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Maths-

General

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Question

$ABC text is an equilateral triangle end text. straight K text is a point on end text BC text such that end text BK colon KC equals 2 colon 1 text . Prove that end text 9 AK squared equals 7 AC squared$

hint Hint:

Hint :- let the side of equilateral triangle be x , then we get height of equilateral triangle as $left parenthesis x square root of 3 right parenthesis divided by 2$ .Now , Find length of DE and apply pythagoras theorem in Right angle triangle ADE to find length of AD in terms of AB.

The correct answer is: Hence proved

Aim :- Prove that $9 A K squared equals 7 A C squared$
Explanation(proof ) :-
BK:KC = 2:1 KC :( KC+BK) = 1:(1+2) KC :CB = 1:3
In equilateral triangle all sides are equal
Let , AB = BC = CA = x
Given CK = $1 third$ BC = $X over 3$ ( BC is trisected at K)
In equilateral triangle ,the altitudes and medians coincide
so,AE is median also i.e E is midpoint of BC
CE = $X over 2$
From diagram ,KE = CE - CK = $x over 2 minus x over 3 equals x over 6$
We get height of equilateral triangle when side x is $fraction numerator x square root of 3 over denominator 2 end fraction$

Applying pythagoras theorem on triangle ADE
We get, $A K squared equals E K squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared$
$A K squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 9 x squared$
Substitute x by AC as x = AC
$A K squared equals 7 over 9 A C squared not stretchy rightwards double arrow 9 A K squared equals 7 A C squared$
Hence proved

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