Question

Perimeter of a square whose length of diagonal is 7√2 m is
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Answer 1

Answer:

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28m

Step-by-step explanation:

Let the side of the square be x

Diagonal of square

According to pythagorean theorem $p^{2}+b^{2} = h^{2}$

Similarly, here $x^{2} + x^{2} = (7\sqrt{2})^{2}$

$2x^{2}$ $= 98$

$x = \sqrt{\frac{98}{2} }$

$x=7$

Perimeter f square = 7×4 = 28m

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\textsf{ \to Diagonal of a square is \sf 7\sqrt2 m.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\textsf{ \to The Perimeter of the square. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Given that the diagonal of a square is 7√2 cm. and we need to find the perimeter of the square . Let's find the side of the square in order to find its perimeter.

$\underline{\purple{\bf Figure :- }}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\thicklines\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){4}}\put(4,4){\line(-1,0){4}}\put(0,4){\line(0,-1){4}}\put(0.01,0){\line(1,1){4}}\put(0,-0.3){ \bf A } \put(4,-0.3){ \bf B } \put(4,4.3){ \bf C } \put(0,4.3){ \bf D }\put(1,2){ \sf 7\sqrt{2} cm }\put(2, - 0.3){ \bf a } \put(4.3,2){ \bf a } \put(3.8,0){\framebox(0.2,0.2)}\end{picture}$

• In ∆ ABC ,

$\sf\implies AB^2 + BC^2 = AC^2 \qquad\bigg\lgroup \bf{\red{ By \ Pythagoras\ Theorem}}\bigg\rgroup \\\\\sf\implies a^2 + a^2 = (7\sqrt2)^2 \\\\\sf\implies 2a^2 = 49\times 2 \\\\\sf\implies 2a^2 = 98 \\\\\sf\implies a^2=\dfrac{98}{2} \\\\\sf\implies a^2 = 49 \\\\\sf\implies \boxed{\pink{\sf Side = 7 m. }}$

Hence perimeter of square = 4(side) = 4× 7m = 28 m.

$\underline{\blue{\sf Hence \ the \ perimeter\ of the \ square \ is \ \textsf{\textbf{ 28cm.}}}}$