Similar Figures

You should be familiar with Congruent and Similar Triangles.

In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. This means that we can obtain one figure from the other through a process of expansion or contraction, possibly followed by translation, rotation or reflection. If the objects also have the same size, then they are congruent.

For a general shape, it can be tricky to show that two items are similar. We have to check all corresponding angles and ratios of side lengths before we can reach a conclusion. For example, a $1 \times 2$ rectangle is not similar to a $2 \times 3$ rectangle, even though they both have 4 right angles, since their side lengths have different ratios. Likewise, a square of side length 1 is not similar with a rhombus of side length 1 and vertex angle $60 ^ \circ$, even through they all have side lengths of 1.

For triangles, due to the cosine rule and sine rule, we have much more options of determining if they are congruent or similar triangles.

Show that all circles are similar to each other.

---

If we lay the circles over each other such that their centers coincide, then the expansion by factor $\frac{R_2} {R_1}$ $($where $R_1$ and $R_2$ are the radii of the two circles, respectively$)$ maps the first circle to the second. $_\square$

Note: In a similar fashion, we can show that all regular polygons with the same number of edges are similar to each other. Why doesn't this work for rectangles?

The following result holds for similar figures $F_1$ and $F_2$ with a corresponding side length ratio of $R:$

• The ratio of their perimeters and diagonals will be in the same ratio $R$.

• The ratio of their areas will be $R^2$.

If the sides of two rectangles are in the ratio $3 : 5$, what is the ratio of their areas?

---

The ratio of their areas would be $3^2 : 5^2 = 9 : 25$. $_\square$

If the area of a square is increased 21% percent, what is the increase in the perimeter of the same square (in percentage)?

---

Let $L$ denote the original length of the square and let $L'$ denote the new length of the square after the area has been increased. Then, $\frac{ L'^2}{L^2} = 1.21 = 1.10^2$. So the length $L$ was increased by 10%. Since the perimeter of the square is proportional to its side length, the the perimeter of the square also increased by 10%. $_\square$