In the given figure- - ABC is right- angled at C- Let BC - a - CA - b - AB - c and CD - p- where CD 1- p 2-1- a 2-1- b 2

(i) Area of ΔABC = 1/2AB × CD = 1/2cp. Also, area of ΔABC = 1/2BC × AC = 1/2ab. ∴1/2cp = 1/2ab ⇒ cp = ab (ii) cp = ab ⇒ p = ab/c ⇒ p^2 = a^2b^2/c^2 ⇒ 1/p^ ...

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Standard X

Mathematics

Addition of Rational Expressions

In the given ...

Question

In the given figure, ΔABC is right- angled at C. Let BC = a, CA = b, AB = c and CD = p, where CD $\frac{1}{p^{2}}$ = $\frac{1}{a^{2}}$ + $\frac{1}{b^{2}}$

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Solution

(i) Area of ΔABC = $\frac{1}{2}$ AB × CD = $\frac{1}{2}$ cp.

Also, area of ΔABC = $\frac{1}{2}$ BC × AC = $\frac{1}{2}$ ab.

∴ $\frac{1}{2}$ cp = $\frac{1}{2}$ ab ⇒ cp = ab

(ii) cp = ab ⇒ p = $\frac{a b}{c}$

⇒ $p^{2}$ = $\frac{a^{2} b^{2}}{c^{2}}$

⇒ $\frac{1}{p^{2}}$ - $\frac{c^{2}}{b^{2} b^{2}}$ - $\frac{a^{2} + b^{2}}{a^{2} b^{2}}$ [ ∵ $c^{2}$ = $a^{2}$ + $b^{2}$ ]

⇒ $\frac{1}{p^{2}}$ - $\frac{1}{a^{2}}$ + $\frac{1}{b^{2}}$

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In the given figure, ΔABC is right- angled at C. Let BC = a, CA = b, AB = c and CD = p, where CD 1p2 = 1a2 + 1b2

(i) Area of ΔABC = 12AB × CD = 12cp. Also, area of ΔABC = 12BC × AC = 12ab. ∴12cp = 12ab ⇒ cp = ab (ii) cp = ab ⇒ p = abc ⇒ p2 = a2b2c2 ⇒ 1p2 - c2b2b2 - a2+b2a2b2 [ ∵ c2 = a2 + b2] ⇒1p2 - 1a2 + 1b2

In the given figure, ΔABC is right- angled at C. Let BC = a, CA = b, AB = c and CD = p, where CD $\frac{1}{p^{2}}$ = $\frac{1}{a^{2}}$ + $\frac{1}{b^{2}}$