Question

Sohe the following subquestions. Any 4)
1) Find the side of a square whose diagonal is 10cm.​

Answer 1

Answer:

Let □ ◻ABCD be the given square. l(diagonal AC) = 10 cm Let the side of the square be ‘x’ cm. In ∆ABC, ∠B = 90° [Angle of a square] 

∴ AC2 = AB2 + BC2 [Pythagoras theorem] 

∴ 102 = x2 + x2 

∴ 100 = 2x2 

∴ x2 = 100/2

∴x2 = 50 

∴ x = √ 50 50 [Taking square root of both sides] =  √ 25 × 2 25×2  =  5 √ 2 52

∴side of square is  5 √ 2 52 cm. = 4 x  5 √ 2 52

 ∴ Perimeter of square = 20 √ 2 2 cm

Hope this helps you!

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Answer 2

Answer:

5√2 cm

Step-by-step explanation:

Here, the question stated that the diagonal of a square measures 10 cm and we're asked to calculate the length of its side.

Let us suppose the side of the square as x cm.

We know, diagonal of a square is product of √2 and the side of the square. That is,

$\qquad\quad\bigstar\: \: \underline{\boxed{\pmb{\sf{Diagonal\: of\: square=\sqrt{2}\times side}}}}$

Here, we're given with the diagonal of square that is 10 cm. So, substituting the value,

$:\implies\quad\sf{\sqrt{2}\times x=10}$

Transposing √2 from LHS to RHS. It's arithmetic operator will get change.

$:\implies\quad\sf{x=\dfrac{10}{\sqrt{2}}}$

Simplifying.

$:\implies\quad\underline{\boxed{\pmb{\frak{x=5\sqrt{2}}}}}$

❝ Therefore, the side of the square is 5√2 cm. ❞

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More Formulas:

$\twoheadrightarrow\quad\sf{Perimeter\: of\: square=4\times side}$

$\twoheadrightarrow\quad\sf{Area\: of\: square={(side)}^{2}}$