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Answer 1

77. From the Figure : We can notice that AED is a Right Angled Triangle

⇒ h² + x ² = p²

⇒ h² = p² - x²

From the Figure : We can Notice that AEC is a Right Angled Triangle

⇒ $h^2 + (x + \frac{a}{2})^2 = b^2$

⇒ $h^2 = b^2 - (x + \frac{a}{2})^2$

Equating Both h² Equations, We get :

⇒ $p^2 - x^2 = b^2 - (x + \frac{a}{2})^2$

⇒ $p^2 - x^2 = b^2 - (x^2 + \frac{a^2}{4} + ax)$

⇒ $p^2 - x^2 = b^2 - x^2 - \frac{a^2}{4} - ax$

⇒ $b^2 - p^2 - ax = \frac{a^2}{4}$

⇒ k = 4

78. Given : D is the Mid-Point of the Side BC

⇒ $BD = \frac{a}{2}$

But : BD = BE + ED

⇒ $BE + ED = \frac{a}{2}$

⇒ $BE + x = \frac{a}{2}$

⇒ $BE = \frac{a}{2} - x$

We can Notice that ABE is a Right Angled Triangle

⇒ $h^2 + (\frac{a}{2} - x)^2 = c^2$

But : h² = p² - x²

⇒ $p^2 - x^2 + (\frac{a^2}{4} + x^2 - ax) = c^2$

⇒ $c^2 + ax - \frac{a^2}{4} = p^2$

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Answer 2

Step-by-step explanation:

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