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A triangular window of a building is shown above. Its diagram represents a $\triangle ABC$ with $\angle A = 90^\circ$ and $AB = AC$. Points P and R trisect AB and $PQ || RS || AC$.
Based on the above, answer the following questions :
(i) Show that $\triangle BPQ \sim \triangle BAC$.
(ii) Prove that $PQ = \frac{1}{3} AC$.
(iii) (a) If $AB = 3 \text{ m}$, find length BQ and BS. Verify that $BQ = \frac{1}{2} BS$.
OR
(iii) (b) Prove that $BR^2 + RS^2 = \frac{4}{9} BC^2$.
Based on the above, answer the following questions :
(i) Show that $\triangle BPQ \sim \triangle BAC$.
(ii) Prove that $PQ = \frac{1}{3} AC$.
(iii) (a) If $AB = 3 \text{ m}$, find length BQ and BS. Verify that $BQ = \frac{1}{2} BS$.
OR
(iii) (b) Prove that $BR^2 + RS^2 = \frac{4}{9} BC^2$.
Show SolutionHide Solution↓
(i) In $\triangle BAC$ and $\triangle BPQ, PQ || AC \Rightarrow \angle BQP = \angle BCA$ and $\angle B$ is common $\therefore \triangle BPQ \sim \triangle BAC$ (1 mark)
(ii) Since, $\triangle BPQ \sim \triangle BAC \Rightarrow \frac{PQ}{AC} = \frac{BP}{BA} = \frac{1}{3} \Rightarrow PQ = \frac{1}{3} AC$ ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) $\frac{BP}{BA} = \frac{PQ}{AC}$ (corresponding sides of similar triangles) $\Rightarrow BP = PQ$ (as $BA = AC$) $\therefore PQ = \frac{1}{3} \times 3 = 1 \text{ m}$. Hence, $BQ = \sqrt{2} \text{ m}$ (1 mark)
getting $BS = 2\sqrt{2} \text{ m} \Rightarrow \frac{1}{2} BS = BQ$ (Hence verified) ($\frac{1}{2} + \frac{1}{2}$ marks)
OR
(iii) (b) $BR^2 + RS^2 = (\frac{2}{3} AB)^2 + (\frac{2}{3} AC)^2 = \frac{4}{9} (AB^2 + AC^2) = \frac{4}{9} BC^2$ ($\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ marks)
(ii) Since, $\triangle BPQ \sim \triangle BAC \Rightarrow \frac{PQ}{AC} = \frac{BP}{BA} = \frac{1}{3} \Rightarrow PQ = \frac{1}{3} AC$ ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) $\frac{BP}{BA} = \frac{PQ}{AC}$ (corresponding sides of similar triangles) $\Rightarrow BP = PQ$ (as $BA = AC$) $\therefore PQ = \frac{1}{3} \times 3 = 1 \text{ m}$. Hence, $BQ = \sqrt{2} \text{ m}$ (1 mark)
getting $BS = 2\sqrt{2} \text{ m} \Rightarrow \frac{1}{2} BS = BQ$ (Hence verified) ($\frac{1}{2} + \frac{1}{2}$ marks)
OR
(iii) (b) $BR^2 + RS^2 = (\frac{2}{3} AB)^2 + (\frac{2}{3} AC)^2 = \frac{4}{9} (AB^2 + AC^2) = \frac{4}{9} BC^2$ ($\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ marks)