Geometry - Similarity (of triangles, polygons)

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD = 3 \text{ cm}$, $DB = 5 \text{ cm}$, and $AE = 4.5 \text{ cm}$, find the length of $AC$.

1. Since $DE \parallel BC$, by the Basic Proportionality Theorem, we have: $\frac{AD}{DB} = \frac{AE}{EC}$
2. Substitute the given values: $\frac{3}{5} = \frac{4.5}{EC}$
3. Solve for $EC$: $3 \times EC = 5 \times 4.5$ $3 \times EC = 22.5$ $EC = \frac{22.5}{3} = 7.5 \text{ cm}$
4. Find $AC$ by adding the segments: $AC = AE + EC = 4.5 + 7.5 = 12 \text{ cm}$

We use the Basic Proportionality Theorem which states that a line parallel to one side of a triangle divides the other two sides proportionally. After finding the lower segment $EC$, we add it to the upper segment $AE$ to get the total length of side $AC$.

The areas of two similar triangles $\triangle PQR$ and $\triangle XYZ$ are $64 \text{ cm}^2$ and $121 \text{ cm}^2$ respectively. If $QR = 12 \text{ cm}$, find the length of $YZ$.

1. We know that for similar triangles: $\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle XYZ)} = \left(\frac{QR}{YZ}\right)^2$
2. Substitute the known values: $\frac{64}{121} = \left(\frac{12}{YZ}\right)^2$
3. Take the square root of both sides: $\sqrt{\frac{64}{121}} = \frac{12}{YZ}$ $\frac{8}{11} = \frac{12}{YZ}$
4. Solve for $YZ$ using cross-multiplication: $8 \times YZ = 12 \times 11$ $8 \times YZ = 132$ $YZ = \frac{132}{8} = 16.5 \text{ cm}$

This problem applies the property that the ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides. By taking the square root of the area ratio, we find the linear scale factor and use it to calculate the missing side length.

1. In $\triangle AXY$ and $\triangle ABC$: $\angle A = \angle A$ (Common) $\angle AXY = \angle ABC$ (Corresponding angles as $XY \parallel BC$) Therefore, $\triangle AXY \sim \triangle ABC$ by AA similarity.

2. Find the scale factor $k$ of the sides: $AB = AX + XB = 2 + 4 = 6\text{ cm}$ $k = \frac{AX}{AB} = \frac{2}{6} = \frac{1}{3}$

3. Use the area ratio property: $\frac{\text{Area}(\triangle AXY)}{\text{Area}(\triangle ABC)} = k^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$ $\frac{5}{\text{Area}(\triangle ABC)} = \frac{1}{9}$ $\text{Area}(\triangle ABC) = 5 \times 9 = 45\text{ cm}^2$

4. Calculate the area of the trapezium: $\text{Area}(XYCB) = \text{Area}(\triangle ABC) - \text{Area}(\triangle AXY)$ $\text{Area}(XYCB) = 45 - 5 = 40\text{ cm}^2$

Since the line segment is parallel to the base, the smaller triangle at the top is similar to the whole triangle. The ratio of their areas is the square of the ratio of their corresponding side lengths. The area of the trapezium is the difference between the large and small triangle areas.

Given scale factor $k = \frac{1}{200}$.

(i) For Area: $\frac{\text{Area of Model}}{\text{Actual Area}} = k^2$ $\frac{0.5}{\text{Actual Area}} = \left(\frac{1}{200}\right)^2 = \frac{1}{40000}$ $\text{Actual Area} = 0.5 \times 40000 = 20000\text{ m}^2$

(ii) For Volume: $\frac{\text{Volume of Model}}{\text{Actual Volume}} = k^3$ $\frac{0.02}{\text{Actual Volume}} = \left(\frac{1}{200}\right)^3 = \frac{1}{8000000}$ $\text{Actual Volume} = 0.02 \times 8000000 = 160000\text{ m}^3$

Scaling applies to all dimensions. Linear dimensions scale by $k$, area by $k^2$, and volume by $k^3$. Here, the ship is the object and the model is the image.