Question

Please help me 7th march is my maths exam solve this question

In an equilateral triangle ABC , D is a point on side BC, such that BD =1/3 BC. Prove that 9AD square = 7 AB square.

Answer 1

Answer:

                                                                                                                               

Given = BD = 1/3 BC

To prove = 9AD^2 = 7AB^2

CONSTRUCTION = We draw AE ⊥ BC

Proof :- Let the sides of the equilateral triangle be a and AE be the altitude of

Δ ABC

∴ BE = EC = BC/2 = a/2

And AE = a √3/2

ALSO,

BD = 1/3BC

∴ DE = BE-BD

⇒ a/2 -a/3 = a/6

BY APPLYING PYTHAGORAS THEOREM

AD^2 = AE^2+DC^2

AD^2 = (a √3/2)^2+ (a/6)

AD^2 = (3a^2/4) + (a/36)

AD^2 =  28a^2/36

AD^2 = 7AB^2/9

9AD^2 = 7AB^2

                                                                                                                               

HOPE IT HELPS YOU

Answer 2

Diagram in attachment .

Given that ABC is an equilateral triangle .

All sides are equal .

Construct AE .

∠AED = ∠AEC = 90° .

In Δ AEB and Δ AEC ,

AB = AC

AE = AE

∠AEB = ∠AEC

Δ AEB ≅ Δ AEC [ RHS ]

⇒ EB = EC [ cpct ]

By Pythagoras theorem :

In Δ AEB ,

AE² = AB² - EB²

⇒ AE² = AB² - ( BC/2 )²

⇒ AE² = AB² - ( AB/2 )² [ ∵ AB = BC ]

⇒ AE² = AB² - AB²/4

⇒ AE² = 3 AB²/4

By Pythagoras theorem :

In Δ AED :

AD² = AE² + ED²

⇒ AD² = 3 AB²/4 + ED²

ED = BC - 1/2 BC - 1/3 BC [ ∵ BE = EC ]

⇒ ED = 1/2 BC - 1/3 BC

⇒ ED = 1/6 BC

⇒ AD² = 3 AB²/4 + ( BC/6 )²

⇒ AD² = 3 AB²/4 + ( AB/6 )² [ ∵ AB = BC ]

⇒ AD² = 3 AB²/4 + AB²/36

⇒ AD² = ( 27 AB² + AB² ) / 36

⇒ 36 AD² = 28 AB²

⇒ 9 AD² =7 AB² .

Hence proved !