Area and Similarity

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Learning goals

Choose a positive length scale factor by dividing an image length by its corresponding original length
Explain perimeter and area scaling using corresponding lengths and triangle partitions
Calculate areas of similar polygons from corresponding lengths or perimeters
Recover a positive length scale factor from an area ratio when similarity is known
Distinguish equal area, similarity, and congruence using geometric evidence

Keep the direction of the comparison

Similar figures have the same shape, with corresponding lengths multiplied by one positive number. The dilation lesson established length scaling and angle preservation; here we use those results to study area.

We call the figure before resizing the original and the resized figure the image. The length scale factor is the image length divided by its corresponding original length:

k=\frac{\text{image length}}{\text{original length}}.

We require $k>0$ . Keep image over original when calculating a scale factor. If a given area ratio is written in the order original:image, reverse it before using the image-over-original formula. A factor greater than $1$ enlarges; a factor between $0$ and $1$ reduces. A factor of $1$ preserves size.

We write $A_{\text{original}}$ and $A_{\text{image}}$ for the two areas, and use corresponding subscripts on $P$ for the perimeters.

Worked example 1 Scale every boundary side

Problem. A triangle has side lengths $4.5$ , $6$ , and $7.5$ cm. A similar image has a $6$ cm side corresponding to the original $4.5$ cm side. Find the length scale factor taking the original to its image, and find both perimeters.

Schematic; corresponding sides determine one length factor.

Solution. Divide corresponding lengths in image-over-original order:

\begin{aligned} k&=\frac{6}{4.5},\\ k&=\frac43. \end{aligned}

The original perimeter is the sum of its three sides:

\begin{aligned} P_{\text{original}}&=4.5+6+7.5,\\ P_{\text{original}}&=18\text{ cm}. \end{aligned}

The image sides are $6$ , $8$ , and $10$ cm, so its perimeter is $24$ cm. This is also the original perimeter multiplied by $\tfrac43$ .

Adding scaled sides scales their sum. For any polygon with original side lengths $s_1,\ldots,s_n$ , the image perimeter is

\begin{aligned} P_{\text{image}}&=ks_1+\cdots+ks_n,\\ P_{\text{image}}&=k(s_1+\cdots+s_n),\\ P_{\text{image}}&=kP_{\text{original}}. \end{aligned}

Check your understanding

Similar polygons have corresponding side lengths 15 cm in the original and 9 cm in the image. Which factor multiplies original lengths to give image lengths and also multiplies perimeter?

5/3

3/5

9/25

25/9

Count the two changes on a grid

Worked example 2 Three times the rows and columns

Problem. A rectangle is $3$ units by $2$ units. Its similar image is made with length scale factor $3$ . Find the image dimensions and compare the two areas.

The grid shows one whole-number example of squared area scaling.

Solution. The image dimensions are $9$ units and $6$ units. The original has $6$ unit squares; the image has $54$ unit squares.

\begin{aligned} \frac{A_{\text{image}}}{A_{\text{original}}}&=\frac{54}{6},\\ \frac{A_{\text{image}}}{A_{\text{original}}}&=9. \end{aligned}

There are three times as many columns and three times as many rows, giving an area factor of $3^2$ .

This grid explains an integer enlargement of this rectangle. It does not show that every polygon can fill its own enlargement with identical small copies. For noninteger factors and other polygon shapes, triangles give a general argument.

Why every triangle contributes the squared factor

The triangle area formula uses a chosen base and its perpendicular height. Similarity scales both of those lengths, even when the height meets an extended base line.

Area scaling from triangles to polygons#

First take a triangle with base $10$ cm and perpendicular height $7$ cm. Its area is $35$ square cm. Under length scale factor $2$ , the base becomes $20$ cm and the height $14$ cm, so

\begin{aligned} A_{\text{image}}&=\frac12\cdot20\cdot14,\\ A_{\text{image}}&=140\text{ cm}^2. \end{aligned}

The area factor is $4$ because both the base and height were multiplied by $2$ .

A perpendicular height is a length, so it scales with the base.

Now let the original base and height be any positive lengths $b$ and $h$ . Similarity carries the base line and its perpendicular altitude to a base line and perpendicular altitude. Thus the corresponding lengths are $kb$ and $kh$ . Rigid motions preserve these lengths and areas, so a turn or reflection does not affect the argument.

\begin{aligned} A_{\text{image}}&=\frac12(kb)(kh),\\ A_{\text{image}}&=k^2\left(\frac12bh\right),\\ A_{\text{image}}&=k^2A_{\text{original}}. \end{aligned}

Nothing used the specific base $10$ , height $7$ , or factor $2$ . The reasoning applies to every nondegenerate triangle and every positive length scale factor.

For a convex pentagon, draw the two diagonals from one vertex to its nonadjacent vertices. Convexity keeps both diagonals inside. They meet only at that vertex, and their three triangular regions fill the pentagon with no overlapping interiors. Carry those exact diagonals into the image, rather than choosing a different partition.

Schematic corresponding partitions, drawn at independent scales; the labels describe a general factor k.

For a numerical partition, suppose the three original triangle areas are $6$ , $13$ , and $17$ square units. Under factor $\tfrac12$ , the image triangle areas are $\tfrac64$ , $\tfrac{13}4$ , and $\tfrac{17}4$ . Adding gives $9$ square units, one quarter of the original $36$ square units.

In general, call those three areas $a,b,c$ . Each image triangle has the squared area factor. Area addition gives

\begin{aligned} A_{\text{image}}&=k^2a+k^2b+k^2c,\\ A_{\text{image}}&=k^2(a+b+c),\\ A_{\text{image}}&=k^2A_{\text{original}}. \end{aligned}

A concave outline needs cuts that stay inside the region.

For this L-shaped polygon, extend the horizontal notch edge leftward inside the region. The cut separates two rectangles, and one diagonal in each rectangle gives four triangles. Their interiors do not overlap, and together they fill the L.

The same area argument works for any finite nonoverlapping triangle partition. Here is a way to make such a partition even when the outline has a notch.

In the six-vertex example below, choose vertical cuts. None follows an edge, and no cut contains two vertices. The six vertex lines enclose five strips. In the middle strip, the notch separates the polygon into an upper piece $U$ and a lower piece $L$ .

Dashed vertex cuts isolate connected pieces; dotted diagonals divide the four trapezoids.

Follow the top and bottom edges of $U$ across that strip. Neither edge ends inside the strip, so those same two edges bound $U$ all the way from one cut to the next. Together with the two parallel cuts, they form a trapezoid. The separate piece $L$ is bounded by a different pair of edges and the same cuts, so it is another trapezoid.

Across the whole figure, three connected pieces are already triangles and four are trapezoids. A dotted diagonal inside each trapezoid divides it into two triangles. The final count is

3+2\cdot4=11.

For any simple polygon, choose parallel cuts through its vertices. Turn the cut direction if it would follow an edge or put two vertices on one cut. Between consecutive cuts there is no vertex where a bounding edge could end or change. The edges cannot cross there because the polygon is simple. Thus each connected piece stays between two fixed edges and the parallel cuts, making a convex triangle or trapezoid.

Divide every trapezoid by an interior diagonal, just as in the example. Keep the pieces of positive area; their interiors are disjoint and they fill the polygon. Shared cut segments add no area. Similarity carries these exact cuts and triangles to the image, so the same coverage and nonoverlap hold there. Applying the triangle area factor to each corresponding piece and adding proves the polygon area factor.

This proves the area rule for simple polygons with positive area, convex or concave. A self-crossing outline is outside these hypotheses. If a region has polygonal holes, apply the result to the outer region and each hole, then subtract their areas as in Composite and Shaded Figures.

For similar simple polygons, with positive areas and positive length scale factor,

\frac{A_{\text{image}}}{A_{\text{original}}}=k^2.

Check your understanding

Why can the area rule for triangles be added across a polygon partition?

A Every triangle in a polygon is congruent B Every enlarged polygon is tiled by copies of the original C Corresponding pieces each use the same squared factor and have nonoverlapping interiors D The number of sides equals the area factor

Use length information to compare areas

Worked example 3 Reduce a polygon

Problem. A simple polygon has area $112\text{ cm}^2$ and perimeter $48$ cm. A similar image has length scale factor $\tfrac34$ . Find its area and perimeter.

Use the stated measurements; these outlines are schematic.

Solution. Perimeter uses one factor:

\begin{aligned} P_{\text{image}}&=\frac34\cdot48,\\ P_{\text{image}}&=36\text{ cm}. \end{aligned}

Area uses the square of that factor:

\begin{aligned} A_{\text{image}}&=\left(\frac34\right)^2\cdot112,\\ A_{\text{image}}&=63\text{ cm}^2. \end{aligned}

Both measurements decrease, but not by the same fraction. The perimeter is $\tfrac34$ of its original value; the area is $\tfrac9{16}$ of its original value.

Check your understanding

A polygon of area 75 square cm is reduced to a similar image with length factor 2/5. What is the image area?

A 30 square cm B 12 square cm C 187.5 square cm D 6 square cm

Reverse an area ratio

When similarity is given or proved, an area ratio determines the positive length scale factor. For example, an image with $\tfrac9{16}$ of the original area has $\tfrac34$ of every corresponding length, because $\tfrac34$ squared is $\tfrac9{16}$ . In general,

k=\sqrt{\frac{A_{\text{image}}}{A_{\text{original}}}}.

Use the positive square root because lengths and the length scale factor are positive. An area ratio is dimensionless only after both areas have been expressed in the same square unit.

Worked example 4 An area ratio determines a height

Problem. Two triangles are similar. The original has area $75\text{ cm}^2$ , height $10$ cm, and perimeter $40$ cm. The image has area $147\text{ cm}^2$ . Find the corresponding image height and perimeter.

Schematic; given similarity permits the area-to-length calculation.

Solution. Keep image over original, then take the positive square root:

\begin{aligned} k^2&=\frac{147}{75},\\ k^2&=\frac{49}{25},\\ k&=\frac75. \end{aligned}

The corresponding height is

\begin{aligned} h_{\text{image}}&=\frac75\cdot10,\\ h_{\text{image}}&=14\text{ cm}. \end{aligned}

The image perimeter is

\begin{aligned} P_{\text{image}}&=\frac75\cdot40,\\ P_{\text{image}}&=56\text{ cm}. \end{aligned}

Check your understanding

Similar figures have area ratio 81:25 in the order original:image. What multiplies an original length to give its image length?

81/25

9/5

25/81

5/9

Equal area, similarity, and congruence

Equal area records how much region is enclosed. Similarity records shape. Congruence records shape and size.

Worked example 5 Equal areas with different shapes

Problem. Compare a $3$ cm by $8$ cm rectangle with a $4$ cm by $6$ cm rectangle. Do they have equal areas? Are they similar or congruent? Justify each conclusion.

Compare products and corresponding side ratios.

Solution. The first area is $3\cdot8$ , or $24\text{ cm}^2$ . The second is $4\cdot6$ , also $24\text{ cm}^2$ .

For similar rectangles, the ratio of longer side to shorter side must agree. Here those ratios are $\tfrac83$ and $\tfrac64$ , which differ. The rectangles are not similar, so they are not congruent either.

Thus equal area alone implies neither similarity nor congruence. Congruent polygons do have equal areas, by the area postulate. Similar polygons need not have equal areas: their area factor depends on the length factor.

There is a useful conclusion when both conditions are supplied. If two similar polygons with positive area also have equal areas, their squared factor is $1$ . The positive length factor must then be $1$ , so their similarity uses only size-preserving motions and they are congruent. Dropping the similarity hypothesis would invalidate that conclusion.

Check your understanding

Two simple polygons with positive area are known to be similar and to have equal areas. What follows?

A They are congruent B They have different perimeters C They have length factor 2 D There is not enough information to compare corresponding lengths

Area and Similarity

Learning goals

Keep the direction of the comparison

Count the two changes on a grid

Why every triangle contributes the squared factor

Use length information to compare areas

Reverse an area ratio

Equal area, similarity, and congruence

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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