Practice · An area rule with one unknown in it is already an equation · Sensim Math

Variant 1 easy

The rhombus at the right has an area of $\frac{15}{12}\ \text{cm}^2$ . If one of its diagonals is $1\frac{4}{8}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

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

They differ by 1/6 cm.

Answer:

\frac{1}{6}

cm

4 · Reviewdoes it hold up?

Checking the area with both diagonals: $\frac{3}{2} \times \frac{5}{3} \div 2 = \frac{5}{4}$ , the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{5}{16}$ 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

The rhombus at the right has an area of $\frac{8}{16}\ \text{cm}^2$ . If one of its diagonals is $1\frac{3}{9}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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 \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.

1\frac{1}{3} - \frac{3}{4} = \frac{7}{12}

They differ by 7/12 cm.

Answer:

\frac{7}{12}

cm

4 · Reviewdoes it hold up?

Checking the area with both diagonals: $\frac{4}{3} \times \frac{3}{4} \div 2 = \frac{1}{2}$ , the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{1}{8}$ 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

The rhombus at the right has an area of $\frac{17}{18}\ \text{cm}^2$ . If one of its diagonals is $1\frac{8}{9}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

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

They differ by 8/9 cm.

Answer:

\frac{8}{9}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{17}{72}$ 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

The rhombus at the right has an area of $\frac{14}{25}\ \text{cm}^2$ . If one of its diagonals is $1\frac{2}{5}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

1\frac{2}{5} - \frac{4}{5} = \frac{3}{5}

They differ by 3/5 cm.

Answer:

\frac{3}{5}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{7}{50}$ 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

The rhombus at the right has an area of $\frac{32}{16}\ \text{cm}^2$ . If one of its diagonals is $1\frac{4}{8}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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 \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.

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

They differ by 1 and 1/6 cm.

Answer:

1\frac{1}{6}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{1}{2}$ 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

The rhombus at the right has an area of $\frac{37}{18}\ \text{cm}^2$ . If one of its diagonals is $1\frac{3}{9}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

3\frac{1}{12} - 1\frac{1}{3} = 1\frac{3}{4}

They differ by 1 and 3/4 cm.

Answer:

1\frac{3}{4}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{37}{72}$ 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

The rhombus at the right has an area of $\frac{40}{16}\ \text{cm}^2$ . If one of its diagonals is $1\frac{3}{9}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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 \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.

3\frac{3}{4} - 1\frac{1}{3} = 2\frac{5}{12}

They differ by 2 and 5/12 cm.

Answer:

2\frac{5}{12}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{5}{8}$ 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

The rhombus at the right has an area of $\frac{44}{36}\ \text{cm}^2$ . If one of its diagonals is $1\frac{2}{6}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

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

They differ by 1/2 cm.

Answer:

\frac{1}{2}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{11}{36}$ 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

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

Your answer

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.

\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.

\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.

\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.

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

They differ by 1 and 3/8 cm.

Answer:

1\frac{3}{8}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{45}{128}$ 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

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

Your answer

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.

\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.

\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.

\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.

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

They differ by 1/3 cm.

Answer:

\frac{1}{3}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{7}{32}$ 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

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

Your answer

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.

\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.

\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.

\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.

1\frac{3}{10} - 1\frac{1}{5} = \frac{1}{10}

They differ by 1/10 cm.

Answer:

\frac{1}{10}

cm

4 · Reviewdoes it hold up?

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

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{39}{200}$ 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

The rhombus at the right has an area of $\frac{30}{50}\ \text{cm}^2$ . If one of its diagonals is $1\frac{6}{12}\ \text{cm}$ long, what is the difference between the lengths of the two diagonals, in $\text{cm}$ ?

Your answer

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.

\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.

\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.

\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.

1\frac{1}{2} - \frac{4}{5} = \frac{7}{10}

They differ by 7/10 cm.

Answer:

\frac{7}{10}

cm

4 · Reviewdoes it hold up?

Checking the area with both diagonals: $\frac{3}{2} \times \frac{4}{5} \div 2 = \frac{3}{5}$ , the area we were given.

Another way: Working with the halves instead -- each diagonal cut at the centre -- gives four right triangles of $\frac{3}{20}$ 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.

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

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