Source-matched teaching notes, visual explanations, graduated worked examples and independent examination-style practice.
Use the key facts confidently, select an appropriate method, communicate each working step, and check the reasonableness of the final result.
Subtopic 1
Similar shapes are exact enlargements or reductions of one another: corresponding angles are equal, and corresponding sides have a constant ratio. Their orientation may differ because a shape can also be turned or reflected.
For a change from shape A to shape B, \(k=\frac{\text{length in B}}{\text{corresponding length in A}}\). Multiply every length in A by \(k\). To return to A, divide by \(k\). A reduction has \(0<k<1\).
Triangles: two equal pairs of corresponding angles establish similarity (AA); use angle facts or parallel lines to justify this before setting up a proportion.
Always put corresponding sides in the same order in your ratios. Similarity is not the same as congruence: congruent shapes have scale factor 1. Do not assume two rectangles are similar just because both contain right angles: their side ratios must also match.
Study the reason for each step; in an examination the method and relevant geometric explanation can earn marks even if a later calculation goes wrong.
A rectangle measures \(3\text{ cm}\) by \(8\text{ cm}\). A second measures \(6\text{ cm}\) by \(16\text{ cm}\). Are they similar?
Compare matching sides: \(6/3=2\) and \(16/8=2\). Both ratios match and corresponding angles are right angles, so yes; \(k=2\).
A triangle side of \(7\text{ cm}\) corresponds to \(21\text{ cm}\). Find the image of a \(5\text{ cm}\) side.
\(k=21/7=3\). Required side \(=5\times3=\boxed{15\text{ cm}}\).
An \(18\text{ cm}\) side reduces to \(7.2\text{ cm}\). What happens to a corresponding \(12\text{ cm}\) side?
\(k=7.2/18=0.4\); hence \(12\times0.4=\boxed{4.8\text{ cm}}\).
Triangles \(ABC\) and \(PQR\) have \(\angle A=\angle P=52^\circ\) and \(\angle B=\angle Q=71^\circ\). Explain why they are similar.
Two corresponding angle pairs are equal. The third is \(180-52-71=57^\circ\) in each. The triangles are similar by AA.
Two similar right triangles have matching vertical sides \(8\) and \(14\text{ cm}\). The first has base \(12\text{ cm}\). Find the second base.
\(k=14/8=7/4\). Base \(=12\times7/4=\boxed{21\text{ cm}}\).
A \(1.6\text{ m}\) pole casts a \(2.4\text{ m}\) shadow. At the same moment a tower casts a \(18\text{ m}\) shadow. Find the tower height.
The Sun gives equal angles and both objects are vertical, so triangles are similar. \(h/18=1.6/2.4\), hence \(h=18\times1.6/2.4=\boxed{12\text{ m}}\).
In triangle \(ABC\), \(D\) lies on \(AB\), \(E\) on \(AC\), and \(DE\parallel BC\). Given \(AD=6\), \(AB=15\), \(DE=8\), find \(BC\).
Corresponding angles from the parallel lines are equal, hence \(ADE\sim ABC\). \(AD/AB=DE/BC\). Thus \(6/15=8/BC\), giving \(BC=8\times15/6=\boxed{20}\).
Subtopic 2
When all linear dimensions are multiplied by \(k\), each unit of area involves two lengths and changes by \(k^2\); each unit of volume involves three lengths and changes by \(k^3\). Surface areas follow the square rule; capacities and volumes follow the cube rule.
\[\text{Length ratio}=k,\qquad\text{Area ratio}=k^2,\qquad\text{Volume ratio}=k^3.\]
If the area ratio is known, recover the length factor with \(\sqrt{\text{area ratio}}\). If the volume ratio is known, use \(\sqrt[3]{\text{volume ratio}}\).
Convert units before comparing: \(1\text{ m}^2=10\,000\text{ cm}^2\), and \(1\text{ m}^3=1\,000\,000\text{ cm}^3\). Do not multiply area or volume by the length factor just once.
Study the reason for each step; in an examination the method and relevant geometric explanation can earn marks even if a later calculation goes wrong.
A square is enlarged by length scale factor \(3\). How many times as great is its area?
Area factor \(=3^2=\boxed{9}\).
A cube has all its edges doubled. How does volume change?
Volume factor \(=2^3=\boxed{8}\).
Similar shapes have areas \(20\text{ cm}^2\) and \(180\text{ cm}^2\). Find the small-to-large length factor.
Area ratio \(180/20=9\), so \(k=\sqrt9=\boxed{3}\).
Two similar containers hold \(24\text{ L}\) and \(81\text{ L}\). Find the length ratio from the first to the second.
Volume ratio \(81/24=27/8\); \(k=\sqrt[3]{27/8}=\boxed{3/2}\).
A model is built to length scale \(1:15\), and its exposed surface area is \(80\text{ cm}^2\). Find corresponding real surface area in \(\text{m}^2\).
Area factor \(15^2=225\). Actual area \(=80\times225=18\,000\text{ cm}^2=\boxed{1.8\text{ m}^2}\).
Two similar cylindrical vessels have heights \(12\) and \(18\text{ cm}\). The smaller holds \(640\text{ mL}\). Find the larger capacity.
Length factor \(=18/12=1.5\); volume factor \(1.5^3=3.375\). Capacity \(=640\times3.375=\boxed{2160\text{ mL}}\).
The curved surfaces of two similar solids are \(54\) and \(150\text{ cm}^2\). If the smaller radius is \(6\text{ cm}\), find the larger radius.
Length factor \(k=\sqrt{150/54}=\sqrt{25/9}=5/3\). Larger radius \(6\times5/3=\boxed{10\text{ cm}}\).
Independent practice
Questions in each subtopic progress from confidence-building to developed methods and original exam-style problems. The problems are original practice, not copied past-paper questions. Answers stay closed until selected.
Work through the basics first; save the examination-style questions for last.
Lengths \(4\) and \(10\text{ cm}\) correspond. Find the scale factor from smaller to larger.
Step-by-step solution: \(k=10/4=\boxed{2.5}\).
An image uses scale factor \(3\). Find the image of a \(7\text{ cm}\) segment.
Step-by-step solution: \(7\times3=\boxed{21\text{ cm}}\).
A \(30\text{ cm}\) side is reduced using \(k=0.2\). Find its new length.
Step-by-step solution: \(30\times0.2=\boxed{6\text{ cm}}\).
Rectangles \(2\times5\) and \(4\times10\) have matching corners. Are they similar?
Step-by-step solution: \(4/2=2\) and \(10/5=2\). Both ratios and all corresponding angles match; yes.
Rectangles \(3\times5\) and \(6\times12\) have matching corners. Are they similar?
Step-by-step solution: \(6/3=2\) but \(12/5=2.4\); the multipliers differ, so no.
A model is \(9\text{ cm}\) high and its real object is \(1.8\text{ m}\) high. Find the model-to-real length scale factor.
Step-by-step solution: Convert \(1.8\text{ m}=180\text{ cm}\). Hence \(k=180/9=\boxed{20}\).
Similar triangles have matching sides \(10\) and \(15\text{ cm}\). If another side on the first is \(8\text{ cm}\), find its match.
Step-by-step solution: \(k=15/10=1.5\); new length \(=8\times1.5=\boxed{12\text{ cm}}\).
A pair of similar polygons have perimeters \(26\) and \(65\text{ cm}\). Find the scale factor from small to large.
Step-by-step solution: Perimeter scales like length, so \(k=65/26=\boxed{2.5}\).
An angle in one triangle is \(63^\circ\); its corresponding angle in a similar triangle is unknown. Find it.
Step-by-step solution: Corresponding angles in similar figures are equal, so \(\boxed{63^\circ}\).
Similar triangles have corresponding sides \(x,12\) and \(9,18\text{ cm}\). Find \(x\).
Step-by-step solution: Keep matching sides aligned: \(x/9=12/18\), so \(x=9\times12/18=\boxed{6\text{ cm}}\).
A vertical mast has height \(9\text{ m}\) and shadow \(12\text{ m}\). A building has shadow \(44\text{ m}\). Find its height.
Step-by-step solution: By similar right triangles, \(h/44=9/12\). Thus \(h=44\times9/12=\boxed{33\text{ m}}\).
For similar triangles \(PQR\) and \(STU\), \(PQ=4\), \(QR=7\), \(ST=14\). Find \(TU\).
Step-by-step solution: \(k=ST/PQ=14/4=3.5\). Thus \(TU=7\times3.5=\boxed{24.5}\).
In \(ABC\), \(DE\parallel BC\). \(AD=5\), \(DB=3\), and \(DE=9\). Find \(BC\).
Step-by-step solution: \(AB=5+3=8\). Similarity gives \(DE/BC=AD/AB=5/8\). So \(BC=9\times8/5=\boxed{14.4}\).
A photo \(12\times18\text{ cm}\) is enlarged to width \(27\text{ cm}\). What height preserves similarity?
Step-by-step solution: The width multiplier is \(27/18=1.5\); new height \(=12\times1.5=\boxed{18\text{ cm}}\).
A triangle has sides \(5,7,9\); a second has sides \(15,21,27\). Determine whether the triangles are similar.
Step-by-step solution: All corresponding side ratios are \(15/5=21/7=27/9=3\), so the triangles are similar by SSS.
Work through the basics first; save the examination-style questions for last.
Similar triangles use length factor \(2\). Find the area factor.
Step-by-step solution: \(2^2=\boxed{4}\).
Similar boxes use length factor \(4\). Find the volume factor.
Step-by-step solution: \(4^3=\boxed{64}\).
A region has area \(6\text{ cm}^2\); it is enlarged by \(k=3\). Find the new area.
Step-by-step solution: \(6\times3^2=6\times9=\boxed{54\text{ cm}^2}\).
A similar solid has initial volume \(10\text{ cm}^3\), scaled by \(k=2\). Find its volume.
Step-by-step solution: \(10\times2^3=\boxed{80\text{ cm}^3}\).
Two similar shapes have areas \(12\) and \(48\text{ cm}^2\). Find the length factor.
Step-by-step solution: Area factor \(48/12=4\), so \(k=\sqrt4=\boxed{2}\).
Similar solids have volumes \(25\) and \(200\text{ cm}^3\). Find the length factor.
Step-by-step solution: Volume factor \(200/25=8\), so \(k=\sqrt[3]8=\boxed{2}\).
The smaller surface area is \(90\text{ cm}^2\) and length ratio small:large is \(2:5\). Find the larger surface area.
Step-by-step solution: Area factor \((5/2)^2=25/4\); area \(90\times25/4=\boxed{562.5\text{ cm}^2}\).
A similar container is \(3/4\) the original length. The original holds \(128\text{ L}\). Find its new capacity.
Step-by-step solution: Volume factor \((3/4)^3=27/64\). Capacity \(128\times27/64=\boxed{54\text{ L}}\).
A model has area \(48\text{ cm}^2\) on scale \(1:25\). Find the actual area in \(\text{m}^2\).
Step-by-step solution: \(48\times25^2=30\,000\text{ cm}^2=\boxed{3\text{ m}^2}\).
Two similar tanks have capacities \(27\text{ L}\) and \(216\text{ L}\). A smaller height is \(18\text{ cm}\). Find the larger height.
Step-by-step solution: Volume factor \(216/27=8\), length factor \(\sqrt[3]8=2\). Larger height \(18\times2=\boxed{36\text{ cm}}\).
The lengths of similar solids have ratio \(3:5\), and smaller volume is \(81\text{ cm}^3\). Find larger volume.
Step-by-step solution: Volume ratio \(27:125\), larger volume \(81\times125/27=\boxed{375\text{ cm}^3}\).
A small spherical ornament has radius \(4\text{ cm}\) and surface area \(64\pi\text{ cm}^2\). A similar one has radius \(10\text{ cm}\). Find its area.
Step-by-step solution: \(k=10/4=2.5\); new area \(64\pi(2.5)^2=\boxed{400\pi\text{ cm}^2}\).
A model room at scale \(1:20\) has volume \(200\text{ cm}^3\). Find the true volume in \(\text{m}^3\).
Step-by-step solution: Actual volume \(200\times20^3=1\,600\,000\text{ cm}^3=\boxed{1.6\text{ m}^3}\).
Two similar solids have total surface areas in ratio \(49:81\). Find their volumes ratio.
Step-by-step solution: Length ratio \(\sqrt{49}:\sqrt{81}=7:9\), hence volume ratio \(7^3:9^3=\boxed{343:729}\).