
To solve the problem, we need to find the length of segment $AQ$ using the concept of similar triangles formed by a line parallel to one side of a triangle.
1. Understanding the Geometry:
In the given triangle, $PQ \parallel BC$, so triangles $APQ$ and $ABC$ are similar by Basic Proportionality Theorem (Thales' Theorem).
2. Given Information:
$AP = 3 \, \text{cm}$
$BP = 2 \, \text{cm}$
$CQ = 3 \, \text{cm}$
3. Use of Similar Triangles:
Since $PQ \parallel BC$, we write the similarity condition:
$ \frac{AP}{PB} = \frac{AQ}{QC} $
4. Substituting the Known Values:
$ \frac{3}{2} = \frac{AQ}{3} $
5. Cross Multiplying to Solve for $AQ$:
$ 3 \times 3 = 2 \times AQ $
$ 9 = 2AQ \Rightarrow AQ = \frac{9}{2} = 4.5 \, \text{cm} $
Final Answer:
The length of $AQ$ is $ 4.5 \, \text{cm} $.
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In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.