Answers

The Brainliest Answer!
2014-09-03T13:50:01+05:30
Given that
parallel line DE divides ΔABC into two parts of equal area.

Therefore, 
Area of ΔADE = 1/2 * Area of ΔABC

 \frac{\Delta ADE}{\Delta ABC} =  \frac{1}{2}

ΔADE ~ ΔABC

We know that, the ratio of the area of two similar triangles is equal to the ratio of squares of the respective altitudes.
Therefore, 
\frac{\Delta ADE}{\Delta ABC} = \frac{AM^2}{AL^2}

\frac{\Delta ADE}{\Delta ABC} = \frac{AM^2}{(AM+ML)^2}

\frac{1}{2} = \frac{AM^2}{(AM+ML)^2}

\frac{2}{1} = \frac{(AM+ML)^2}{AM^2}

 \sqrt{2} = = \frac{(AM+ML)}{AM}

 \sqrt{2} = 1+ \frac{ML}{AM}

 \frac{ML}{AM} =  \sqrt{2} - 1

 \frac{AM}{ML} =  \frac{1}{ \sqrt{2} -1}

AM : ML = 1 :  \sqrt{2} -1}
1 5 1
2014-09-03T13:51:40+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
See diagram.

The area of triangle Δ AEF = Quadrilateral EBCF.

\frac{1}{2} EF * AD\ = \frac{1}{2} (EF + BC) DG \\ \\ EF * AD = (EF + BC) DG \\ \\ \frac {AD}{DG} = \frac {EF + BC}{EF} = 1 + \frac{BC}{EF}\ \ -- Equation 1 \\ \\ \\ As\ ABC\ and\ AEF\ are\ similar,\ \\ \\ \frac{BC}{EF}\ =\ \frac{AG}{AD}\ =\ \frac{(AD+DG)}{AD}\ =\ 1\ +\ \frac{DG}{AD}\ \ --\ Equation 2 \\ \\ Call\ \frac{AD}{DG}\ =\ x\ =\ Ratio\ of\ two\ parts\ of\ the\ Altitude. \\

\\ \\ From\ equations\ 1\ and\ 2\ we\ get \\ \\ x = 1 + \frac{BC}{EF} = 1 + 1 + \frac{1}{x} = 2 + \frac{1}{x} \\ \\ x^2 - 2\ x - 1 = 0 \\ \\ x = 1 +- \sqrt2 => x = (\sqrt2 - 1),\ as\ x\ is\ positive. \\ \\ \ Ratio: \sqrt{2} - 1 \\


1 4 1
I have asked this question from nstse book and it says that the answer is 1:(√2+1)
Please verify the answer.
it is same: √2-1 : 1 = (√2-1)(√2+1) : 1*(√2+1) = (√2)sqare - 1square : (√2+1) = 1 : (√2+1) . my answer is same as ur answer
hope u have understood that. thanks