Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \).
Step 1: Understanding the Concept:
The Basic Proportionality Theorem (BPT), also known as Thales's Theorem, states that 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.
Step 2: Key Formula or Approach:
According to the Basic Proportionality Theorem, since DE \(||\) BC, we have:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 3: Detailed Explanation:
Given:
\[\begin{array}{rl} \bullet & \text{In \(\triangle\) ABC, DE \(||\) BC.} \\ \bullet & \text{AD = 1.8 cm} \\ \bullet & \text{DB = 5.4 cm} \\ \bullet & \text{EC = 7.2 cm} \\ \end{array}\]
We need to find the length of AE.
Using the BPT, we set up the proportion:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Substitute the given values into the equation:
\[ \frac{1.8}{5.4} = \frac{AE}{7.2} \]
First, simplify the ratio on the left side:
\[ \frac{1.8}{5.4} = \frac{18}{54} = \frac{1}{3} \]
Now the equation becomes:
\[ \frac{1}{3} = \frac{AE}{7.2} \]
To solve for AE, multiply both sides by 7.2:
\[ AE = \frac{1}{3} \times 7.2 \]
\[ AE = 2.4 \text{ cm} \]
Step 4: Final Answer:
The length of AE is 2.4 cm.
In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ.
In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Activity:
2AX = 3BX (Given)
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]
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