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Maths-

General

Easy

Question

A farmer has a square plot of land, where he wants to grow five different crops at a time. One-half of the area in the middle, he wants to grow wheat but in rest of the four equal triangular parts, he wants to grow different crops. p and q are the mid-points of sides CA and CB respectively of a triangular part ABC right-angled at c . Prove that $4 open parentheses A Q squared plus B P squared close parentheses equals 5 A B squared$ .

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $4 open parentheses A Q squared plus B P squared close parentheses equals 5 A B squared$
Hint :- Applying pythagoras theorem to both triangles AQC and PBC. find the AQ² and BP² add them and substitute proper conditions to prove the condition .
Explanation(proof ) :-
Given, p is midpoint of CA i.e PC = ½ AC
Q is mid point of BC i.e QC = ½ BC
Applying pythagoras theorem to triangle AQC we get ,
$A Q squared equals A C squared plus Q C squared$
As we know QC = ½ BC we get $A Q squared equals A C squared plus fraction numerator B C squared over denominator A end fraction$ — Eq1

Applying pythagoras theorem to triangle PBC we get ,
$B P squared equals P C squared plus B C squared$
As we know PC = ½ AC we get $B P squared equals B C squared plus fraction numerator A C squared over denominator A end fraction$ — Eq2
Applying pythagoras theorem to triangle ABC we get ,
$A B squared equals A C squared plus B C squared$ — Eq3
Adding Eq1 and Eq2 we get ,
$A Q squared plus B P squared equals A C squared plus fraction numerator B C squared over denominator 4 end fraction plus B C squared plus fraction numerator A C squared over denominator 4 end fraction$
$not stretchy rightwards double arrow A Q squared plus B P squared equals 5 over 4 open parentheses A C squared plus B C squared close parentheses$
Substitute Eq 3 in the above , we get
$A Q squared plus B P squared equals 5 over 4 open parentheses A B squared close parentheses not stretchy rightwards double arrow 4 open parentheses A Q squared plus B P squared close parentheses equals 5 A B squared$
Hence proved

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