Question

find the perimeter whose length and diagnol are 24cm and 25cm respectively​

Answer 1

$\large\red{\underline{\mathfrak{Question:}}}$

$\dashrightarrow$ Find the perimeter of rectangle whose length and diagonal are 24 cm and 25 cm respectively.

$\large\red{\underline{\mathfrak{To\:Find:}}}$

$\longrightarrow$ Perimeter of triangle = ?

$\large\red{\underline{\mathfrak{Given:}}}$

$\: \: \: \: \: \: \: \: \: :\implies \tt{Length \: = \: 24 \: cm}$

$\: \: \: \: \: \: \: \: \: :\implies \tt{Diagonal \: = \: 25 \: cm}$

$\large\red{\underline{\mathfrak{Solution:}}}$

$\longrightarrow$ Let the Breadth be b.

Applying Pythagoras Theorem of rectangle:

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

Let us consider that,

$\sf{AC = 25 , AB = 24 \: and \: BC = b}$

$\: \: \: \: \: \: \: \: \: \: :\implies\sf{(25)^2 = (24)^2 + (b)^2}$

$\: \: \: \: \: \: \: \: \: \: :\implies\sf{625 = 576 + (b)^2}$

$\: \: \: \: \: \: \: \: \: \: :\implies\sf{625 - 576 = (b)^2}$

$\: \: \: \: \: \: \: \: \: \: :\implies\sf{49 = (b)^2}$

$\: \: \: \: \: \: \: \: \: \: :\implies\sf{\therefore\:b = 7\:cm}$

$\sf{Now,\: the \: perimete \: of \: rectangle} = 2(l + b)$

Length = 24 cm and breadth = 7 cm

$\: \: \: \: \: \: \: : \implies \tt{2(24 + 7)}$

$\: \: \: \: \: \: \: : \implies \tt{2(31)}$

$\: \: \: \: \: \: \: : \implies \tt{62\:cm}$

Thus,The perimeter of rectangle is 62 cm.