In the following figure, $A E ⊥ B C$ , D is the mid point of BC, then x is equal to:

\frac{1}{a} [b^{2} - d^{2} - \frac{a^{2}}{4}]

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\frac{h + d}{3}

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\frac{c + d - h}{2}

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\frac{a^{2} + b^{2} + d^{2} - c^{2}}{4}

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Solution

The correct option is A $\frac{1}{a} [b^{2} - d^{2} - \frac{a^{2}}{4}]$
In $Δ A E D, d^{2} = x^{2} + h^{2} ⟹ d^{2} - x^{2} = h^{2}$

In $Δ A E C, b^{2} = ({x + \frac{a}{2})}^{2} + h^{2} ⟹ b^{2} - {(x + \frac{a}{2})}^{2} = h^{2}$

Substituting for $h^{2}$ in the second equation,
$⟹ b^{2} - {(x + \frac{a}{2})}^{2} = d^{2} - x^{2} ⟹ b^{2} - d^{2} = x^{2} + \frac{a^{2}}{4} + a x - x^{2} ⟹ x = \frac{b^{2} - d^{2} - \frac{a^{2}}{4}}{a}$

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Similar questions

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

I f x + i y = \sqrt{\frac{a + i b}{c + i d}}, t h e n (x^{2} + y^{2})^{2}

2 A^{⊝} + B_{2} \to 2 B^{⊝} + A_{2}; 2 C^{⊝} + B_{2} \to N o r e a c t i o n; 2 D^{⊝} + A_{2} \to 2 A^{⊝} + D_{2}

a^{2} + b^{2} = c^{2} + d^{2}

a^{2} + d^{2} - a d = b^{2} + c^{2} + b c

\frac{a b + c d}{a d + b c}

a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2}

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