Question

In a ∆ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find the AB and PQ.

Answer 1

Hint- Proceed the solution of this question, first with the help of basic proportionality theorem find the unknown side that we can easily so that we can prove two corresponding triangles similar. Hence once we show a triangle similar then easily with the help of properties of a similar triangle we can find the desired sides of the triangle.

Complete step-by-step solution -

In the question it is given that Δ ABC, AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm. Also, PQ ∥ BC. And we were required to find: AB and PQ.

By using Thales Theorem, which is a Basic proportionality theorem,
we have [As it’s given that PQ ∥ BC]

\[ \Rightarrow \dfrac{{{\text{AP}}}}{{{\text{PB}}}} = \dfrac{{{\text{AQ}}}}{{{\text{QC}}}}\]
\[ \Rightarrow \dfrac{{2.4}}{{{\text{PB}}}} = \dfrac{2}{3}\]
On cross multiplication
\[ \Rightarrow {\text{PB}} = \dfrac{{2.4 \times 3}}{2} = 3.6{\text{ cm}}\]
Now finding, AB = AP + PB
AB = 2.4 + 3.6 ⇒ AB = 6 cm
Now, considering Δ APQ and Δ ABC
We have,
⇒∠A = ∠A (common)
⇒ ∠APQ = ∠ABC (Corresponding angles are equal, PQ||BC and AB being a transversal)
We know that the mathematical definition for similar triangles states that the triangles (Δ APQ and Δ ABC) have proportional corresponding sides and all the corresponding angles are the same. The Angle-Angle (AA) criterion tells us that two triangles are similar if two corresponding angles are equal to each other.

Thus, Δ APQ and Δ ABC are similar to each other by AA criteria.
Now, we know that Corresponding parts of similar triangles are proportional.
\[ \Rightarrow \dfrac{{{\text{AP}}}}{{{\text{AB}}}} = \dfrac{{{\text{PQ}}}}{{{\text{BC}}}}\]
On cross multiplication

\[ \Rightarrow {\text{PQ}} = \dfrac{{{\text{AP}} \times {\text{BC}}}}{{{\text{AB}}}}\]
\[ \Rightarrow {\text{PQ}} = \dfrac{{2.4 \times 6}}{6}\]
\[\therefore {\text{PQ = }}2.4{\text{ cm}}\]

Note- In this particular system we should know that Thales theorem states that
If two lines AB and CD as shown in below figure. When AC and BD intersect at a point E, and when E is either (1) a point interior to both segment AC and segment BD or else (2) exterior to both segments, then the triangles AEB and CED (1) share the angle ∠AEB = angle ∠CED or else (2) these two angles are vertical angles. Then for such a figure as shown below

Line AB is parallel to line CD if and only if
Triangle AEB is similar to triangle CED if and only if
$\dfrac{{{\text{EC}}}}{{{\text{EA}}}} = \dfrac{{{\text{ED}}}}{{{\text{EB}}}}$