Question

In triangle ABC,angleA=90degree,AD perpendicular to BC and angleB=45degree,AB=x,express the value of AD in terms of x.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

in ∆ ABC,

angleC = 180°-(angleA+angleB)

= 180°-(90°+45°)

= 180°- 135°

= 45°

hence ,

angleB = angleC

so, ∆ABC is an isosceles triangle.

so, AB = AC = x

BC² = AB²+AC²

or, BC² = x²+x²

or, BC = √2x²

= x√2

now,

in ∆ ABD & ∆ACD,

angleADB = angle ADC = 90°. (given)

AB = AC. (isosceles∆ property)

AD = AD. (common)

so, ∆ABD = ∆ACD. ( by RHS rule)

so, BD = DC ( by CPCT)

now,

BD = 1/2×√2

= 1/(√2×√2)×√2

= √2/(√2×√2)

= 1/√2

now,

in ∆ ABD,

AB²= BD²+AD² (Pythagoras theorum)

(x)²= (1/√2)²+AD²

AD² = x²-(1/2)

AD = √{x²-(1/2)}

AD = √{(2x²-1)/2}

ans.

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