Question

In the figure, angle BAC= 90°, AD is perpendicular to BC, B-D-C, BD= 12 and DC= 3 then the length of AD is​

Answer 1

Solution :-

since ∠ BAC = 90° , ∆BAC is a right angled ∆ .

also, given that,

→ BD = 12

→ DC = 3

→ AD ⟂ BC .

so,

→ AD² = BD * DC { By Altitude on hypotenuse theorem. }

→ AD² = 12 * 3

→ AD² = 36

→ AD = √36

→ AD = 6 (Ans.)

Hence, length of AD is 6 unit .

Extra :- Proof of Altitude on hypotenuse theorem :-

In ∆ABC and ∆DAC we have,

→ ∠BAC = ∠ADC (90°)

→ ∠BCA = ∠ACD (common)

so,

→ ∆ABC ~ ∆DAC (By AA similarity.) ------ Eqn.(1)

similarly, In ∆ABC and ∆DBA we have,

→ ∠BAC = ∠BDA (90°)

→ ∠CBA = ∠ABD (common)

so,

→ ∆ABC ~ ∆DBA (By AA similarity.) ---- Eqn.(2)

from Eqn.(1) and Eqn.(2),

→ ∆DAC ~ ∆DBA

therefore,

→ DA/DC = DB/DA

→ DA * DA = DC * DB

→ DA² = DC * DB

→ AD² = BD * CD (Proved)

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Answer 2

Given :   in ΔABC, ∠BAC = 90°, seg AD ⊥ seg BC and B-D-C,

BD = 12, DC = 3,

To find :  length of AD.

Solution:

Compare ΔADB  and ΔCDA

∠ADB = ∠CDA = 90°

∠ABD = ∠CAD =  90° -∠C

=> ΔADB  ≈ ΔCDA   ( Using AA)

Ratio of corresponding sides of similar triangles is Equal.

AD/CD  = BD/AD

=> AD² = BD * CD

=> AD² =  12 * 3

=>   AD² =  36

=> AD = 6  

Length of AD = 6 cm

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