6 medium answer: 9 cm

The figure shows a square with side length $18\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $18\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 18 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 18 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 18 cm.

BD = 18 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 18 / 2 = 9 cm.

18 \div 2 = 9 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 18 cm.

Answer: 9 cm

4 · Reviewdoes it hold up?

BM = 9 cm is half the 18 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 9 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 9 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 7 medium answer: 10 cm

The figure shows a square with side length $20\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $20\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 20 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 20 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 20 cm.

BD = 20 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 20 / 2 = 10 cm.

20 \div 2 = 10 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 20 cm.

Answer: 10 cm

4 · Reviewdoes it hold up?

BM = 10 cm is half the 20 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 10 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 10 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 8 hard answer: 12 cm

The figure shows a square with side length $24\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $24\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 24 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 24 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 24 cm.

BD = 24 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 24 / 2 = 12 cm.

24 \div 2 = 12 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 24 cm.

Answer: 12 cm

4 · Reviewdoes it hold up?

BM = 12 cm is half the 24 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 12 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 12 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 9 hard answer: 13 cm

The figure shows a square with side length $26\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $26\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 26 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 26 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 26 cm.

BD = 26 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 26 / 2 = 13 cm.

26 \div 2 = 13 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 26 cm.

Answer: 13 cm

4 · Reviewdoes it hold up?

BM = 13 cm is half the 26 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 13 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 13 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!

Variant 10 hard answer: 15 cm

The figure shows a square with side length $30\,\text{cm}$ containing an inscribed circle. Four points on the circle are connected to form a second, inner square $ABCD$ . Find the length, in centimeters, of segment $BM$ .

[Figure] A large square with side length $30\,\text{cm}$ has a circle inscribed in it, touching all four sides. The four points where the circle meets the sides (the midpoints of the large square's sides) are joined to form the inner square $ABCD$ . The two diagonals $AC$ and $BD$ of the inner square meet at the center point $M$ .

Show solution

1 · Understandwhat's really being asked

A big square has side 30 cm with a circle drawn inside touching all four sides. The four touch points (the midpoints of the big square's sides) are joined to make an inner square ABCD. Its diagonals AC and BD cross at the center M. I need the length of BM.

Givens

The outer square has side length 30 cm.
The inscribed circle touches all four sides at the midpoints of the outer square's sides.
Those four midpoints are A (top), B (right), C (bottom), D (left), forming inner square ABCD.
Diagonals AC and BD of the inner square meet at center M.

Unknowns

The length of segment BM.

Constraints

A (top midpoint) and C (bottom midpoint) lie on the vertical center line, so AC spans the full height of the outer square.
B (right midpoint) and D (left midpoint) lie on the horizontal center line, so BD spans the full width of the outer square.
A square's diagonals are equal and bisect each other at M.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Read the picture carefully: the inner square's diagonals are just the center lines of the outer square, so each diagonal equals the outer side. Then halve it because the diagonals bisect each other.

3 · Execute2 carry out the plan

1See that diagonal BD is the full width of the outer square

#1 Draw a Diagram 4.G.A.2

B is the midpoint of the right side and D is the midpoint of the left side. The segment BD runs straight across the middle of the outer square, so its length equals the outer square's side, 30 cm.

BD = 30 \text{ cm}

Connecting opposite side-midpoints of a square gives a segment as long as the side itself.

2Halve the diagonal to get BM

#7 Identify Subproblems 4.G.A.2

The diagonals of the inner square bisect each other at center M, so M is the midpoint of BD. Therefore BM = BD / 2 = 30 / 2 = 15 cm.

30 \div 2 = 15 \text{ cm}

A square's diagonals cut each other exactly in half, so BM is half of 30 cm.

Answer: 15 cm

4 · Reviewdoes it hold up?

BM = 15 cm is half the 30 cm width, which matches that M is the exact center of the figure (also the circle's center, radius 15 cm). The size and centimeter unit are sensible.

Another way: Subproblem view (tool 7): the diagonal BD equals the outer side and also equals the circle's diameter; half of that diameter is the radius 15 cm = BM, confirming the answer.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square diagonal properties (equal, perpendicular, bisecting) and that midpoint-to-midpoint equals a side.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning with the square's side length to get the diagonal and half-diagonal lengths.

💡Takeaway. Joining opposite side-midpoints of a square gives a line as long as the side, and the diagonals split in half -- so BM is just the side divided by 2!