Question

1.Find the perimeter of the rectangle whose One side measure 20cm and the diagonal is 29m. ​

Answer 1

82 cm answer of your question..

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Answer 2

$\:\:\:\: \large \underline {\sf{Given\::}}$

→ Side of rectangle , AB = 20 cm

→ Diagonal of rectangle , AC = 29 cm

$\:\:\:\: \large \underline {\sf{To\:Find\::}}$

→ Perimeter of the given rectangle , ABCD

$\:\:\:\: \large \underline {\sf{Solution\::}}$

• First we have to find another side of rectangle , BC for that we'll need to apply Pythagoras Theorem :

→ $\sf {AC}^{2}\: =\: {AB}^{2}\:+\:{BC}^{2}$

→ $\sf {29}^{2}\: =\: {20}^{2}\:+\:{BC}^{2}$

→ $\sf 841 \: =\: 400 \:+\:{BC}^{2}$

→ $\sf {BC}^{2}\: =\: 841\:-\:400$

→ $\sf {BC}^{2}\: =\: 441$

→ $\sf BC \: =\: \sqrt{441}$

→ $\sf BC \: =\: 21\:cm$

• Value of another side , BC is 21 cm

• Now , let us find perimeter of the given rectangle :

$\large{\boxed{\sf{Perimeter\: =\: 2(l\:+\:b)}}}$

• Here , length ( l ) is 21 cm and breadth ( b ) is 20 cm .

→ $\large{\boxed{\sf{Perimeter\: =\: 2(21\:+\:20)}}}$

→ $\large{\boxed{\sf{Perimeter\: =\: 2(41)}}}$

→ $\large{\boxed{\sf{Perimeter\: =\: 82\:cm}}}$

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