Math, asked by anishprofessional, 9 months ago

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Answered by brainlyaryan12

4

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✪ $\huge{\red{\underline{\overline{\mathbf{Question}}}}}$ ✪

→ In a ∆ABC , $\angle$ B=90° and D is the mid point of BC. Prove that :

(i) AC² = AD² + 3CD²

(ii) BC² = 4( AD² - AB² )

✪ $\huge{\green{\underline{\overline{\mathbf{Answer}}}}}$ ✪

⇒Given:

⇒ $\angle$ B=90°
⇒BD = CD
⇒ $BC=2BD=2DC$

⇒To Prove:

⇒(i) AC² = AD² + 3CD²
(ii) BC² = 4( AD² - AB² )

$\huge{\pink{\overbrace{\underbrace{\red{Solution:-}}}}}$

⭐Solving For 1st-

⇒ $AC^2=AB^2+BC^2$

⇒ $AC^2=AD^2-BD^2+BC^2$ $\small{[AB^2=AD^2-BD^2]}$

⇒ $AC^2=AD^2-BD^2+(2BD)^2$ [BC=2BD]

⇒ $AC^2=AD^2-BD^2+4BD^2$

⇒ $AC^2=AD^2+3BD^2$

⇒ $AC^2=AD^2+3CD^2$ [BD=CD]

$\large{\orange{\fbox{\blue{Hence\;Proved}}}}$

⭐Solving for 2nd -

⇒ $AD^2=AB^2+BD^2$

⇒ $BD^2=AD^2-AB^2$

⇒ $\big(\frac{BC}{2}\big)^2=AD^2-AB^2$ [ $BD=\frac{BC}{2}$ ]

⇒ $\frac{BC^2}{4}=AD^2-AB^2$

⇒ $BC^2=4(AD^2-AB^2)$

$\large{\orange{\fbox{\blue{Hence\;Proved}}}}$

≿━━━━━━━━━༺❀༻━━━━━━━━━≾

Formulas Used :-

Pythagoras Theorem-

⇒ $H^2=P^2+B^2$

≿━━━━━━━━━༺❀༻━━━━━━━━━≾

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Answered by rajaryan25dec

2

Answer with Step-by-step explanation:

Given: ∠B = 90ᵒ

D is midpoint of BC, BD = CD = 1/2BC

To Prove: (i) AC² = AD² + 3CD²

(ii) BC² = 4(AD² - AB²)

Proof: (i) In ▲ABD, ∠B = 90ᵒ

By using Pythagoras theorem,

H² = B² + P²

AD² = BD² + AB² ----------------(1)

In ▲ABC, ∠B = 90ᵒ

By using Pythagoras Theorem

AC² = BC² + AB²

AC² = (2CD)² + AB²

AC² = 4CD² + AB² ---------------(2)

Now, by subtracting (1) from (2)

AC² - AD² = 4CD² + AB² - (BD² + AB²)

AC² - AD² = 4CD² + AB² - CD² + AB² {BC = CD}

AC² - AD² = 3CD²

AC² = AD² + 3CD²

Hence Proved.

(ii) From equation (1)

AD² = BD² + AB²

AD² = (1/2 BC)² + AB² {Given: BD = CD = 1/2BC}

AD² = 1/4 BC² + AB²

AD² - AB² = 1/4 BC²

1/4 BC² = AD² - AB²

BC² = 4(AD² - AB²)

Hence Proved.

Hope it helps.

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