State true or false:

In $△ A B C$ , $A B = B C = C A = 2 a$ and $A D$ is perpendicular to side $B C$ , then $A D = 2 a \sqrt{3}$

True

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False

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Solution

The correct option is B False
In $△$ s $A B D$ and $A C D$ ,
$∠ A D B = ∠ A D C$ (Both $90^{o}$ )
$A D = A D$ (Common)
$A B = A C$ (Given)
Hence, $△ A B D \sim △ A C D$ (SSA rule)
Hence, $B D = C D$ (By cpct)
Now, in triangle ABD,
$A B^{2} = A D^{2} + B D^{2}$ (From Pythagoras theorem)
$(2 a)^{2} = A D^{2} + a^{2}$
$A D^{2} = 3 a^{2}$
$A D = a \sqrt{3}$

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Similar questions

△ A B C

A B = B C = C A = 2 a

△ A B C = a^{2} \sqrt{2}

△ A B C

A D

B C

A C^{2} = A B^{2} + B C^{2} - 2 B C \times B D

A B C D E

A B = A E, B C = E D

∠ A B C = ∠ A E D .

A C = A D

Δ A B C, i f A B = B C = C A = 2 a

A D ⊥ B C t h e n

AD = a \sqrt{3}

A r e a (∆ A B C) = \sqrt{3} a^{2}

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