Wrong answer (Question:837675.0)

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bhumi

· started a discussion

· 1 Months ago

Wrong answer (Question:837675.0)

Answer should be option 3

Question:

In a right angle triangle ABC, right angled at B., if $B D ⊥ C A$ , such that BD = x $'$ , AC = y, AB = z BC=x then-

Options:

$\frac{1}{x^{' 2}} = \frac{1}{x^{2}} + \frac{1}{y^{2}}$

$\frac{1}{x^{' 2}} = \frac{y^{2}}{z^{2} (y^{2} - z^{2})}$

$\frac{1}{x^{' 2}} = \frac{1}{x^{2}} + \frac{1}{z^{2}}$

$\frac{1}{x^{' 2}} = \frac{1}{y^{2}} + \frac{1}{z^{2}}$

Solution:

Area of triangle

= $\frac{1}{2}$ × x × z = $\frac{1}{2}$ × x $'$ × y

= xz = x $'$ y.

Now, x² + z² = y²

${(\frac{x' y}{z})}^{^{2}} + z^{2} = y^{2}$

$x'^{2} = \frac{z^{2}}{y^{2}} (y^{2} - z^{2}) \Rightarrow \frac{1}{x'^{2}} = \frac{y^{2}}{z^{2} (y^{2} - z^{2})}$

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Wrong answer (Question:837675.0)

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