Question

Plz solve ASAP
Maths
Class 10

Answer 1

Let's

=> AB = a

=> AD = h

=> AC = b

=> BC = l
__________

now, 

» Area of triangle = 1/2 ×base×height

=>  area = A = 1/2 × a × b

=> b = 2A / a _____[Eq(1)]

» Again it can also be written as

=> Area  = A = 1/2 × h × l

=> l = 2A / h _____[Eq(2)]

_______

Now in ΔABC,

=> a² + b² = l²

» substituting values of a and b from (1) and (2) in this equation we get,

 => a2²+ (2A/a)² = (2A/h)²

=> a² + 4A² / a² = 4A² / h²

=> (a^4 + 4A²) / a² = 4A² / h²

=> h² = 4 A² a² / (a^4 + 4A²)

=> h = 2×A×a / √(a^4 + 4A²)
______________[ANSWER]

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_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Heya!
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♦Given that =>
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◾A is the Area of Right Triangle .

◾AB = 'a '

◾AD = 'h' < Height >

◾Let AC = 'b '

◾Let BC = 'c '



♦Area of a ∆ = ½ × Base ( B ) × Height ( H )

=>Thus , Area of ∆ =

=> ½ × a × b

=> A = ab/2


◾b = 2A/ a ...................( 1 )

✴Also ,

◾c = 2A / h ..................( 2 )


♦Applying Pythagoras theorem to ∆ABC
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=> ( H ) ² = ( P )² + ( B ) ²

=> a² + b² = c²


◾Putting values of b and c from Equation ( 1 ) and ( 2)
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=> a² + ( 2A / a ) ² = ( 2A / h ) ²

=> a² + 4A ² / a² = 4A² / h²

=> a⁴ + 4A² / a² = 4A² / h²



◀From the Equation ,
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H² => a²• 4A² / a⁴ + 4A²

H => 2A•a / √a⁴ + 4A²


✴Hence Proved ✴
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