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In a triangle ABC angel B=45 …

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In a triangle ABC angel B=45 .Prove that AC^2=AB^2+BC^2-4AREA(triangle ABC).

Posted by Nithin Goud 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Construction: Draw AD {tex}\perp{/tex} BC
Proof: In right {tex}\triangle{/tex}ADC,
By using pythagoras theorem, we get
AC² = AD² + CD² ...(i)
In right {tex}\triangle{/tex}ADB,
By using pythagoras theorem, we get
AB² = AD² + BD²
{tex}\Rightarrow{/tex}AB²- BD² = AD²
Putting in (i), we get
AC² = AB²- BD² + CD²
{tex}\Rightarrow{/tex} AC² = AB² + CD²- BD²
= AB²+ (CD - BD)(CD + BD) = AB² + (CD - BD) BC
= AB² + (BC - BD - BD).BC = AB² + BC²- 2BC.BD
= AB² + BC²- 2.AD.BC [ {tex}\because{/tex} AD = BD]
= AB² + BC²- 4ar({tex}\triangle{/tex}ABC).

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CBSE > Class 10 > Mathematics