Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 18 cm
Option 3: 4 cm
Option 4: 6 cm
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Correct Answer: 18 cm
Solution : According to the question, Given: AC = 9 cm Area of $\triangle$ABC = 40 cm 2 Area of $\triangle$ADC = 10 cm 2 $\triangle$BAC ∼ $\triangle$ADC ⇒ $\frac{\text{Area of }\triangle ABC}{\text{Area of } \triangle ADC}=\frac{AB^{2}}{AD^{2}}=\frac{BC^{2}}{AC^{2}}$ (In similar triangles, the ratio of their area is the square of the ratio of corresponding sides.) ⇒ $\frac{40}{10}=\frac{BC^{2}}{(9)^{2}}$ ⇒ $\frac{40}{10}\times 81=BC^{2}$ ⇒ $BC=18$ cm Hence, the correct answer is 18 cm.
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