ΔABC ∼ ΔPQR, ar (ΔABC) = 16 cm2 and ar (ΔPQR) = 25 cm2. If BC = 20 cm, then QR is equal to :
25 cm
This problem involves similar triangles and their areas. When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Understanding Similar Triangles and Area Ratio
Two triangles are considered similar if their corresponding angles are equal and their corresponding sides are in proportion. A key property of similar triangles is how their areas relate to their side lengths.
The theorem states that if ΔABC ∼ ΔPQR, then the ratio of their areas is given by:
$$ \frac{\text{ar}(\Delta \text{ABC})}{\text{ar}(\Delta \text{PQR})} = \left(\frac{\text{AB}}{\text{PQ}}\right)^2 = \left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{\text{CA}}{\text{RP}}\right)^2 $$
In this problem, we are given the areas of the two similar triangles and the length of a side in the first triangle (ΔABC). We need to find the length of the corresponding side in the second triangle (ΔPQR).
Applying the Area Ratio Theorem to find QR
We are given:
- ΔABC ∼ ΔPQR
- ar (ΔABC) = 16 cm$^2$
- ar (ΔPQR) = 25 cm$^2$
- BC = 20 cm
We need to find the length of QR, which is the side in ΔPQR corresponding to BC in ΔABC.
Using the area ratio theorem for similar triangles:
$$ \frac{\text{ar}(\Delta \text{ABC})}{\text{ar}(\Delta \text{PQR})} = \left(\frac{\text{BC}}{\text{QR}}\right)^2 $$
Substitute the given values into the equation:
$$ \frac{16}{25} = \left(\frac{20}{\text{QR}}\right)^2 $$
To solve for QR, first take the square root of both sides of the equation:
$$ \sqrt{\frac{16}{25}} = \sqrt{\left(\frac{20}{\text{QR}}\right)^2} $$
$$ \frac{\sqrt{16}}{\sqrt{25}} = \frac{20}{\text{QR}} $$
$$ \frac{4}{5} = \frac{20}{\text{QR}} $$
Now, we can cross-multiply to solve for QR:
$$ 4 \times \text{QR} = 5 \times 20 $$
$$ 4 \times \text{QR} = 100 $$
Divide both sides by 4:
$$ \text{QR} = \frac{100}{4} $$
$$ \text{QR} = 25 $$
So, the length of side QR is 25 cm.
Step-by-Step Calculation Summary
Here are the steps we followed:
- Identified that the triangles are similar.
- Recalled the theorem about the ratio of areas of similar triangles being the square of the ratio of corresponding sides.
- Set up the equation using the given areas and side length: $\frac{16}{25} = \left(\frac{20}{\text{QR}}\right)^2$.
- Took the square root of both sides: $\frac{4}{5} = \frac{20}{\text{QR}}$.
- Solved for QR using cross-multiplication: $4 \times \text{QR} = 5 \times 20$.
- Calculated the final value: $\text{QR} = 25$ cm.
| Given Information | Value |
|---|---|
| ar (ΔABC) | 16 cm$^2$ |
| ar (ΔPQR) | 25 cm$^2$ |
| BC | 20 cm |
| Relationship | ΔABC ∼ ΔPQR |
| Step | Calculation | Result |
|---|---|---|
| Area Ratio | $\frac{\text{ar}(\Delta \text{ABC})}{\text{ar}(\Delta \text{PQR})} = \frac{16}{25}$ | $\frac{16}{25}$ |
| Side Ratio Squared | $\left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{20}{\text{QR}}\right)^2$ | $\left(\frac{20}{\text{QR}}\right)^2$ |
| Equation | $\frac{16}{25} = \left(\frac{20}{\text{QR}}\right)^2$ | $\frac{16}{25} = \left(\frac{20}{\text{QR}}\right)^2$ |
| Square Root | $\sqrt{\frac{16}{25}} = \frac{4}{5}$ | $\frac{4}{5}$ |
| Equation (Simplified) | $\frac{4}{5} = \frac{20}{\text{QR}}$ | $\frac{4}{5} = \frac{20}{\text{QR}}$ |
| Solve for QR | $4 \times \text{QR} = 5 \times 20$ | $4 \times \text{QR} = 100$ |
| Final QR Value | $\text{QR} = \frac{100}{4}$ | 25 cm |
Conclusion on Finding QR
Based on the property of similar triangles, the length of the corresponding side QR in ΔPQR is found to be 25 cm.
Revision Table: Similar Triangle Concepts
| Concept | Description | Property Example (ΔABC ∼ ΔPQR) |
|---|---|---|
| Similar Triangles | Triangles with equal corresponding angles and proportional corresponding sides. | ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R AND $\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{CA}}{\text{RP}}$ |
| Ratio of Areas | The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. | $\frac{\text{ar}(\Delta \text{ABC})}{\text{ar}(\Delta \text{PQR})} = \left(\frac{\text{BC}}{\text{QR}}\right)^2$ |
| Ratio of Perimeters | The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides. | $\frac{\text{Perimeter}(\Delta \text{ABC})}{\text{Perimeter}(\Delta \text{PQR})} = \frac{\text{BC}}{\text{QR}}$ |
Additional Information on Similar Triangles
Similar triangles are a fundamental concept in geometry with many applications. Understanding the relationship between their sides, angles, perimeters, and areas is crucial.
- Angle-Angle (AA) Similarity: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
- Side-Side-Side (SSS) Similarity: If the corresponding sides of two triangles are proportional, then the triangles are similar.
- Side-Angle-Side (SAS) Similarity: If two sides of a triangle are proportional to two sides of another triangle, and the included angles are congruent, then the triangles are similar.
- Similar triangles are used in scaling, mapping, and solving problems involving distances and heights indirectly (e.g., using shadows).
- The ratio of altitudes, medians, and angle bisectors of similar triangles is also equal to the ratio of their corresponding sides.
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