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
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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