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Home » In the given figure, ∠BAC = and AD⊥BC. Then, (a) BC.CD = (b) AB.AC = (c) BD.CD = (d) AB.AC =
In the given figure, ∠BAC = {90^0} and AD⊥BC. Then, (a) BC.CD = B{C^2} (b) AB.AC = B{C^2} (c) BD.CD = A{D^2} (d) AB.AC = A{D^2}
CBSE | Class 10 | Exercise MCQ | Maths | RS Aggarwal | Triangles
In the given figure, ∠BAC = {90^0} and AD⊥BC. Then, (a) BC.CD = B{C^2} (b) AB.AC = B{C^2} (c) BD.CD = A{D^2} (d) AB.AC = A{D^2}

Correct Answer: (c) BD.CD = A{D^2}

Explanation:

 

 

 

In ∆ BDA and ∆ADC,

∠???????????? = ∠???????????? = 900

∠???????????? = 900 − ∠????????????

=> 900 − (900 − ∠????????????)

=> 900 − 900 + ∠????????????

=> ∠????????????

Applying AA similarity,

∆???????????? − ∆????????????.

\begin{array}{l} \frac{{BD}}{{AD}} = \frac{{AD}}{{CD}}\\ \end{array}

????????2 = ????????. ????????

 

i

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