Question

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Answer 1

Given :
ABCD is a square whose are sides are of length (x) units in which two segments are down such that AE = 12 units, FC = 9 units and FE = 3units with which AE perpendicular to FE and CF perpendicular to CG.


Construction :
Extend AE to G and Draw CG perpendicular to AG such that CG = 3units and
EG = 9units.


Solution :
In triangle right triangle ACG,
AG = AE + EG
= 12 + 9
= 21 units.
FE = CG = 3units
By Applying Pythagoras theorem,

AC^2 = AG^2 + CG^2
=> AC^2 = (21^2) + (3^2)
=> AC^2 = 441 + 9
=> AC^2 = 450
=> AC = 15√2 units.


Now,
If side of square is of x units, then diagonal is x√2 units by Pythagoras theorem.

=> AC = Diagonal of square = x√2
=> 15√2 = x√2
=> x = 15 units.

Therefore, length of square side is 15 units.

Answer 2

$\large{\mathit{HELLO \: \: DEAR,}}$

$\mathit{CONSTRUCTION:- JOIN\:\: BD\:\: , BD\:\: IS\:\: A \:\:DIAGONAL\:\: OF \:\:SQUARE,} \\ \\ \mathit{NOW,} \\ \\ \mathit{IN \: \: \triangle \: \: BOD,} \\ \\ \mathit{(BD)^2 = (BO)^2 + (DO)^2} \\ \\ \mathit{(BD)^2 = (3)^2 + (21)^2} \\ \\ \mathit{(BD)^2 = 9 + 441} \\ \\ \mathit{(BD)^2 = 450} \\ \\ \mathit{(BD) = 15\sqrt{2}} \\ \\ \\ \mathit{WE \: \: KNOW \: \: THAT:-} \\ \\ \mathit{DIAGONAL \: \: OF \: \: SQUARE = SIDE\sqrt{2} = X\sqrt{2}} \\ \\ \mathit{BD = X\sqrt{2}} \\ \\ \mathit{15\sqrt{2} = X\sqrt{2}} \\ \\ \mathit{X = 15}$


$\large{\mathit{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$