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Area and Similarity: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A direct enlargement

A simple polygon has area $18\text{ cm}^2$ . A similar image is made using length scale factor $5/2$ to take the original to its image. Find its area.

Schematic; use the stated measurements.

Text description of this figure

Original and image polygons, schematic. Original: Area 18 square cm. Image: Length factor 5/2. Similarity is supplied in the text.
1. Hint 1
  Area depends on how two independent length dimensions change.
2. Hint 2
  Square the length scale factor before multiplying the original area.
Answer

$112.5\text{ cm}^2$ , or $\tfrac{225}{2}\text{ cm}^2$ .

Full solution

The area factor is the square of the length factor.
$k^2=\frac{25}{4}.$

Multiply the original area by that factor.
$A_{\text{image}}=\frac{25}{4}\cdot18.$ $A_{\text{image}}=112.5\text{ cm}^2.$

Answer

$112.5\text{ cm}^2$ , or $\tfrac{225}{2}\text{ cm}^2$ .

Key idea

Area uses the square of a positive length scale factor.
Problem 2 Corresponding measurements

Two simple polygons are similar. An original side of $22$ cm corresponds to an image side of $14$ cm. The original perimeter is $66$ cm and its area is $242\text{ cm}^2$ .

Find the image perimeter.

Find the image area.

Schematic; use the stated measurements.

Text description of this figure

Original and image polygons, schematic. Original: Side 22 cm. Image: Matching side 14 cm. Similarity is supplied in the text.
1. Hint 1
  Keep image measurements over original measurements.
2. Hint 2
  Perimeter uses the resulting factor once; area uses its square.
Answer

Image perimeter:
$42\text{ cm}.$
Image area:
$98\text{ cm}^2.$

Full solution

The length scale factor taking the original to its image is
$k=\frac{14}{22}.$ $k=\frac7{11}.$

For perimeter, multiply once.
$P_{\text{image}}=\frac7{11}\cdot66.$ $P_{\text{image}}=42\text{ cm}.$

For area, multiply by the squared factor.
$A_{\text{image}}=\frac{49}{121}\cdot242.$ $A_{\text{image}}=98\text{ cm}^2.$

Answer

Image perimeter:
$42\text{ cm}.$
Image area:
$98\text{ cm}^2.$

Key idea

An ordered side ratio determines perimeter and area factors when similarity is known.
Problem 3 From areas to a side

Two triangles are similar, with area ratio $36:49$ in the order original:image. An original side is $18$ cm. Find the corresponding image side and explain why similarity is needed.

Schematic; use the stated measurements.

Text description of this figure

Two similar triangles, schematic. Their area ratio, in the order original:image, is 36:49. The diagram does not encode the unknown length scale factor.
1. Hint 1
  Areas determine a corresponding length ratio only for similar shapes.
2. Hint 2
  Take the positive square root, keeping image area over original area.
Answer

Image side: $21$ cm. Similarity is needed to link corresponding lengths to the area ratio.

Full solution

Image area divided by original area is $49/36$ .
Similarity gives the positive length factor
$k=\sqrt{\frac{49}{36}}.$ $k=\frac76.$

Multiply the original side by that factor.
$\text{image side}=\frac76\cdot18.$ $\text{image side}=21\text{ cm}.$

Area by itself measures the enclosed region and does not fix its shape.
The given similarity supplies the common factor for corresponding lengths.

Answer

Image side: $21$ cm. Similarity is needed to link corresponding lengths to the area ratio.

Key idea

An inverse area calculation depends on an established similarity.
Problem 4 The outer portion

In nondegenerate triangle $ABC$ , $D$ lies between $A$ and $B$ , $E$ lies between $A$ and $C$ , and $DE$ is parallel to $BC$ . The ratio $AD:AB$ is $3:5$ .

The area of $ABC$ is $200\text{ cm}^2$ . Find the area inside $ABC$ but outside triangle $ADE$ .

Schematic; use the stated measurements.

Text description of this figure

Triangle ABC with D on AB and E on AC. DE is parallel to BC. The drawing is schematic; the length ratio is supplied in the problem.
1. Hint 1
  Parallel lines connect the inner triangle to the whole triangle.
2. Hint 2
  AA gives similarity, so square the inner-to-whole length ratio.
3. Hint 3
  Subtract the inner area from the whole area.
Answer

$128\text{ cm}^2$ .

Full solution

The triangles share angle A.
Parallel DE and BC give a second pair of equal angles, so ADE is similar to ABC by AA.

Let $A_{\triangle ADE}$ denote the inner triangle area.
The inner-to-whole length factor is $3/5$ , so
$A_{\triangle ADE}=\frac9{25}\cdot200.$ $A_{\triangle ADE}=72\text{ cm}^2.$

The inner triangle lies inside ABC because D and E lie on its sides.
Subtract its area.
$A_{\text{outside}}=200-72.$ $A_{\text{outside}}=128\text{ cm}^2.$

Answer

$128\text{ cm}^2$ .

Key idea

A parallel cut makes a similar triangle whose area fraction is a squared length fraction.
Problem 5 A resized emblem

A flat polygonal emblem is enlarged uniformly by length scale factor $1.5$ . The original face has area $54\text{ cm}^2$ and its boundary has length $30$ cm.

Find the enlarged face area.

Find the length of trim needed to follow its boundary once, with no allowance for joins.

Schematic; use the stated measurements.

Text description of this figure

Original and image polygons, schematic. Original: Area 54; P = 30 cm. Image: Length factor 1.5. Similarity is supplied in the text.
1. Hint 1
  Uniform enlargement gives similar faces.
2. Hint 2
  Use separate factors for the face area and the boundary length.
Answer

Face area:
$121.5\text{ cm}^2.$
Trim length:
$45\text{ cm}.$

Full solution

The face area uses the squared factor.
$A_{\text{image}}=1.5^2\cdot54.$ $A_{\text{image}}=121.5\text{ cm}^2.$

The trim follows the perimeter and uses the length factor once.
$P_{\text{image}}=1.5\cdot30.$ $P_{\text{image}}=45\text{ cm}.$

Answer

Face area:
$121.5\text{ cm}^2.$
Trim length:
$45\text{ cm}.$

Key idea

A uniform enlargement changes covering area and boundary length by different factors.
Problem 6 A preserved fraction

A simple polygon of positive area is partitioned into two polygonal pieces with nonoverlapping interiors. Both pieces have positive areas. A similarity carries the whole polygon and the partition to an image with positive length factor $k$ .

Prove that the first piece occupies the same fraction of the total area in the original and in the image.
1. Hint 1
  Compare the area of the piece to the area of the whole in each figure.
2. Hint 2
  Both numerator and denominator acquire the same area factor.
3. Hint 3
  The common factor can be canceled because it is positive.
Answer

The first piece occupies the same fraction of the total area in the original and in the image.

Full solution

Let $A_1$ be the first piece area and $A$ the whole original area.
The original fraction is $A_1/A$ .

The similarity carries each piece to its image without creating an overlap or a gap.
Every area has factor $k^2$ , so the image fraction is
$\frac{k^2A_1}{k^2A}.$

Since $k>0$ and $A>0$ , cancellation is valid.
$\frac{k^2A_1}{k^2A}=\frac{A_1}{A}.$

Schematic; use the stated measurements.

Answer

The first piece occupies the same fraction of the total area in the original and in the image.

Key idea

The same area factor on a part and its whole preserves their area fraction.
Problem 7 Check the calculation

Similar polygons have original area $128\text{ cm}^2$ and image area $392\text{ cm}^2$ . The original perimeter is $48$ cm. Examine the following calculation.

Step 1.
$k^2=\frac{392}{128}.$ $k^2=\frac{49}{16}.$
Step 2.
$k=\frac{49}{16}.$
Step 3.
$P_{\text{image}}=48\cdot\frac{49}{16}.$
Identify the first invalid step, repair it, and find the image perimeter.
1. Hint 1
  Distinguish the length factor from its square.
2. Hint 2
  Check which step first stops following from the preceding information.
3. Hint 3
  Use the positive square root before scaling perimeter.
Answer

Step 2 is first invalid. Correct length factor:
$k=\tfrac74.$
Image perimeter:
$84\text{ cm}.$

Full solution

Step 1 correctly uses image area over original area and simplifies the ratio.
Step 2 incorrectly treats $k^2$ as $k$ .

Take the positive square root instead.
$k=\sqrt{\frac{49}{16}}.$ $k=\frac74.$

Then scale the original perimeter by the length factor.
$P_{\text{image}}=48\cdot\frac74.$ $P_{\text{image}}=84\text{ cm}.$

Answer

Step 2 is first invalid. Correct length factor:
$k=\tfrac74.$
Image perimeter:
$84\text{ cm}.$

Key idea

Recover the length factor before applying an area ratio to a perimeter.
Problem 8 Equal area and shape

A rectangle is $5$ cm by $12$ cm. Construct a numerical counterexample to the statement that any rectangle with the same area must be similar to it. Give positive whole-number side lengths for a second rectangle, verify equal areas, and show that the shapes are not similar.
1. Hint 1
  Equal products can come from different pairs of side lengths.
2. Hint 2
  Compare longer-side to shorter-side ratios to test rectangle similarity.
Answer

One possible second rectangle is $3$ cm by $20$ cm. Both areas are $60\text{ cm}^2$ , but the side ratios $12/5$ and $20/3$ differ. Other valid counterexamples are accepted.

Full solution

The given rectangle has area
$A=5\cdot12.$ $A=60\text{ cm}^2.$

A $3$ cm by $20$ cm rectangle also has area
$A=3\cdot20.$ $A=60\text{ cm}^2.$

For similarity, longer sides must correspond and shorter sides must correspond.
The ratios $12/5$ and $20/3$ are unequal, so these rectangles are not similar.
Equal area therefore does not guarantee similarity.

Schematic; use the stated measurements.

Answer

One possible second rectangle is $3$ cm by $20$ cm. Both areas are $60\text{ cm}^2$ , but the side ratios $12/5$ and $20/3$ differ. Other valid counterexamples are accepted.

Key idea

A counterexample satisfies the given condition while failing the proposed conclusion.
Problem 9 A percentage comparison

A similar image of a polygon has $44\%$ more area than the original. Find the percentage increase in its perimeter. Explain why using $44\%$ as the perimeter increase would not follow.
1. Hint 1
  Translate the area increase into an area multiplying factor.
2. Hint 2
  The length factor is the positive square root of that factor.
3. Hint 3
  Subtract one from the perimeter factor before converting to a percentage.
Answer

The perimeter increases by $20\%$ . The $44\%$ increase describes area, which uses the square of the length factor.

Full solution

An area increase of $44\%$ gives area factor $1.44$ .
Similarity gives
$k=\sqrt{1.44}.$ $k=1.2.$

Perimeter uses factor $1.2$ , so it becomes $120\%$ of its original value.
Its increase is $20\%$ .

The area factor includes the change in two dimensions.
A perimeter adds lengths and uses the length factor only once.

Answer

The perimeter increases by $20\%$ . The $44\%$ increase describes area, which uses the square of the length factor.

Key idea

Convert an area percentage to a factor before taking its positive square root.
Problem 10 Recover and transfer

Two simple polygons are similar. The original area is $63\text{ cm}^2$ and the image area is $175\text{ cm}^2$ . The image perimeter is $95$ cm. A diagonal of the original is $9$ cm long.

Find the original perimeter.

Find the corresponding image diagonal.

Schematic; use the stated measurements.

Text description of this figure

Original and image polygons, schematic. Original: Area 63; diagonal 9 cm. Image: Area 175; P = 95 cm. Similarity is supplied in the text.
1. Hint 1
  First recover the positive factor that multiplies original lengths to give image lengths.
2. Hint 2
  To return from an image perimeter to the original, divide by that factor.
3. Hint 3
  To move an original diagonal to its image, multiply by that factor.
Answer

Original perimeter:
$57\text{ cm}.$
Image diagonal:
$15\text{ cm}.$

Full solution

The area ratio is image over original.
$k^2=\frac{175}{63}.$ $k^2=\frac{25}{9}.$ $k=\frac53.$

The known perimeter belongs to the image.
Divide by $5/3$ to recover the original.
$P_{\text{original}}=95\cdot\frac35.$ $P_{\text{original}}=57\text{ cm}.$

The known diagonal belongs to the original.
Corresponding diagonals are lengths and use the same factor.
$d_{\text{image}}=\frac53\cdot9.$ $d_{\text{image}}=15\text{ cm}.$

Answer

Original perimeter:
$57\text{ cm}.$
Image diagonal:
$15\text{ cm}.$

Key idea

After recovering a length factor, check which direction each requested measurement travels.

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