Question 6

Let $$\triangle ABC$$ be a right-angled triangle, $$\angle ABC = 90^\circ$$, $$AB = 5$$ cm, $$BC = 12$$ cm. Squares ABPD and BCEQ are drawn in the exterior of the $$\triangle ABC$$. DR is drawn perpendicular to line AC at R and ES is perpendicular to the line AC at S. Let $$DR = h_1$$ and $$ES = h_2$$, then $$\frac{h_1 + h_2}{AC} =$$

Solution

Take $$B = (0,0)$$, $$A = (0,5)$$, $$C = (12,0)$$, so $$AC = 13$$ and the line AC is $$5x + 12y - 60 = 0$$. The exterior squares put $$D = (-5,5)$$ and $$E = (12,-12)$$, so $$h_1 = \frac{|-25 + 60 - 60|}{13} = \frac{25}{13}$$ and $$h_2 = \frac{|60 - 144 - 60|}{13} = \frac{144}{13}$$. Their sum is $$\frac{169}{13} = 13$$, so $$\frac{h_1 + h_2}{AC} = \frac{13}{13} = 1$$.