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Properties of Isosceles and Equilateral Triangles

In the follow...

Question

In the following figure, AE $⊥$ BC, 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 $\frac{1}{a} [b^{2} - d^{2} - \frac{a^{2}}{4}]$
Applying Pythagoras theorem in $△ A E C$ ,
$h^{2} = b^{2} - (x + \frac{a}{2})^{2}$ ...........(1)

Applying Pythagoras theorem in $△ A E D$
$h^{2} = d^{2} - x^{2}$ ...........(2)

Equating (1) and (2)
We get, $x = \frac{1}{a} [b^{2} - d^{2} - \frac{a^{2}}{4}]$

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b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{2}

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}

In the given figure, D is the midpoint of side BC and AE $⊥$ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

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

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

(iii) $(b^{2} + c^{2}) = 2 p^{2} + \frac{1}{2} a^{2}$

(iv) $(b^{2} - c^{2}) = 2 a x$

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}

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