Question 91
ABC is a right angled triangle in which ∠B = 90°. If BD ⊥ AC, AB = 3 cm and BC = 4 cm, then what is the value of BD (in cm)?
Solution
In $$\triangle$$ ABC,
=> $$(AC)^2=(AB)^2+(BC)^2$$
=> $$(AC)^2=(3)^2+(4)^2$$
=> $$(AC)^2=9+16=25$$
=> $$AC=\sqrt{25}=5$$ cm
Now, area of $$\triangle$$ ABC = $$\frac{1}{2}\times(AB)(BC)$$ = $$\frac{1}{2}\times(AC)(BD)$$
=> $$BD = \frac{3\times4}{5}$$
=> $$BD=\frac{12}{5}$$ cm
=> Ans - (A)
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