Question

find the perimeter of a rectangle with the diagonal 13cm and length of one side is 12cm
a. 33
b.34
c.35
d.36​

Answer 1

Step-by-step explanation:

The perimeter of a rectangle is 34

Answer 2

$\huge{\underline{\sf{Given:-}}}$

$\small{\orange{\sf{→Diagonal\: of \:rectangle\: => 13cm}}}$

$\small{\orange{\sf{Length\:of\:one\: side \:=> 12cm}}}$

$\huge{\underline{\sf{To \:Find:-}}}$

$\small{\orange{\sf{→Perimeter\: of \:rectangle \:=> ?}}}$

$\huge{\underline{\sf{Solution:-}}}$

ᴛᴏ ғɪɴᴅ ᴛʜᴇ ᴘᴇʀɪᴍᴇᴛᴇʀ, ᴡᴇ ɴᴇᴇᴅ ᴛᴏ ғɪɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ.

ʟᴇᴛ ᴀʙᴄᴅ ʀᴇᴘᴇʀᴇsᴇɴᴛ ᴛʜᴇ ɢɪᴠᴇɴ ʀᴇᴄᴛᴀɴɢʟᴇ ᴡɪᴛʜ ᴅɪᴀɢᴏɴᴀʟ ʙᴅ = 13ᴄᴍ

ᴡᴇ ᴋɴᴏᴡ ᴇᴀᴄʜ ᴀɴɢʟᴇ ᴏғ ᴀ ʀᴇᴄᴛᴀɴɢʟᴇ ɪs 90°

sᴏ, ʜᴇʀᴇ ᴡᴇ ᴄᴀɴ ᴜsᴇ ᴘʏᴛʜᴀɢᴏʀᴀs ᴛʜᴇᴏʀᴇᴍ ᴛᴏ ғɪɴᴅ ᴛʜᴇ ᴀɴᴏᴛʜᴇʀ sɪᴅᴇ ɪ.ᴇ ʙʀᴇᴀᴅᴛʜ

$\angle{C} => 90^0$

Now

ɪɴ ᴛʀɪᴀɴɢʟᴇ ʙᴅᴄ ,ʙᴅ ɪs ᴛʜᴇ ʜʏᴘᴏᴛᴇɴᴜsᴇ

→ By Pythagoras theorem

$\blue{BD^2 = DC^2 + BC^2}$

$\blue{13^2 = 12^2 + BC^2}$

$\blue{BC^2 = 169 - 144}$

$\blue{BC^2 = 25}$

$\blue{BC = \sqrt{25}}$

$\green{BC = 5cm}$

Thus, The other side is 5cm i.e breadth.

Then

$\orange{Perimeter → 2(l+b)}$

$\pink{=> 2(12+5)}$

$\pink{=> 2(17)}$

$\pink{=> 34cm}$

Hence, The perimeter of the rectangle is 34cm