Triangles Question - CBSE_NCERT_EXEMPLAR_CH06

Question

In $\Delta ABC$, $P$ and $Q$ are points on $AB$ and $AC$ respectively such that $PQ \parallel BC$. If $AP = 3	ext{ cm}, PB = 6	ext{ cm}, AQ = 4	ext{ cm}$ and $QC = 8	ext{ cm}$, prove that $\Delta ABC \sim \Delta APQ$ and find $PQ$ if $BC = 12	ext{ cm}$.

Accepted Answer

Stepwise Solution:

$AB = AP + PB = 3 + 6 = 9 ext{ cm}$. $AC = AQ + QC = 4 + 8 = 12 ext{ cm}$. [0.5 Mark]

$\dfrac{AP}{AB} = \dfrac{3}{9} = \dfrac{1}{3}$ and $\dfrac{AQ}{AC} = \dfrac{4}{12} = \dfrac{1}{3}$. Also $\angle A = \angle A$ (common). [1.0 Mark]

By SAS similarity criterion, $\Delta APQ \sim \Delta ABC$. [0.5 Mark]

$\dfrac{PQ}{BC} = \dfrac{AP}{AB} \Rightarrow \dfrac{PQ}{12} = \dfrac{1}{3} \Rightarrow PQ = 4 ext{ cm}$. [1.0 Mark]

Marking Scheme:

• Calculating full side lengths $AB=9, AC=12$: 0.5 Mark
• Proving SAS similarity $\Delta APQ \sim \Delta ABC$: 1.5 Marks
• Calculating $PQ = 4 ext{ cm}$: 1.0 Mark

Question:

Step-by-Step Solution

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