In the given figure, if PX-5 cm, XR = 3 cm, QR-7.2 cm and XY || PQ, then the length of RY is 
We are tasked with finding the length of \( RY \) in the given figure, where \( XY \parallel PQ \), \( PX = 5 \, \text{cm} \), \( XR = 3 \, \text{cm} \), and \( QR = 7.2 \, \text{cm} \).
Step 1: Use the Basic Proportionality Theorem (BPT).
The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Here, \( XY \parallel PQ \), so:
\[ \frac{PX}{XR} = \frac{QY}{RY}. \]
Step 2: Substitute the known values.
From the problem, \( PX = 5 \, \text{cm} \), \( XR = 3 \, \text{cm} \), and \( QR = QY + RY = 7.2 \, \text{cm} \). Let \( RY = x \), so \( QY = 7.2 - x \). Substituting into the proportion:
\[ \frac{PX}{XR} = \frac{QY}{RY} \implies \frac{5}{3} = \frac{7.2 - x}{x}. \]
Step 3: Solve for \( x \).
Cross-multiply to eliminate the fractions:
\[ 5x = 3(7.2 - x). \]
Expand the terms:
\[ 5x = 21.6 - 3x. \]
Rearrange to isolate \( x \):
\[ 5x + 3x = 21.6 \implies 8x = 21.6 \implies x = \frac{21.6}{8} = 2.7 \, \text{cm}. \]
Final Answer: The length of \( RY \) is \( \mathbf{2.7 \, \text{cm}} \), which corresponds to option \( \mathbf{(1)} \).
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.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).