← An area rule with one unknown in it is already an equation · Work Backwards to Recover a Start Value

An area rule with one unknown in it is already an equation · 12 practice problems

6.EE.B.66.EE.B.76.G.A.1

Generated variants — 12

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

Variant 1 easy answer: 16\frac{1}{6} cm

The rhombus at the right has an area of 1512 cm2\frac{15}{12}\ \text{cm}^2. If one of its diagonals is 148 cm1\frac{4}{8}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 4/8 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 15/12 cm2 and one diagonal 1 and 1/2 cm. We want how much the two diagonals differ by.

Givens
  • The area is 15/12 cm2.
  • One diagonal is 1 and 1/2 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
32×□÷2=54\frac{3}{2} \times \square \div 2 = \frac{5}{4}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
32×□=52\frac{3}{2} \times \square = \frac{5}{2}
The diagonals multiply to 2 and 1/2.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
52÷32=123\frac{5}{2} \div \frac{3}{2} = 1\frac{2}{3}
The other diagonal is 1 and 2/3 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
123−112=161\frac{2}{3} - 1\frac{1}{2} = \frac{1}{6}
They differ by 1/6 cm.
Answer: 16\frac{1}{6} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 32×53÷2=54\frac{3}{2} \times \frac{5}{3} \div 2 = \frac{5}{4}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 516\frac{5}{16} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 2 easy answer: 712\frac{7}{12} cm

The rhombus at the right has an area of 816 cm2\frac{8}{16}\ \text{cm}^2. If one of its diagonals is 139 cm1\frac{3}{9}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 3/9 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 8/16 cm2 and one diagonal 1 and 1/3 cm. We want how much the two diagonals differ by.

Givens
  • The area is 8/16 cm2.
  • One diagonal is 1 and 1/3 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
43×□÷2=12\frac{4}{3} \times \square \div 2 = \frac{1}{2}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
43×□=1\frac{4}{3} \times \square = 1
The diagonals multiply to 1.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
1÷43=341 \div \frac{4}{3} = \frac{3}{4}
The other diagonal is 3/4 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
113−34=7121\frac{1}{3} - \frac{3}{4} = \frac{7}{12}
They differ by 7/12 cm.
Answer: 712\frac{7}{12} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 43×34÷2=12\frac{4}{3} \times \frac{3}{4} \div 2 = \frac{1}{2}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 18\frac{1}{8} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 3 easy answer: 89\frac{8}{9} cm

The rhombus at the right has an area of 1718 cm2\frac{17}{18}\ \text{cm}^2. If one of its diagonals is 189 cm1\frac{8}{9}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 8/9 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 17/18 cm2 and one diagonal 1 and 8/9 cm. We want how much the two diagonals differ by.

Givens
  • The area is 17/18 cm2.
  • One diagonal is 1 and 8/9 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
179×□÷2=1718\frac{17}{9} \times \square \div 2 = \frac{17}{18}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
179×□=179\frac{17}{9} \times \square = \frac{17}{9}
The diagonals multiply to 1 and 8/9.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
179÷179=1\frac{17}{9} \div \frac{17}{9} = 1
The other diagonal is 1 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
189−1=891\frac{8}{9} - 1 = \frac{8}{9}
They differ by 8/9 cm.
Answer: 89\frac{8}{9} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 179×1÷2=1718\frac{17}{9} \times 1 \div 2 = \frac{17}{18}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 1772\frac{17}{72} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 4 easy answer: 35\frac{3}{5} cm

The rhombus at the right has an area of 1425 cm2\frac{14}{25}\ \text{cm}^2. If one of its diagonals is 125 cm1\frac{2}{5}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 2/5 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 14/25 cm2 and one diagonal 1 and 2/5 cm. We want how much the two diagonals differ by.

Givens
  • The area is 14/25 cm2.
  • One diagonal is 1 and 2/5 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
75×□÷2=1425\frac{7}{5} \times \square \div 2 = \frac{14}{25}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
75×□=2825\frac{7}{5} \times \square = \frac{28}{25}
The diagonals multiply to 1 and 3/25.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
2825÷75=45\frac{28}{25} \div \frac{7}{5} = \frac{4}{5}
The other diagonal is 4/5 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
125−45=351\frac{2}{5} - \frac{4}{5} = \frac{3}{5}
They differ by 3/5 cm.
Answer: 35\frac{3}{5} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 75×45÷2=1425\frac{7}{5} \times \frac{4}{5} \div 2 = \frac{14}{25}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 750\frac{7}{50} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 5 medium answer: 1161\frac{1}{6} cm

The rhombus at the right has an area of 3216 cm2\frac{32}{16}\ \text{cm}^2. If one of its diagonals is 148 cm1\frac{4}{8}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 4/8 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 32/16 cm2 and one diagonal 1 and 1/2 cm. We want how much the two diagonals differ by.

Givens
  • The area is 32/16 cm2.
  • One diagonal is 1 and 1/2 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
32×□÷2=2\frac{3}{2} \times \square \div 2 = 2
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
32×□=4\frac{3}{2} \times \square = 4
The diagonals multiply to 4.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
4÷32=2234 \div \frac{3}{2} = 2\frac{2}{3}
The other diagonal is 2 and 2/3 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
223−112=1162\frac{2}{3} - 1\frac{1}{2} = 1\frac{1}{6}
They differ by 1 and 1/6 cm.
Answer: 1161\frac{1}{6} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 32×83÷2=2\frac{3}{2} \times \frac{8}{3} \div 2 = 2, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 12\frac{1}{2} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 6 medium answer: 1341\frac{3}{4} cm

The rhombus at the right has an area of 3718 cm2\frac{37}{18}\ \text{cm}^2. If one of its diagonals is 139 cm1\frac{3}{9}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 3/9 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 37/18 cm2 and one diagonal 1 and 1/3 cm. We want how much the two diagonals differ by.

Givens
  • The area is 37/18 cm2.
  • One diagonal is 1 and 1/3 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
43×□÷2=3718\frac{4}{3} \times \square \div 2 = \frac{37}{18}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
43×□=379\frac{4}{3} \times \square = \frac{37}{9}
The diagonals multiply to 4 and 1/9.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
379÷43=3112\frac{37}{9} \div \frac{4}{3} = 3\frac{1}{12}
The other diagonal is 3 and 1/12 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
3112−113=1343\frac{1}{12} - 1\frac{1}{3} = 1\frac{3}{4}
They differ by 1 and 3/4 cm.
Answer: 1341\frac{3}{4} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 43×3712÷2=3718\frac{4}{3} \times \frac{37}{12} \div 2 = \frac{37}{18}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 3772\frac{37}{72} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 7 medium answer: 25122\frac{5}{12} cm

The rhombus at the right has an area of 4016 cm2\frac{40}{16}\ \text{cm}^2. If one of its diagonals is 139 cm1\frac{3}{9}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 3/9 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 40/16 cm2 and one diagonal 1 and 1/3 cm. We want how much the two diagonals differ by.

Givens
  • The area is 40/16 cm2.
  • One diagonal is 1 and 1/3 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
43×□÷2=52\frac{4}{3} \times \square \div 2 = \frac{5}{2}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
43×□=5\frac{4}{3} \times \square = 5
The diagonals multiply to 5.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
5÷43=3345 \div \frac{4}{3} = 3\frac{3}{4}
The other diagonal is 3 and 3/4 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
334−113=25123\frac{3}{4} - 1\frac{1}{3} = 2\frac{5}{12}
They differ by 2 and 5/12 cm.
Answer: 25122\frac{5}{12} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 43×154÷2=52\frac{4}{3} \times \frac{15}{4} \div 2 = \frac{5}{2}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 58\frac{5}{8} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 8 medium answer: 12\frac{1}{2} cm

The rhombus at the right has an area of 4436 cm2\frac{44}{36}\ \text{cm}^2. If one of its diagonals is 126 cm1\frac{2}{6}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 2/6 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 44/36 cm2 and one diagonal 1 and 1/3 cm. We want how much the two diagonals differ by.

Givens
  • The area is 44/36 cm2.
  • One diagonal is 1 and 1/3 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
43×□÷2=119\frac{4}{3} \times \square \div 2 = \frac{11}{9}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
43×□=229\frac{4}{3} \times \square = \frac{22}{9}
The diagonals multiply to 2 and 4/9.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
229÷43=156\frac{22}{9} \div \frac{4}{3} = 1\frac{5}{6}
The other diagonal is 1 and 5/6 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
156−113=121\frac{5}{6} - 1\frac{1}{3} = \frac{1}{2}
They differ by 1/2 cm.
Answer: 12\frac{1}{2} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 43×116÷2=119\frac{4}{3} \times \frac{11}{6} \div 2 = \frac{11}{9}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 1136\frac{11}{36} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 9 hard answer: 1381\frac{3}{8} cm

The rhombus at the right has an area of 4532 cm2\frac{45}{32}\ \text{cm}^2. If one of its diagonals is 118 cm1\frac{1}{8}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 1/8 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 45/32 cm2 and one diagonal 1 and 1/8 cm. We want how much the two diagonals differ by.

Givens
  • The area is 45/32 cm2.
  • One diagonal is 1 and 1/8 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
98×□÷2=4532\frac{9}{8} \times \square \div 2 = \frac{45}{32}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
98×□=4516\frac{9}{8} \times \square = \frac{45}{16}
The diagonals multiply to 2 and 13/16.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
4516÷98=212\frac{45}{16} \div \frac{9}{8} = 2\frac{1}{2}
The other diagonal is 2 and 1/2 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
212−118=1382\frac{1}{2} - 1\frac{1}{8} = 1\frac{3}{8}
They differ by 1 and 3/8 cm.
Answer: 1381\frac{3}{8} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 98×52÷2=4532\frac{9}{8} \times \frac{5}{2} \div 2 = \frac{45}{32}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 45128\frac{45}{128} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 10 hard answer: 13\frac{1}{3} cm

The rhombus at the right has an area of 4248 cm2\frac{42}{48}\ \text{cm}^2. If one of its diagonals is 116 cm1\frac{1}{6}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 1/6 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 42/48 cm2 and one diagonal 1 and 1/6 cm. We want how much the two diagonals differ by.

Givens
  • The area is 42/48 cm2.
  • One diagonal is 1 and 1/6 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
76×□÷2=78\frac{7}{6} \times \square \div 2 = \frac{7}{8}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
76×□=74\frac{7}{6} \times \square = \frac{7}{4}
The diagonals multiply to 1 and 3/4.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
74÷76=112\frac{7}{4} \div \frac{7}{6} = 1\frac{1}{2}
The other diagonal is 1 and 1/2 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
112−116=131\frac{1}{2} - 1\frac{1}{6} = \frac{1}{3}
They differ by 1/3 cm.
Answer: 13\frac{1}{3} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 76×32÷2=78\frac{7}{6} \times \frac{3}{2} \div 2 = \frac{7}{8}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 732\frac{7}{32} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 11 hard answer: 110\frac{1}{10} cm

The rhombus at the right has an area of 3950 cm2\frac{39}{50}\ \text{cm}^2. If one of its diagonals is 115 cm1\frac{1}{5}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 1/5 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 39/50 cm2 and one diagonal 1 and 1/5 cm. We want how much the two diagonals differ by.

Givens
  • The area is 39/50 cm2.
  • One diagonal is 1 and 1/5 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
65×□÷2=3950\frac{6}{5} \times \square \div 2 = \frac{39}{50}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
65×□=3925\frac{6}{5} \times \square = \frac{39}{25}
The diagonals multiply to 1 and 14/25.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
3925÷65=1310\frac{39}{25} \div \frac{6}{5} = 1\frac{3}{10}
The other diagonal is 1 and 3/10 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
1310−115=1101\frac{3}{10} - 1\frac{1}{5} = \frac{1}{10}
They differ by 1/10 cm.
Answer: 110\frac{1}{10} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 65×1310÷2=3950\frac{6}{5} \times \frac{13}{10} \div 2 = \frac{39}{50}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 39200\frac{39}{200} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.
Variant 12 hard answer: 710\frac{7}{10} cm

The rhombus at the right has an area of 3050 cm2\frac{30}{50}\ \text{cm}^2. If one of its diagonals is 1612 cm1\frac{6}{12}\ \text{cm} long, what is the difference between the lengths of the two diagonals, in cm\text{cm}?

1 6/12 cm
Show solution
1 · Understandwhat's really being asked

A rhombus has area 30/50 cm2 and one diagonal 1 and 1/2 cm. We want how much the two diagonals differ by.

Givens
  • The area is 30/50 cm2.
  • One diagonal is 1 and 1/2 cm.
Unknowns
  • The difference between the two diagonals.
Constraints
  • A rhombus's area is its two diagonals multiplied and halved.
2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#11 Work Backwards

The area rule already holds everything: two diagonals, one known, one not. Put the numbers in and it is an equation. Then read the question again -- it asks for the difference, not the diagonal.

3 · Execute4 carry out the plan

1Write the rule with the unknown in it

#13 Convert to Algebra 6.EE.B.6
A rhombus's area is the two diagonals multiplied, halved.
32×□÷2=35\frac{3}{2} \times \square \div 2 = \frac{3}{5}
One equation, one unknown.

2Double both sides

#13 Convert to Algebra 6.EE.B.7
That clears the halving and leaves the product of the two diagonals.
32×□=65\frac{3}{2} \times \square = \frac{6}{5}
The diagonals multiply to 1 and 1/5.

3Divide by the known diagonal

#11 Work Backwards 6.G.A.1
Dividing by a fraction multiplies by its reciprocal.
65÷32=45\frac{6}{5} \div \frac{3}{2} = \frac{4}{5}
The other diagonal is 4/5 cm.

4Answer what was asked

#7 Identify Subproblems 6.G.A.1
The question wants the difference, not the diagonal we just found.
112−45=7101\frac{1}{2} - \frac{4}{5} = \frac{7}{10}
They differ by 7/10 cm.
Answer: 710\frac{7}{10} cm
4 · Reviewdoes it hold up?

Checking the area with both diagonals: 32×45÷2=35\frac{3}{2} \times \frac{4}{5} \div 2 = \frac{3}{5}, the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of 320\frac{3}{20} cm2 each, and the same two diagonals.

Standardsmin grade 6
  • 6.EE.B.6 Use variables to represent numbers and write expressions — Naming the missing diagonal so the rule becomes an equation.
  • 6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving that equation for the diagonal.
  • 6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the rhombus area rule and taking the difference.
💡Takeaway. Give the missing length a name and the formula turns into an equation. Then check what the question actually asked for.