In a triangle PQR, point X is on PQ and point Y is on PR such that XP = 1.5 units, XQ = 6 units, PY = 2 units and YR = 8 units. Which of the following are correct? 1. QR = 5XY 2. QR is parallel to XY 3. Triangle PYX is similar to triangle PRQ Select the correct answer using the code given below.
1, 2 and 3
Understanding the Triangle Geometry Problem
The question describes a triangle PQR with specific points X on side PQ and Y on side PR. We are given the lengths of segments XP, XQ, PY, and YR. We need to determine the correctness of three statements regarding the relationship between triangle PYX and triangle PRQ, the parallelism of lines XY and QR, and the length relationship between XY and QR.
First, let's calculate the total lengths of sides PQ and PR using the given segment lengths:
- Length of PQ = XP + XQ = 1.5 + 6 = 7.5 units.
- Length of PR = PY + YR = 2 + 8 = 10 units.
Now, let's analyze each statement.
Statement 3: Triangle PYX is similar to triangle PRQ
To check for similarity between triangle PYX and triangle PRQ, we can use the Side-Angle-Side (SAS) similarity criterion. This criterion states that if two sides in one triangle are proportional to two sides in another triangle, and the included angle (the angle between the two sides) are congruent, then the triangles are similar.
Let's consider the sides around the common angle \(\angle P\):
- Side PX in triangle PYX and side PQ in triangle PRQ. The ratio is: \(\frac{PX}{PQ} = \frac{1.5}{7.5} = \frac{15}{75} = \frac{1}{5}\)
- Side PY in triangle PYX and side PR in triangle PRQ. The ratio is: \(\frac{PY}{PR} = \frac{2}{10} = \frac{1}{5}\)
We observe that the ratios of the corresponding sides adjacent to angle P are equal: \(\frac{PX}{PQ} = \frac{PY}{PR} = \frac{1}{5}\). Also, the angle \(\angle P\) is common to both triangles (it is the included angle). Therefore, by the SAS similarity criterion, triangle PYX is similar to triangle PRQ.
Statement 3 is correct.
Statement 2: QR is parallel to XY
Since we have established that triangle PYX is similar to triangle PRQ (\(\triangle PYX \sim \triangle PRQ\)), the corresponding angles of these similar triangles are equal.
Specifically, the corresponding angles are:
- \(\angle PXY = \angle PQR\) (corresponding angles)
- \(\angle PYX = \angle PRQ\) (corresponding angles)
- \(\angle P = \angle P\) (common angle)
Consider the line segments XY and QR intersected by the transversal PQ. The angles \(\angle PXY\) and \(\angle PQR\) are corresponding angles. Since \(\angle PXY = \angle PQR\), this indicates that the line segment XY is parallel to the line segment QR.
Alternatively, consider the line segments XY and QR intersected by the transversal PR. The angles \(\angle PYX\) and \(\angle PRQ\) are corresponding angles. Since \(\angle PYX = \angle PRQ\), this also indicates that the line segment XY is parallel to the line segment QR.
This is also a consequence of the Converse of the Basic Proportionality Theorem (BPT). If a line segment divides two sides of a triangle proportionally (as XY divides PQ and PR such that PX/PQ = PY/PR), then the line segment is parallel to the third side.
Statement 2 is correct.
Statement 1: QR = 5XY
Because triangle PYX is similar to triangle PRQ (\(\triangle PYX \sim \triangle PRQ\)), the ratio of their corresponding sides is constant. This constant ratio is the similarity ratio we found earlier:
\(\frac{PX}{PQ} = \frac{PY}{PR} = \frac{1}{5}\)
The third pair of corresponding sides are XY in triangle PYX and QR in triangle PRQ. Therefore, the ratio of these sides must also be equal to the similarity ratio:
\(\frac{XY}{QR} = \frac{1}{5}\)
We can rearrange this equation to find the relationship between QR and XY:
\(QR = 5 \times XY\)
So, the length of QR is 5 times the length of XY.
Statement 1 is correct.
Summary of Findings
Based on our analysis:
- Statement 1 (QR = 5XY) is correct.
- Statement 2 (QR is parallel to XY) is correct.
- Statement 3 (Triangle PYX is similar to triangle PRQ) is correct.
All three statements are correct.
| Statement | Analysis | Correctness |
|---|---|---|
| 1. QR = 5XY | Ratio of corresponding sides from similar triangles \(\triangle PYX \sim \triangle PRQ\) is 1:5. Thus, XY/QR = 1/5, leading to QR = 5XY. | Correct |
| 2. QR is parallel to XY | Similarity \(\triangle PYX \sim \triangle PRQ\) implies corresponding angles (\(\angle PXY\) & \(\angle PQR\), \(\angle PYX\) & \(\angle PRQ\)) are equal. Equal corresponding angles prove parallelism. | Correct |
| 3. Triangle PYX is similar to triangle PRQ | Using SAS criterion: PX/PQ = 1.5/7.5 = 1/5 and PY/PR = 2/10 = 1/5. Included angle \(\angle P\) is common. Proportional sides and equal included angle prove similarity. | Correct |
Revision Table: Key Geometric Concepts
| Concept | Definition/Criterion | Relevance to Problem |
|---|---|---|
| Triangle Similarity | Triangles are similar if their corresponding angles are equal and corresponding sides are proportional. Criteria include AA, SAS, SSS. | Used to prove \(\triangle PYX \sim \triangle PRQ\) using SAS. |
| SAS Similarity Criterion | If two sides of one triangle are proportional to two sides of another triangle, and the included angles are equal, the triangles are similar. | Directly applied to check if \(\triangle PYX \sim \triangle PRQ\). |
| Parallel Lines | Lines that never intersect. In geometry, parallelism can be shown by properties of angles formed by a transversal (e.g., equal corresponding angles, equal alternate interior angles, supplementary consecutive interior angles). | Proved XY || QR using the equality of corresponding angles derived from triangle similarity. |
| Basic Proportionality Theorem (BPT) / Thales's Theorem | If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. (Converse is also true: If a line divides two sides in the same ratio, it is parallel to the third side). | The converse supports the finding that XY || QR based on PX/XQ = PY/YR (which implies PX/PQ = PY/PR). |
Additional Information: Proportional Sides and Parallelism
In this problem, the points X and Y divide the sides PQ and PR such that:
- \(\frac{XP}{XQ} = \frac{1.5}{6} = \frac{15}{60} = \frac{1}{4}\)
- \(\frac{PY}{YR} = \frac{2}{8} = \frac{1}{4}\)
Since \(\frac{XP}{XQ} = \frac{PY}{YR} = \frac{1}{4}\), by the converse of the Basic Proportionality Theorem (BPT), the line segment XY is parallel to QR. This confirms Statement 2.
Furthermore, when a line segment like XY is parallel to the base QR of triangle PQR, it creates a smaller triangle PYX that is similar to the larger triangle PRQ. This is because the parallel lines ensure that \(\angle PXY = \angle PQR\) and \(\angle PYX = \angle PRQ\) (corresponding angles), and \(\angle P\) is common. This confirms Statement 3 by the Angle-Angle (AA) similarity criterion.
For similar triangles, the ratio of corresponding sides is constant. The ratio of the segments on the sides PX/PQ = PY/PR. Since XP = 1.5 and XQ = 6, PQ = 1.5 + 6 = 7.5. So PX/PQ = 1.5/7.5 = 1/5. Since PY = 2 and YR = 8, PR = 2 + 8 = 10. So PY/PR = 2/10 = 1/5. The ratio of the bases must also be the same: XY/QR = 1/5. This gives QR = 5 * XY, confirming Statement 1.
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