Practice · Triangle side equals sum of two radii · Sensim Math

Variant 1 answer: 12 cm

In the figure on the right, the perimeter of triangle ABC is $24\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $9\ \text{cm}$ and side AC is labeled $9\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 24 cm, and sides AB and AC are 9 cm and 9 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 24 cm.
AB = 9 cm and AC = 9 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

24 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 24 by 2.

24 \div 2 = 12

Undoing a doubling is just dividing by 2.

Answer: 12 cm

Review

The radius total (12 cm) is exactly half the 24 cm perimeter, which is right because each radius is counted twice. As a check, BC = 24 - 9 - 9 = 6 cm, and 9 + 9 + 6 = 24 confirms the side lengths are consistent.

Find BC first: 24 - 9 - 9 = 6. Then add the three side equations another way; the radius total stays 12 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 2 answer: 18 cm

In the figure on the right, the perimeter of triangle ABC is $36\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $12\ \text{cm}$ and side AC is labeled $12\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 36 cm, and sides AB and AC are 12 cm and 12 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 36 cm.
AB = 12 cm and AC = 12 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

36 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 36 by 2.

36 \div 2 = 18

Undoing a doubling is just dividing by 2.

Answer: 18 cm

Review

The radius total (18 cm) is exactly half the 36 cm perimeter, which is right because each radius is counted twice. As a check, BC = 36 - 12 - 12 = 12 cm, and 12 + 12 + 12 = 36 confirms the side lengths are consistent.

Find BC first: 36 - 12 - 12 = 12. Then add the three side equations another way; the radius total stays 18 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 3 answer: 30 cm

In the figure on the right, the perimeter of triangle ABC is $60\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $22\ \text{cm}$ and side AC is labeled $21\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 60 cm, and sides AB and AC are 22 cm and 21 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 60 cm.
AB = 22 cm and AC = 21 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

60 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 60 by 2.

60 \div 2 = 30

Undoing a doubling is just dividing by 2.

Answer: 30 cm

Review

The radius total (30 cm) is exactly half the 60 cm perimeter, which is right because each radius is counted twice. As a check, BC = 60 - 22 - 21 = 17 cm, and 22 + 21 + 17 = 60 confirms the side lengths are consistent.

Find BC first: 60 - 22 - 21 = 17. Then add the three side equations another way; the radius total stays 30 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 4 answer: 16 cm

In the figure on the right, the perimeter of triangle ABC is $33\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $11\ \text{cm}$ and side AC is labeled $12\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 33 cm, and sides AB and AC are 11 cm and 12 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 33 cm.
AB = 11 cm and AC = 12 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

33 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 33 by 2.

33 \div 2 = 16

Undoing a doubling is just dividing by 2.

Answer: 16 cm

Review

The radius total (16 cm) is exactly half the 33 cm perimeter, which is right because each radius is counted twice. As a check, BC = 33 - 11 - 12 = 10 cm, and 11 + 12 + 10 = 33 confirms the side lengths are consistent.

Find BC first: 33 - 11 - 12 = 10. Then add the three side equations another way; the radius total stays 16 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 5 answer: 14 cm

In the figure on the right, the perimeter of triangle ABC is $28\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $10\ \text{cm}$ and side AC is labeled $11\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 28 cm, and sides AB and AC are 10 cm and 11 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 28 cm.
AB = 10 cm and AC = 11 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

28 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 28 by 2.

28 \div 2 = 14

Undoing a doubling is just dividing by 2.

Answer: 14 cm

Review

The radius total (14 cm) is exactly half the 28 cm perimeter, which is right because each radius is counted twice. As a check, BC = 28 - 10 - 11 = 7 cm, and 10 + 11 + 7 = 28 confirms the side lengths are consistent.

Find BC first: 28 - 10 - 11 = 7. Then add the three side equations another way; the radius total stays 14 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 6 answer: 25 cm

In the figure on the right, the perimeter of triangle ABC is $50\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $18\ \text{cm}$ and side AC is labeled $16\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 50 cm, and sides AB and AC are 18 cm and 16 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 50 cm.
AB = 18 cm and AC = 16 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

50 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 50 by 2.

50 \div 2 = 25

Undoing a doubling is just dividing by 2.

Answer: 25 cm

Review

The radius total (25 cm) is exactly half the 50 cm perimeter, which is right because each radius is counted twice. As a check, BC = 50 - 18 - 16 = 16 cm, and 18 + 16 + 16 = 50 confirms the side lengths are consistent.

Find BC first: 50 - 18 - 16 = 16. Then add the three side equations another way; the radius total stays 25 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 7 answer: 22 cm

In the figure on the right, the perimeter of triangle ABC is $44\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $9\ \text{cm}$ and side AC is labeled $9\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 44 cm, and sides AB and AC are 9 cm and 9 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 44 cm.
AB = 9 cm and AC = 9 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

44 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 44 by 2.

44 \div 2 = 22

Undoing a doubling is just dividing by 2.

Answer: 22 cm

Review

The radius total (22 cm) is exactly half the 44 cm perimeter, which is right because each radius is counted twice. As a check, BC = 44 - 9 - 9 = 26 cm, and 9 + 9 + 26 = 44 confirms the side lengths are consistent.

Find BC first: 44 - 9 - 9 = 26. Then add the three side equations another way; the radius total stays 22 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 8 answer: 20 cm

In the figure on the right, the perimeter of triangle ABC is $40\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $14\ \text{cm}$ and side AC is labeled $13\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 40 cm, and sides AB and AC are 14 cm and 13 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 40 cm.
AB = 14 cm and AC = 13 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

40 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 40 by 2.

40 \div 2 = 20

Undoing a doubling is just dividing by 2.

Answer: 20 cm

Review

The radius total (20 cm) is exactly half the 40 cm perimeter, which is right because each radius is counted twice. As a check, BC = 40 - 14 - 13 = 13 cm, and 14 + 13 + 13 = 40 confirms the side lengths are consistent.

Find BC first: 40 - 14 - 13 = 13. Then add the three side equations another way; the radius total stays 20 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 9 answer: 24 cm

In the figure on the right, the perimeter of triangle ABC is $48\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $17\ \text{cm}$ and side AC is labeled $17\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 48 cm, and sides AB and AC are 17 cm and 17 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 48 cm.
AB = 17 cm and AC = 17 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

48 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 48 by 2.

48 \div 2 = 24

Undoing a doubling is just dividing by 2.

Answer: 24 cm

Review

The radius total (24 cm) is exactly half the 48 cm perimeter, which is right because each radius is counted twice. As a check, BC = 48 - 17 - 17 = 14 cm, and 17 + 17 + 14 = 48 confirms the side lengths are consistent.

Find BC first: 48 - 17 - 17 = 14. Then add the three side equations another way; the radius total stays 24 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Variant 10 answer: 15 cm

In the figure on the right, the perimeter of triangle ABC is $30\ \text{cm}$ . What is the sum of the radii of the three circles, in centimeters?

Three circles of different sizes each touch one another, and the centers of the circles are joined to form triangle ABC. The top vertex is A, the bottom-left vertex is B, and the bottom-right vertex is C. Side AB is labeled $11\ \text{cm}$ and side AC is labeled $12\ \text{cm}$ . Each side of the triangle equals the sum of the radii of the two circles that meet along that side.

Show solution

Understand

Three circles of different sizes each touch one another. Joining their centers makes triangle ABC. Each side equals the sum of the two touching circles' radii. The triangle's perimeter is 30 cm, and sides AB and AC are 11 cm and 12 cm. I need the total of the three radii.

Givens

Three mutually touching circles; centers joined form triangle ABC (A top, B lower-left, C lower-right).
Each side equals the sum of the radii of the two circles meeting along it.
Perimeter of triangle ABC is 30 cm.
AB = 11 cm and AC = 12 cm.

Unknowns

The sum of the radii of the three circles, in cm.

Constraints

Two touching circles have center distance equal to the sum of their radii.
Each radius is counted in exactly two sides (each circle touches the other two).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Each side is a sum of two radii. Adding all three sides counts every radius twice, so the perimeter is double the radius total. The given AB and AC are not even needed for the total, which the radius-sum pattern makes clear.

Execute

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

Touching circles have center distance equal to the sum of their radii, so AB = rA + rB, AC = rA + rC, BC = rB + rC.

AB+AC+BC = (r_A+r_B)+(r_A+r_C)+(r_B+r_C)

Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.

#9 Solve an Easier Related Problem 3.OA.D.9

Adding all three sides counts every radius exactly two times, because each circle touches the other two. So the perimeter equals twice the total of the three radii.

30 = 2 \times (r_A+r_B+r_C)

The repeated 'each radius twice' pattern turns the whole perimeter into a simple doubling.

#9 Solve an Easier Related Problem 3.OA.A.2

Since the perimeter is double the radius total, divide 30 by 2.

30 \div 2 = 15

Undoing a doubling is just dividing by 2.

Answer: 15 cm

Review

The radius total (15 cm) is exactly half the 30 cm perimeter, which is right because each radius is counted twice. As a check, BC = 30 - 11 - 12 = 7 cm, and 11 + 12 + 7 = 30 confirms the side lengths are consistent.

Find BC first: 30 - 11 - 12 = 7. Then add the three side equations another way; the radius total stays 15 regardless of the individual radii.

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each triangle side as the sum of two touching circles' radii.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius is counted twice when adding the sides, so the perimeter is double the radius total.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to find the sum of the radii.

💡 This only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!

Triangle side equals sum of two radii · 10 practice problems

Generated variants — 10

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