Prove that the perpendicular bisector of the hypotenuse of 30-60-90- triangle divides the larger leg into two parts having the ratio 2- 1-

Let the hypotenous = AC = h ∴ AE = EC = h/2 ∵ DE is the perpendicular bisector) In Δ DEA DE/AE = tan30 = 1/√(3) DE = AE/√(3) = h/2√(3) DE/AD = sin 30 AD = DE/s ...

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Standard XII

Mathematics

Distance between Parallel Lines

Prove that th...

Question

Prove that the perpendicular bisector of the hypotenuse of $30^{\circ} - 60^{\circ} - 90^{\circ}$ triangle divides the larger leg into two parts having the ratio 2:1.

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Solution

Let the hypotenous = AC = h
$∴ A E = E C = \frac{h}{2}$
$∵$ DE is the perpendicular bisector)
In $Δ D E A$
$\frac{D E}{A E} = t a n 30 = \frac{1}{\sqrt{3}}$
$D E = \frac{A E}{\sqrt{3}} = \frac{h}{2 \sqrt{3}}$
$\frac{D E}{A D} = s i n 30$
$A D = \frac{D E}{s i n 30} = \frac{\frac{h}{2 \sqrt{3}}}{\frac{1}{2}}$
$= h \sqrt{h}$
In $Δ D E C$
Let $∠ D C E = θ$
$∴ t a n θ = \frac{D E}{E C} = \frac{h}{2 \sqrt{3}} \times \frac{1}{\frac{h}{2}}$
$= \frac{1}{\sqrt{3}}$
$\Rightarrow θ = 30^{\circ}$
$∴ ∠ D C E = ∠ D C B = 30^{\circ}$
$s i n 30 = \frac{D E}{D C} = \frac{\frac{h}{2 \sqrt{3}}}{D C} = \frac{1}{2}$
$D C = \frac{h}{\sqrt{3}}$
In $Δ D C B$
$s i n 30 = \frac{D B}{D C}$
$\frac{1}{2} = \frac{D B}{D C} \Rightarrow D B = \frac{D C}{2} = \frac{h}{2 \sqrt{3}}$
$∴ \frac{A D}{D B} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{2 \sqrt{3}}} = \frac{2}{1} = 2 : 1$

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Prove that the perpendicular bisector of the hypotenuse of 30∘−60∘−90∘ triangle divides the larger leg into two parts having the ratio 2:1.

Prove that the perpendicular bisector of the hypotenuse of $30^{\circ} - 60^{\circ} - 90^{\circ}$ triangle divides the larger leg into two parts having the ratio 2:1.