Question

⭐If A. B and P are three collinear points, B lying between A and P, such that AB= 6cm,and AB * AP=BP^2.Prove that AB^2 +AP^2=3BP^2,and find the length of BP.⭐

CLASS 10 CHAPTER :QUADRATIC EQUATIONS

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Answer 1

Given ,

A , B and P are collinear points .

B lying between A and P ,

such that AB = 6 cm ,

AB × AP = BP² ---( 1 )

To prove : AB² + AP² = 3BP²

proof :

AP = AB + BP

=> AP - AB = BP

Do the square both sides of the

equation , we get

( AP - AB )² = BP²

=> AP² + AB² - 2 ( AP ) (AB ) = BP²

=> AP² + AB² - 2 × BP² = BP² [ from(1)]

=> AP² + AB² = BP² + 2BP²

=> AP² + AB² = 3 BP² --( 2 )

Now ,

ii ) AP = AB + BP

=> AP = 6 + BP

multiply each term with

6 , we get

=> 6AP = 36 + 6BP

=> BP² = 36 + 6BP

[ Since ,

AP × AB = BP²

=> AP × 6 = BP²

=> 6AP = BP² ]

=> BP² - 6BP - 36 = 0

Let BP = x ,

∆ = 6² - 4 × 1 × ( -36 )

= 36 + 144

= 180

By Quadratic Formula ,

x = [ ( - b ± √∆ )/2a

=> x = [ -(-6 )±√180 ]/2

=> x = ( 6 ± 6√5 )/2

=> x = 3 ± 3√5

Therefore ,

x = BP = 3 + 3√5


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Answer 2

Answer:

Correct option is

B

BA+AC=BC

According to figure,

BA, AC & BC  and  BA+AC=BC