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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