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Case Study – 2
On a road leading to a school, there is a triangle (ABC) shaped board on the road. It is divided into two parts by a line DE, which is parallel to BC. On the upper part, it is written ‘DRIVE SLOW' and on the lower part it is written, ‘SCHOOL AHEAD'.
Based on the information given above, answer the following questions :
(i) If AD = $3$ cm, BD = $5$ cm and AE = $4$ cm, then find the length of AC.
(ii) If $\angle$ ADE = $50^{\circ}$ and $\angle$ DAE = $45^{\circ}$, then find $\angle$ ACB.
(iii) (a) If AD = $4$ cm and BD = $6$ cm, then find $\frac{DE}{BC}$.
OR
(iii) (b) If AD = $3$ cm, BD = $6$ cm and AE = $5$ cm, then find $\frac{AB}{AC}$.
On a road leading to a school, there is a triangle (ABC) shaped board on the road. It is divided into two parts by a line DE, which is parallel to BC. On the upper part, it is written ‘DRIVE SLOW' and on the lower part it is written, ‘SCHOOL AHEAD'.
Based on the information given above, answer the following questions :
(i) If AD = $3$ cm, BD = $5$ cm and AE = $4$ cm, then find the length of AC.
(ii) If $\angle$ ADE = $50^{\circ}$ and $\angle$ DAE = $45^{\circ}$, then find $\angle$ ACB.
(iii) (a) If AD = $4$ cm and BD = $6$ cm, then find $\frac{DE}{BC}$.
OR
(iii) (b) If AD = $3$ cm, BD = $6$ cm and AE = $5$ cm, then find $\frac{AB}{AC}$.
Show SolutionHide Solution↓
(i) $\frac{3}{3+5} = \frac{4}{AC}$
$\Rightarrow AC = \frac{32}{3}$ cm
(ii) $\angle ACB = \angle AED = 180^{\circ} – (50^{\circ} + 45^{\circ}) = 85^{\circ}$
(iii) (a) $\triangle ADE \sim \triangle ABC$
$\therefore \frac{DE}{BC} = \frac{AD}{AB} = \frac{4}{10}$ or $\frac{2}{5}$
OR
(b) DE $||$ BC
$\therefore \frac{3}{6} = \frac{5}{EC} \Rightarrow EC = 10$
$\frac{AB}{AC} = \frac{3+6}{5+10} = \frac{9}{15}$ or $\frac{3}{5}$
$\Rightarrow AC = \frac{32}{3}$ cm
(ii) $\angle ACB = \angle AED = 180^{\circ} – (50^{\circ} + 45^{\circ}) = 85^{\circ}$
(iii) (a) $\triangle ADE \sim \triangle ABC$
$\therefore \frac{DE}{BC} = \frac{AD}{AB} = \frac{4}{10}$ or $\frac{2}{5}$
OR
(b) DE $||$ BC
$\therefore \frac{3}{6} = \frac{5}{EC} \Rightarrow EC = 10$
$\frac{AB}{AC} = \frac{3+6}{5+10} = \frac{9}{15}$ or $\frac{3}{5}$