E4.4 Similarity

Learning objectives

E4.4.1Calculate lengths of similar shapes.• Calculate lengths of similar shapes.E4.4.2Use the relationships between lengths and areas of similar shapes and• Use the relationships between lengths and areas of similar shapes and lengths, surface areas and volumes of similar solids.E4.4.3Solve problems and give simple explanations involving similarity• Solve problems and give simple explanations involving similarity. Includes use of scale factor, e.g. B A B A Volume of Volume of Lengtho f Lengtho f = ` `j j. Includes showing that two triangles are similar using geometric reasons.

Similar shapes have equal corresponding angles and all corresponding lengths in one constant ratio. That ratio is the linear scale factor.

Match corresponding vertices or sides by angle position and order.
Use one known pair to calculate $k=\frac{\text{target length}}{\text{source length}}$ .
Multiply every source length by $k$ to move to the target shape; divide by $k$ to move back.
Check that every ratio compares corresponding sides in the same direction.

\frac{\text{target side 1}}{\text{source side 1}}=\frac{\text{target side 2}}{\text{source side 2}}=k

Two similar triangles have corresponding sides $4.5$ cm and $9$ cm, so the scale factor from the smaller to the larger is $9\div4.5=2$ . A side corresponding to $3.3$ cm therefore has length $3.3\times2=6.6$ cm in the larger triangle.

Do not pair sides merely because they are drawn in the same orientation. Trace the vertex order or equal angles first. Similar shapes need not be the same size; congruent shapes are the special case $k=1$ .

When similar objects have linear scale factor $k$ , one-dimensional measures scale by $k$ , areas by $k^2$ , and volumes by $k^3$ .

Given relationship	Matching scale factor	Recover linear factor
corresponding lengths	$k$	use the ratio directly
areas or surface areas	$k^2$	take a square root
volumes or capacities	$k^3$	take a cube root

\frac{A_2}{A_1}=k^2,\qquad \frac{V_2}{V_1}=k^3

A small bottle holds $0.4$ L and a similar large bottle holds $1.35$ L. The linear factor from large to small is $\sqrt[3]{0.4/1.35}$ . If the large bottle is $29.7$ cm high, the small height is $29.7\sqrt[3]{0.4/1.35}=19.8$ cm.

Choose the power from the quantity being compared, not from the quantity requested. A volume ratio must be cube-rooted before it can scale a length; an area ratio must be square-rooted. Keep the ratio direction consistent throughout.

To justify similarity, state geometric facts that prove equal corresponding angles or a common scale factor; a visual resemblance is not evidence.

Situation	Sufficient explanation
triangles	two pairs of corresponding angles are equal (AA); the third pair then also matches
shapes or solids	corresponding lengths are all in the same ratio and corresponding angles match
congruent figures	same shape and size, so the similarity scale factor is $1$

Name why angles are equal: corresponding or alternate angles in parallel lines, vertically opposite angles, a common angle, or an applicable established angle property. Then write the triangle correspondence in matching vertex order so the side ratios are paired correctly.

If $BE\parallel CD$ in triangle $ACD$ , then $\angle ABE=\angle ACD$ and $\angle AEB=\angle ADC$ by corresponding angles; $\angle BAE$ is the common angle at $A$ . Therefore triangles $ABE$ and $ACD$ are similar by AA, and their matching sides share one scale factor.

Equal area, one equal angle or one proportional side pair is not enough by itself. Explanations must connect each equality to a valid geometric reason and preserve the correct corresponding order.

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—E4.5 SymmetryRead

E4.4 Similarity

Learning objectives

Find corresponding lengths in similar shapes

Scale length, area and volume correctly

Show that shapes are similar

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